GALOIS THEORY

ANDREW BOOKER AND ABHISHEK SAHA

A good portion of the material that follows is based on lecture notes of Sean Prendiville,Trevor Wooley and Andrey Lazarev.

Contents

1. Introduction 12. Review: Polynomial Rings 23. Field Extensions 64. Ruler and Compass Constructions 125. Some irreducibility criteria 146. Splitting fields and normal extensions 167. The Galois Group of an Extension 208. More on normal extensions 249. Separable extensions 2610. The main theorems of Galois Theory 2911. Galois groups of polynomials 3112. Finite fields 3213. Solubility of polynomials by radicals 3314. Miscellaneous topics 3614.1. The field of complex numbers is algebraically closed 3614.2. The primitive element theorem 36References 37

1. Introduction

Definition (Algebraic over Q). We say α ∈ C is algebraic over Q if there exists a non-zeropolynomial f with rational coefficients such that f(α) = 0.

The numbers 21/2 and 31/3 are algebraic over Q since they satisfy the equations t2 − 2 = 0and t3 − 3 = 0, respectively.

Question. Is α = 21/2 + 31/3 algebraic over Q?

It is actually quite hard to think of a non-zero polynomial f with rational coefficients for whichf(α) = 0. Instead we argue more indirectly. Using the binomial theorem, a little computationshows that for any non-negative integer k the kth power αk lies in the Q-linear span

L = spanQ{

1, 212 , 3

13 , 3

23 , 2

12 3

13 , 2

12 3

23}

Since L is a Q-vector space with a spanning set of size six, we have dimQ L ≤ 6. Therefore theseven elements 1, α, α2, . . . , α6 ∈ L cannot be linearly independent. In particular, there existrationals a0, a1, . . . , a6 not all zero such that

a0 + a1α+ . . .+ a6α6 = 0.

So α is indeed algebraic over Q.

Date: September 9, 2014.

1

Notice that the above argument did not construct an explicit polynomial of which α is a root,instead we exploited the algebraic structure of the vector space L, and the restrictions imposedby this structure yield the existence of the required polynomial. It turns out that not only is La vector space, but it is also a field1, so has much more algebraic structure to exploit. In fact Lcoincides with the smallest2 subfield of C containing α and Q, a subfield we denote3 by Q(α).One could say we have extended the field Q by adjoining α.

Studying such field extensions turns out to be very useful, particularly for proving impos-sibility results, of which there are a number in this course. For example, we show in §4 thatif it is possible to construct a point (x1, x2) in the plane using ruler and compass, then thefield L = Q(x1), obtained from Q by adjoining x1, satisfies dimQ L = 2r for some non-negativeinteger r. Yet we also show that if angle trisection is possible using ruler and compass, thenthere is a constructible point (x1, x2) with dimQ(x1) = 3. Hence angle trisection is not possibleusing only ruler and compass.

Finally, another good reason for studying field extensions Q ⊂ L is that they allow us todefine the Galois group of a polynomial f ∈ Q[t]. We eventually show that one can expressthe roots of f in terms of the operations +,−,×,÷ and n

√applied to the coefficients of f if

and only if the Galois group of f isn’t ‘too non-abelian’. It turns out that f = t5 − 6t+ 3 hasGalois group S5, which is too non-abelian. Hence there is no general formula for the roots of aquintic in terms of radicals (unlike quadratics, cubics and quartics).

2. Review: Polynomial Rings

To re-familiarise with the necessary background material, students should re-read their notesfrom Algebra 2 on rings, integral domains and ideals. Better still, read the first three chaptersof [Gar86]. The essential background is contained in pp.2–4 of Andrey Lazarev’s notes4, but fora more in-depth online reference see pp.40–74 of the notes of Teruyoshi Yoshida5, which containeverything we need and much more besides.

Definition (Polynomial ring over R). Let R be a ring. A polynomial over R in the inde-terminate t is an expression of the form∑

i

aiti = a0 + a1t+ a2t

2 + . . .

where ai ∈ R and the set of i for which ai 6= 0 is finite. We call ai the coefficient of ti. The setof all polynomials over R in the indeterminate t is denoted R[t]. Given two such polynomialsf =

∑i ait

i and g =∑

i biti we define their sum by

f + g :=∑i

(ai + bi)ti (2.1)

and their product by

fg :=∑i

( ∑j+k=i

ajbk

)ti. (2.2)

Example. Let R = Z/12Z. Set f = 4t + 8 and g = 3t2 + 9. Then f + g = 3t2 + 4t + 5 andfg = 0.

Definition (Degree, leading coefficient, monic). Given a non-zero polynomial f =∑

i aiti there

exists a maximal non-negative integer n with an 6= 0. We call n the degree of f , written deg(f),

1It is an instructive exercise to try and prove this. The hard part is showing that β ∈ L \ {0} =⇒ β−1 ∈ L.(Hint: As above, show that the powers of β lie in L, and hence β is algebraic over Q, then multiply this polynomialexpression in β through by an appropriate power of β−1.)

2We give more formal definitions later.3Exercise: Show that any subfield of C not equal to the trivial field {0} must contain Q.4http://www.maths.lancs.ac.uk/~lazarev/galoislecturenotes.pdf5https://www.dpmms.cam.ac.uk/~ty245/2010_Galois_M24/2010_Galois_M24.pdf

2

and call an the leading coefficient of f . By convention, the degree of the zero polynomial is−∞. If the leading coefficient of f equals one, then we say that f is monic.

Notice that for any ring R and any f, g ∈ R[t] we have the inequality

deg(f + g) ≤ max {deg(f),deg(g)} (2.3)

Defining deg(0) = −∞ at first appears curious, however it ensures the validity of the identitydeg(fg) = deg(f) + deg(g), even when fg = 0, provided that R is an integral domain. That theformula can fail otherwise is demonstrated by the above example.

Lemma 1 (Degree product formula for polynomials over domains). Suppose that R is anintegral domain. Then R[t] is also an integral domain. Moreover, for any f, g ∈ R[t] we havethe identity

deg(fg) = deg(f) + deg(g). (2.4)

[The idea: The leading coefficient of fg is the product of the leading coefficients of f and g.]

Proof. If either f or g are zero, then (2.4) follows from the identities deg(f) + (−∞) = (−∞) +deg(g) = −∞. Let us therefore suppose that f =

∑i ait

i and g =∑

j bjtj are non-zero of

degrees m and n respectively. If i and j are non-negative integers with i+j ≥ m+n then eitheri ≥ m or j ≥ n. Hence the only possible non-zero product of the form aibj with i+ j ≥ m+ nis that for which i = m and j = n. In this case ambn 6= 0 since R is an integral domain. Itfollows that fg has degree m+ n with leading coefficient ambn. �

We mainly study R[t] when R = K is a field, so certainly an integral domain.

Convention. Throughout the remainder of these notes K always denotes a field.

Definition (Unit). An element a of a ring R is a unit if it has a multiplicative inverse, i.e.there exists b ∈ R with ab = 1. We write R× for the set of units in R. This set forms a groupunder multiplication.

Examples.

• For a field K, we have K× = K \ {0}.• If R is an integral domain, then by the product degree formula (2.4) we have (R[t])× =R× (exercise).

There is a useful analogy6 in number theory between the rings Z and K[t]. This is made moreconcrete by the rough correspondence between the (logarithm of the) absolute value functionon the integers and the degree function on K[t]. Using this analogy, many results in elementarynumber theory carry over into K[t] without much modification. Our first such result is thepolynomial division algorithm, familiar from AS-level maths, albeit over an arbitrary field, andwith a proof showing that the algorithm does indeed terminate.

Theorem 2 (Division algorithm). Let f, g ∈ K[t] with g non-zero. Then there exist uniquepolynomials q and r in K[t] satisfying

f = gq + r and deg(r) < deg(g). (2.5)

[The idea: Cancel the leading term of f by subtracting g times an appropriately scaled power of t. This decreases the

degree of f . Iterate until we obtain a polynomial of degree less than g.]

Proof. Let m = deg(f) and n = deg(g). We proceed by induction on m = deg(f). If m < n,then we may take q = 0 and r = f . Therefore suppose that m ≥ n. Let a denote the leadingcoefficient of f and b the leading coefficient of g. Then the polynomial f1 = f − ab−1tm−ng isan element of K[t] of degree strictly less than m. Hence by induction there exist q1 and r1 suchthat f1 = gq1 + r1 with deg(r1) < deg(g). Substituting our definition for g1 into this identityand re-arranging we have f = gq + r where q = ab−1tm−n + q1 and r = r1.

6See Terry Tao’s blog post on ‘Dyadic models’, http://terrytao.wordpress.com/2007/07/27/

dyadic-models/.

3

Next let us establish the uniqueness of q and r. Suppose that gq+r = gq′+r′ for polynomialsq, q′, r, r′ ∈ K[t] with deg(r) < deg(g) and deg(r′) < deg(g). First suppose that q − q′ is non-zero. Then by Lemma 1 the polynomial (q − q′)g has degree at least deg(g). Yet by (2.3), thepolynomial r′− r has degree strictly less than deg(g). Since these two polynomials coincide, wehave a contradiction. Therefore q = q′, and subtraction yields that r = r′. �

The utility of the division algorithm is that, just as in the case of Z, it allows us to establisha Euclidean algorithm for finding highest common factors. Let us first make the appropriatedefinitions.

Definition (Divides, common factor, highest common factor, relatively prime, associate). Letf, g ∈ K[t].

• We say that f divides g, and write f |g, if there exists a polynomial h ∈ K[t] such thatg = fh. Otherwise we write f - g.• A polynomial d ∈ K[t] is a common factor of f and g if it divides both f and g.• A polynomial h ∈ K[t] is a highest common factor of f and g if it is a common factor

divisible by every other common factor of f and g.• We say f and g are relatively prime if a highest common factor of f and g is 1.• We say f and g are associate if they divide one another. Equivalently, f = λg for someλ ∈ K×. Notice that if f and g are not both zero then all their highest common factorsare associate.

Theorem 3 (Euclidean algorithm for polynomials). Let f, g ∈ K[t] with g non-zero. Writer0 = f , r1 = g. Given ri−1 and ri with ri 6= 0, define qi+1 and ri+1 via the relation7

ri−1 = riqi+1 + ri+1 with deg(ri+1) < deg(ri). (2.6)

Then for some positive integer I we have rI+1 = 0, in which case a highest common factor of fand g is rI .

[The idea: Each iteration reduces deg(ri) and this number cannot be non-negative indefinitely.]

Proof. Suppose that ri+1 6= 0 for all i ≥ 2. Since deg(ri+1) ≤ deg(ri)− 1 ≤ . . . ≤ deg(r1)− i =deg(g)− i, we have a contradiction when i > deg(g). Hence such an I exists.

Using (2.6) we see that for 1 ≤ i ≤ I, a highest common factor of ri and ri+1 is also a highestcommon factor of ri−1 and ri. Since rI is a highest common factor of rI and rI+1 (the latter is0), we can work back through each of the equations (2.6) to deduce that rI is a highest commonfactor of r0 = f and r1 = g. �

Example. We find a highest common factor of t5 − 3t + 2 and t2 − 2t + 1 in Q[t]. Using thedivision algorithm, we have

t5 − 3t+ 2 = (t3 + 2t2 + 4t+ 4)(t2 − 2t+ 1) + (2t− 2),

t2 − 2t+ 1 =(

12 t−

12

)(2t− 1

).

Therefore 2t− 2 is a highest common factor of t5 − 3t+ 2 and t2 − 2t+ 1, or equivalently, t− 1is a highest common factor.

Theorem 4 (Bezout’s identity for polynomials). Let h be a highest common factor of f and gin K[t]. Then there exist polynomials a, b ∈ K[t] such that

af + bg = h. (2.7)

[The idea: Work back up through the Euclidean algorithm.]

Proof. If g = 0, then h = λf for some λ ∈ K×, in which case we can take a = λ and b = 0. Ifg|f , then h = λg for some λ ∈ K×, in which case we can take a = 0 and b = λ. Let us thereforesuppose that g - f and g 6= 0. We perform the Euclidean algorithm as in Theorem 3 (whosenotation we adopt), initialising with r0 = f and r1 = g. Since g - f , the algorithm terminates

7The existence and uniqueness of such polynomials being guaranteed by the division algorithm.

4

with rI+1 = 0 for some I ≥ 2 and with rI a highest common factor of f and g. The identityri−2 = ri−1qi + ri, valid for all 2 ≤ i ≤ I + 1, shows that each ri is a K[t]-linear combination ofri−1 and ri−2. In particular rI ∈ spanK[t] {rI−1, rI−2} and we have the chain of inclusions

spanK[t] {rI−1, rI−2} ⊂ spanK[t] {rI−2, rI−3} ⊂ . . . ⊂ spanK[t] {r1, r0} .

Hence there exist a, b ∈ K[t] satisfying af + bg = r1. Since r1 and h are associate, we canmultiply this equation through by a constant to obtain (2.7). �

Definition (Irreducible polynomial). Let R be a ring. A polynomial f ∈ R[t] is called irre-ducible if f is not a unit and in every factorisation f = gh, either g or h is a unit.

Examples.

• All polynomials of degree 1 over a field are irreducible.• The polynomial t2 + 1 is irreducible over R, but reducible over C.• If f ∈ R[t] is irreducible with R an integral domain, then the only divisors of f take the

form λ or λf for some λ ∈ R×, i.e. the only divisors of f are units or associates of f .This follows from the product degree formula for polynomials over integral domains.

Irreducible polynomials are analogous to prime numbers.

Lemma 5 (Euclid’s lemma). Suppose that f, g, h ∈ K[t] and that f is irreducible. Then when-ever f |gh, we have f |g or f |h.

[The idea: f - g =⇒ af + bg = 1, then multiply through by h.]

Proof. Let us suppose that f - g. Since the only divisors of f are units or associates of f , wesee that 1 is a highest common factor of f and g. By Bezout’s identity, there exist a, b ∈ K[t]such that af + bg = 1. Multiplying by h and rearranging we obtain afh+ bgh = h. The resultfollows since f divides the left-hand side of this identity. �

Theorem 6 (Unique factorisation). Let f ∈ K[t] with deg(f) ≥ 1. Then f can be writtenas a product of irreducible polynomials in a manner unique up to the order of the factors andmultiplication by constants. Furthermore, if f is monic, then f is a product of monic irreduciblepolynomials, unique up to order.

[The idea: Induct on deg(f) for existence of factorisation. For uniqueness, Euclid’s lemma gives g1 and h1 associate; cancel

and iterate.]

Proof. We first show the existence of such a representation by induction on deg(f). If f isitself irreducible then we are done. Notice that this covers the basis case of our inductionwhen deg(f) = 1. We can therefore suppose that there exist g, h ∈ K[t] with f = gh and1 ≤ deg(g),deg(h) < deg(f). As hypothesis of induction, we may assume that both g and hcan be written as products of irreducible polynomials. Multiplying these products together, weobtain such a representation for f .

Next we establish uniqueness. Suppose that

g1 · · · gr = h1 · · ·hs (2.8)

where all gi and hj are irreducible. Without loss of generality, we may assume that r ≤ s. Sinceg1 divides the product h1 · · ·hs, an iterative application of Euclid’s lemma shows that g1|hj forsome j. Since hj is irreducible and deg(g1) ≥ 1, we must have that g1 = λ1hj for some λ1 ∈ K×.Re-labelling indices if necessary, we may assume that j = 1. Cancelling h1 from either side of(2.8), we obtain that λ1g2 · · · gr = h2 · · ·hs. We then iterate the above argument (re-labellingindices if necessary) to show that for each 1 ≤ i ≤ r there exists λi ∈ K with gi = λihi and

λ1 · · ·λr =∏

r<j≤shj .

If r < s then the right-hand side of this equation has degree strictly greater than zero, acontradiction. Thus r = s and uniqueness follows.

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Suppose that f is monic and f = g1 · · · gr with each gi irreducible. Let λi ∈ K× denotethe leading coefficient of gi, so that gi := λ−1

i gi is a monic irreducible polynomial. We haveλ1 · · ·λr = 1, since both sides of this equation equal the leading coefficient of f . Thereforef = λ1g1 · · ·λrgr = (λ1 · · ·λr)(g1 · · · gr) = g1 · · · gr. If f = h1 · · ·hs with each hi monic andirreducible, then by our previous argument, we have s = r and, after re-ordering, we mayassume that hi and gi are associate for each i. However the only way for two monic polynomialsto be associate is for them to be equal. �

Definition (Zero or root). Let R be a ring. We say that α ∈ R is a zero or root of f ∈ R[t]if f(α) = 0.

Lemma 7 (Zeroes correspond to linear factors). Suppose that α ∈ K and f ∈ K[t]. Thenf(α) = 0 if and only if (t− α)|f .

[The idea: Follow your nose for one direction, apply the division algorithm for the other.]

Proof. (⇒). Suppose that f(α) = 0. By the division algorithm, there exist q, r ∈ K[t] withf = (t − α)q + r and deg(r) < 1. Then r must be a constant. In fact r is the zero constantsince r = f(α)− (α− α)q(α) = 0.

(⇐). Conversely, suppose that (t − α) divides f , so that f = (t − α)g for some g ∈ K[t].Then f(α) = (α− α)g(α) = 0. �

The above lemma is useful for identifying factors of polynomials.

Example. Consider8 K = F11. Then the polynomial f(t) = t11 − t + 7 has no linear factors.Why? One can test factors: f(0) = 7, f(1) = 7, . . .. In fact, f(a) = 7 for all a ∈ F11 (thisfollows from Fermat’s little theorem). Therefore f has no zeroes, hence no linear factors.

Exercise. Show that the polynomial f(t) = t3 − t+ 2 is irreducible over F3.

3. Field Extensions

Recall that in a ring R with a ∈ R, the notation (a) represents the principle ideal generatedby a, or in other words, the set of multiples {ab : b ∈ R}.

Proposition 8. Let K be a field and f ∈ K[t] an irreducible polynomial of degree n. Then thequotient9 ring K[t]/(f) forms a field in which each congruence class has a unique residue of theform

a0 + a1t+ . . .+ an−1tn−1 (ai ∈ K).

In particular K[t]/(f) is isomorphic to Kn as a vector space over K.

Proof. The key observation here is that non-zero residue classes in K[t]/(f) are invertible: Ifg ∈ K[t] is not divisible by f then 1 is a hcf of f and g, hence by Theorem 4, there exista, b ∈ K[t] such that af + bg = 1. In particular bg ≡ 1 mod (f). It follows that K[t]/(f) is afield.

By the division algorithm, every element of this quotient may be written uniquely as apolynomial over K of degree at most n− 1. Thus the set of residues is{

a0 + a1t+ . . .+ an−1tn−1 + (f) : ai ∈ K

},

with each such residue defining a unique congruence class (again by uniqueness of remainder).One can check that the map φ : K[t]/(f)→ Kn given by

a0 + a1t+ . . .+ an−1tn−1 + (f) 7→ (a0, . . . , an−1) (ai ∈ K),

is both linear and invertible. �

Notice that Kn is strictly ‘larger’ than K as an additive group (and vector space) when n > 1.

8Fp is the same set as Z/pZ, but viewed as a field.9Review Ch.3 of [Gar86].

6

Question. Plainly the special residue classes given by a0 ∈ K form a subfield of K[t]/(f)isomorphic to K. How do we formally identify that K lies ‘inside’ K[t]/(f)?

Definition (Field extension). A field extension of a fieldK is a ring monomorphism10 i : K → Lwhere L is also a field.

In practice, we identify the image i(K) with K, since this is a subfield of L isomorphic toK. We then write K for i(K), and talk about the field extension L/K (read ‘L over K’, alsodenoted L : K).

Examples.

(i) The field extension C/R is more formally the monomorphism i : R → C given by x 7→x+ 0 ·

√−1.

(ii) In Proposition 8, the map a0 7→ a0 + (f) gives a field extension K → K[t]/(f).(iii) Let K(t) denote the field of fractions11 of the ring K[t]. Then we have a field extension

given by the map i : K → K(t) defined by λ 7→ λ/1.

Theorem 9 (Field extensions have vector space structure). Suppose that L/K is an extension.Then L forms a vector space over K as follows: If u, v ∈ L are vectors we define their sumu + v using the field addition in L. If λ ∈ K is a scalar and v ∈ L is a vector we define thescalar multiple λv using the field multiplication in L.

Proof. We leave the task of checking that the field axioms are satisfied as an exercise for thereader. �

Thus both C and R(t) (the field of fractions of R[t]) are vector spaces over R.

Definition (Degree, finite extension). The degree of an extension L/K is the dimension ofL as a vector space over K. We denote this by [L : K]. We say L/K is a finite extension if[L : K] <∞, and otherwise say it is infinite.

Examples.

• [C : R] = 2, since C has basis{

1,√−1}

.

• [R : Q] =∞, since otherwise R is isomorphic as a vector space to Qd for some d ∈ N, inwhich case R is countable.• If L = Q[t]/(t2 − 2), then [L : Q] = 2, since L = {λt+ µ : λ, µ ∈ Q} has basis {1, t}.

Note that (t2− 2) is the zero residue class in L, so we can think of t as being essentially√2 or −

√2 (either choice is indistinguishable).

Theorem 10 (Field extension product formula). Suppose that M/L and L/K are field exten-sions. Then M/K is a field extension and

[M : K] = [M : L][L : K]. (3.1)

[The idea: First deal with the infinite case, then show that multiplying a basis for M/L by a basis for L/K gives a basis

for M/K.]

Proof. We first consider the case in which the right-hand side of (3.1) is infinite, so at leastone of [M : L] and [L : K] is infinite. In this case it suffices to show that [M : K] = ∞.If [L : K] = ∞, then M (viewed as a vector space over K) contains an infinite-dimensionalsubspace, so must itself be infinite-dimensional. If [M : L] =∞, then M contains an infinite setwhich is linearly independent over L, so also linearly independent over K. Hence M/K mustagain be infinite dimensional.

We may therefore suppose that the right-hand side of (3.1) is finite, so that both M/L andL/K are finite dimensional. Let x1, . . . , xm denote a basis for M over L, and let y1, . . . , yn

10Injective homomorphism.11See p.3 of http://www.maths.lancs.ac.uk/~lazarev/galoislecturenotes.pdf or p.58 of https://www.

dpmms.cam.ac.uk/~ty245/2010_Galois_M24/2010_Galois_M24.pdf.

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denote a basis for L over K. We claim that the products xiyj , with 1 ≤ i ≤ m and 1 ≤ j ≤ n,constitute a basis for M over K, from which it follows that [M : K] = mn as required.

To see that the K-span of this set encompasses the whole of M , first note that every elementof M is an L-linear combination of the xi. Each coefficient of xi in such a combination is itselfa K-linear combination of the yj . Expanding out brackets shows that every element of M is aK-linear combination of the xiyj .

For linear independence, let us suppose that∑

i,j λijxiyj = 0 with λij ∈ K for all i and j.Re-arranging this sum, we have that

m∑i=1

( n∑j=1

λijyj

)xi = 0.

Since the coefficients of the xi in this sum are elements of L, and the xi are linearly independentover L, we deduce that

n∑j=1

λijyj = 0 (1 ≤ i ≤ m).

The linear independent of the yj over K then gives that λij = 0 for all i and j. �

Note (Tower law). A sequence Kn/Kn−1,Kn−1/Kn−2, . . . ,K1/K0 of field extensions (which wecan write Kn/Kn−1/ . . . /K1/K0) is called a tower of field extensions, and satisfies the towerlaw for field extensions:

[Kn : K0] = [Kn : Kn−1] · · · [K1,K0]. (3.2)

Corollary 11. Suppose that L/K is a field extension for which [L : K] is a prime number.Then the only subfields of L containing K are L and K.

Proof. Say [L : K] = p. If L/M/K, then by the tower law either [L : M ] = 1 or [M : K] = 1.In the first case M = L and in the second M = K. �

Exercise. Show that if (Ki)i∈I is a family of subfields of L, then K :=⋂i∈I Ki is also a subfield

of L.

Definition (K adjoined by the set A). Suppose that L/K is a field extension, and A ⊂ L. Wewrite K(A) for the intersection of all subfields of L that contain K and A. (Note that K(A) isa subfield of L, and indeed is the smallest subfield of L containing both K and A).

• K(A)/K is the extension of K generated by A.• If A = {α1, . . . , αn}, we write K(α1, . . . , αn) for K(A).• An extension L/K is called a simple extension if L = K(α) for some α ∈ L.• We say L/K is a finitely generated extension over K if L = K(A) for some finite setA ⊂ L.

Exercise. Show that for a field extension L/K, if A ⊂ L and β ∈ L then

K(A ∪ {β}) = K(A)(β). (3.3)

Conclude that L/K is finitely generated if and only if there exist a tower of simple field exten-sions Kn/Kn−1/ . . . /K1/K0 with Kn = L and K0 = K.

Example. Consider the field extension K = Q(√

2,√

3) of Q. We claim that K is a simpleextension of Q. In order to see this, write α =

√2 +√

3. Clearly Q(α) ⊂ Q(√

2,√

3). However,we also have the reverse inclusion. First note that α2 = 5 + 2

√6, so

√6 = 1

2(α2 − 5) ∈ Q(α).

Now√

6α = 3√

2 + 2√

3, and we can subtract rational multiples of α from this number to showthat

√2,√

3 ∈ Q(α). [The idea: Letting v = (v1, v2) = (√

2,√

3), we have in effect shown that the fact that Q(α)

is closed under multiplication implies that both α = (1, 1) · v ∈ Q(α) and√

6α = (3, 2) · v ∈ Q(α). We can then use

elementary row operations to show that√

2 = (1, 0) · v ∈ Q(α) and√

3 = (0, 1) · v ∈ Q(α).]

Exercise. Show that for α =√

2 +√

3 we have [Q(α) : Q] = 4. (Hint: Make a guess at asuitable basis then prove it is so - take a look at the proof of the tower law or the introductionfor guessing inspiration.)

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Example. There is a potential ambiguity in the notation K(α), since we use K(t) to denotethe field of fractions of the polynomial ring K[t]. Fortunately, if we set L = K(t) and takeα = t ∈ L, then the simple field extension K(α) coincides with the field of fractions K(t). Inother words, the intersection of all subfields of L = K(t) which contain both K and t is equalto K(t) itself (exercise12).

Definition. Suppose that L/K is a field extension, and that α ∈ L.

• If there is a non-zero polynomial f ∈ K[t] for which f(α) = 0 then we say that α isalgebraic over K.• If there exists no non-zero polynomial f ∈ K[t] for which f(α) = 0 then we say that α

is transcendental over K.

Note. We have seen that L/K is endowed with the structure of a vector space over K. Thenα ∈ L is algebraic over K if the set {αn : n ≥ 0} is linearly dependent, and is transcendental ifthe set {αn : n ≥ 0} is linearly independent.

Fact. The real numbers π, e and 2√

2 are transcendental over Q. In fact the set of transcendentalelements in R has full measure (meaning that the set of elements algebraic over Q has measurezero).

Definition. Suppose that L/K is a field extension and that α ∈ L. We define the evaluationmap Eα : K[t]→ L by Eα(f) = f(α) for each f ∈ K[t].

Note.

• Eα : K[t]→ L is a ring homomorphism.• The element α ∈ L is transcendental ⇐⇒ ker(Eα) = {0} ⇐⇒ Eα is injective.• The element α ∈ L is algebraic ⇐⇒ ker(Eα) 6= {0} ⇐⇒ Eα is not injective.

Proposition 12 (Polynomials over K form a PID). The ring K[t] is a principle ideal domain(PID), so that any ideal I in K[t] takes the form I = (f) = {gf : g ∈ K[t]} for some f ∈ K[t].

Proof. Let I be a non-zero (6= {0}) ideal in K[t] and let f be a polynomial in I of minimal non-negative degree. Given g ∈ I we apply the division algorithm to obtain polynomials q, r ∈ K[t]with g = fq+r and deg(r) < deg(f). Since I is an ideal, we have r = g−fq ∈ I. By minimalityof deg(f), we must have deg(r) = −∞, or equivalently r = 0. Thus f |g for every g ∈ I, fromwhich it follows that I = (f). �

As a consequence, we see that for any α ∈ L which is algebraic over K, the non-zero kernelker(Eα) is generated by some non-zero polynomial mα ∈ K[t], so that ker(Eα) = (mα). Multi-plying through by a non-zero constant, we may assume that mα is monic. If f ∈ K[t] is monicand also satisfies ker(Eα) = (f), then mα|f and f |mα. Thus f and mα are associate with thesame leading coefficient, so must be equal. We can therefore make the following definition.

Definition (Minimal polynomial). Let L/K be a field extension and let α ∈ L be algebraicover K. The minimal polynomial of α over K is the unique monic polynomial satisfyingker(Eα) = (mα).

Theorem 13 (Factorisation of the evaluation map). Suppose that L/K is a field extension andthat α ∈ L is algebraic over K. Then

(i) The minimal polynomial mα of α over K is irreducible in K[t].(ii) The image Eα(K[t]) of the polynomial ring K[t] is the subfield K(α).

12The idea is that if a subfield contains K and t, then closure under addition and multiplication ensures thatall polynomials f ∈ K[t] are contained in this subfield, hence all rational functions f/g with f, g ∈ K[t] and g 6= 0are contained in this subfield.

9

(iii) The evaluation map Eα factorises as i◦ Eα ◦ q, where q : K[t]→ K[t]/(mα) is the quotient

map, Eα : K[t]/(mα) → K(α) is an isomorphism, and i : K(α) → L is the inclusionmapping.

K[t]Eα−−−−→ Lyq xi

K[t]/(mα)Eα−−−−→ K(α)

(iv) Let n = deg(mα). Then the isomorphism Eα between K[t]/(mα) and K(α) is given by

f + (mα) 7→ f(α),

which can be described more explicitly as

a0 + a1t+ . . .+ an−1tn−1 + (mα) 7−→ a0 + a1α+ . . .+ an−1α

n−1, (ai ∈ K).

Proof. We begin with claim (i). Suppose that mα splits over K, say as mα = fg. Then applyingthe evaluation map, we obtain

0 = Eα(mα) = Eα(f)Eα(g) = f(α)g(α).

So f(α) = 0 or g(α) = 0. But if f(α) = 0, then f ∈ (mα), so mα|f and g is necessarily a unit.Similarly if g(α) = 0. Thus mα is irreducible.

Next we prove claim (iii), as it eases our proof of claim (ii). Since mα is irreducible, one findsthatK[t]/(mα) is a field. By the first ring homomorphism theorem13, we have thatK[t]/ ker(Eα)is isomorphic to Im(Eα) via the map

f + ker(Eα) 7→ Eα(f).

Let Eα denote this isomorphism, let q denote the natural quotient map K[t] → K[t]/ ker(Eα)and let i denote the natural inclusion map Im(Eα)→ L. Then we can record our deliberationsin the following commutative diagram

K[t]Eα−−−−→ Lyq xi

K[t]/ ker(Eα)Eα−−−−→ Im(Eα)

To complete the proof of (iii) it therefore remains to prove claim (ii), that K(α) equals

Im(Eα), which by definition can also be written Eα(K[t]). Since Eα is an isomorphism, it mapsthe field K[t]/(mα) to an isomorphic field contained in L. By (ii) we have the equality

Eα

(K[t]/(mα)

)= Eα(K[t]).

Hence Eα(K[t]) is a subfield of L. Furthermore, Eα(k) = k for all k ∈ K and Eα(t) = α.Thus K ∪ {α} is contained in the field Eα(K[t]), so we must have K(α) ⊂ Eα(K[t]). However,since K(α) is closed under addition and multiplication, we clearly have all K-polynomials in α(elements of the form a0 + a1α+ . . .+ arα

r with ai ∈ K) contained in K(α), or in other wordsEα(K[t]) ⊂ K(α). This establishes (ii) and (iii).

Finally, part (iv) follows by combining parts (ii), (iii) and Proposition 8, along with an

investigation of the form of the isomorphism Eα �

Theorem 14. Suppose that L/K is a field extension and that α ∈ L. Then α is algebraic overK if and only if [K(α) : K] <∞, in which case [K(α) : K] = deg(mα).

13See http://en.wikipedia.org/wiki/Isomorphism_theorem#First_isomorphism_theorem_2

10

Proof. Suppose that [K(α) : K] = n < ∞. Then 1, α, . . . , αn must be linearly dependent overK. Hence there exist a0, . . . , an ∈ K not all zero such that a0 + a1α+ . . .+ anα

n = 0. Thus αis algebraic over K.

Conversely, suppose that α is algebraic over K with minimal polynomial mα. Then byTheorem 13 we have

K(α) = Eα

(K[t]/(mα)

),

where Eα : K[t]/(mα) → K(α) is a field isomorphism. But from Proposition 8 we know thatK[t]/(mα) is isomorphic as a K-vector space to Kn, where n = deg(mα). �

Definition. We say that a field extension L/K is algebraic if every element of L is algebraicover K.

Remark 1. A field L being ‘algebraically closed’14 is distinct from being an algebraic extension.For instance C is algebraically closed, but is not an algebraic extension of Q.

Corollary 15. A finite extension L/K is also an algebraic extension. In particular, if α ∈ Lis algebraic over K, then K(α)/K is algebraic.

Proof. Let α ∈ L. Then K(α) is a subspace of L when viewed as a vector space over K. Thus[K(α) : K] ≤ [L : K] <∞. It follows from Theorem 14 that α is algebraic over K. �

We can essentially restrict our attention to algebraic extensions.

Theorem 16. Suppose that L/K is a field extension. The set Lalg of those elements of L whichare algebraic over K is a subfield of L.

Proof. We need to verify that Lalg is closed under addition, multiplication and taking addi-tive/multiplicative inverses. Suppose that α and β are algebraic over K. Since β is algebraicover K, it is also algebraic over K(α). Thus by (3.3), the tower law and Theorem 14, we findthat

[K(α, β) : K] = [K(α)(β) : K]

= [K(α)(β) : K(α)][K(α) : K] <∞.

Now since α + β, αβ, −α and α−1 (assuming α 6= 0) are elements of K(α, β), which is itself afinite extension of K, we see from Corollary 15 that these elements are algebraic over K. �

Exercise. Given a field extension L/K with α ∈ L non-zero and algebraic over K, derive frommα a non-zero polynomial over K of which α−1 is a root. (Hint: If f(α) = 0 then α−1 is a zeroof the rational function f(t−1) ∈ K(t). Clear denominators to obtain an element of K[t].)

In Corollary 15 we established that every finite extension is algebraic. The converse is nottrue in general, for instance Theorem 16 tells us that if A denotes the set of complex numbersalgebraic over Q then A/Q is an algebraic extension, but in §5 we prove that this extension isnot finite. The converse is true however, if we assume in addition that the algebraic extensionis finitely generated. This is the content of the next theorem.

Theorem 17. Suppose that L/K is a field extension. Then the following are equivalent:

(i) [L : K] <∞;(ii) L/K is algebraic and finitely generated;

(iii) There exists finitely many α1, . . . , αn ∈ L all algebraic over K such that L = K(α1, . . . , αn).

Proof. (i) =⇒ (ii): Assuming (i), the extension L/K is algebraic by Corollary 15. Since[L : K] <∞, there exist α1, . . . , αn ∈ L which together form a basis for L over K. As L equalsthe K-linear span of the αi, we certainly have L ⊂ K(α1, . . . , αn). But L is also a subfield ofL containing K ∪ {α1, . . . , αn}, so K(α1, . . . , αn) ⊂ L. Thus L = K(α1, . . . , αn) is also finitelygenerated over K.

14Every polynomial in L[t] of positive degree has a root in L.

11

(ii) =⇒ (iii): Finite generation gives that L = K(α1, . . . , αn) for some αi. Each αi algebraicover K since L/K is algebraic.

(iii) =⇒ (i): This follows from the tower law and Theorem 14, since assuming (iii) andsetting Ki = K({αj : j < i}) we have

[L : K] = [K(α1, . . . , αn) : K]

= [Kn−1(αn) : Kn−1] . . . [K0(α1) : K0] <∞.�

We postpone the proof that [Qalg : Q] = ∞ until we have developed some irreducibilitycriteria in §5. We have already developed enough theory to solve some famous problems.

4. Ruler and Compass Constructions

Let us formalise what it means for a point z = (x, y) ∈ R2 to be constructible using ruler andcompass.

Definition (Constructible from P ). Given a set P of points in the plane R2, we say that a lineL is constructible from P if it passes through two distinct points of P . We say a circle C isconstructible from P if it is centred at a point of P and has a point of P on its circumference.Finally, we say that a point z ∈ R2 is constructible in one step from P if any of the followinghold:

(a) z is the intersection of two distinct lines constructible from P ;(b) z is an intersection of a line and a circle, both constructible from P ;(c) z is an intersection of two distinct circles constructible from P .

We say that z is constructible from P if there exists a finite number of points z1, . . . , zn withzn = z such that each zi is constructible in one step from P ∪ {z1, . . . , zi−1}. We say that z isconstructible using ruler and compass if it is constructible from the set {(0, 0), (1, 0)}.

Let us call an angle constructible if it is found between two constructible lines (lines passingthrough two distinct constructible points).

Exercise. The angle π/3 is constructible.

Exercise. Any constructible angle θ may be bisected by a line using ruler and compass, meaningθ/2 is also constructible. In particular, π/6 is constructible.

In this section we establish a number of impossibility results concerning ruler and compassconstructions, for instance that one cannot trisect all angles using ruler and compass, not evenconstructible angles (we prove that π/9 is not a constructible angle).

The key to all our impossibility results is the following.

Theorem 18. If (x, y) is constructible using ruler and compass, then there exist non-negativeintegers r and s such that [Q(x) : Q] = 2r and [Q(y) : Q] = 2s.

Before proving this theorem, we begin with a lemma.

Notation. If P = {(x1, y1), . . . , (xn, yn)} is a finite set of points in the plane R2, then we defineQ(P ) to be the R-subfield Q(x1, y1, . . . , xn, yn).

Lemma 19. If (x, y) is constructible in one step from P , then x and y satisfy quadratic orlinear polynomials with coefficients in Q(P ).

Proof. There are three cases to consider, of which we only consider the case where (x, y) resultsfrom the intersection of a line L and a circle C, both constructible from P . As L is constructible,there exist distinct points (x1, y1), (x2, y2) ∈ P both lying on the line L. Since (x, y) also lieson L, the numbers x and y satisfy the equation

(x− x1)(y2 − y1) = (y − y1)(x2 − x1).12

Hence there exist a, b, c ∈ Q(P ) with a and b not both zero such that

ax+ by = c. (4.1)

Since C is constructible from P , there exist points (x0, y0), (x3, y3) ∈ P such that C is centredat (x0, y0) and C has (x3, y3) lying on its circumference. Then x and y satisfy the equation

(x− x0)2 + (y − y0)2 = (x3 − x0)2 + (y3 − y0)2.

Clearly the right-hand side above is an element of Q(P ). Expanding brackets, we thereforededuce that there exists λ, µ ∈ Q(P ) such that

x2 + λxy + y2 = µ. (4.2)

If b = 0 in (4.1) then x satisfies a linear equation with coefficients in Q(P ), since we must havea 6= 0 and x = c/a. Therefore suppose b 6= 0. Rearranging (4.1) and substituting for y in (4.2)we deduce that

x2 + λ1

bx(c− ax) +

1

b2(c− ax)2 − µ = 0. (4.3)

The left-hand side of (4.3) cannot be identically zero, since then every point of the unboundedset L would be contained in the bounded set C. Therefore x satisfies a non-zero polynomial ofdegree at most 2 with coefficients in Q(P ). A similar argument works for y. �

Proof of Theorem 18. Since (x, y) is constructible using ruler and compass, there exist a se-quence of points (x0, y0), (x1, y1), . . . , (xn, yn) ∈ R2 such that

• (x0, y0) = (0, 0);• (x1, y1) = (1, 0);• (xn, yn) = (x, y);• For each 1 ≤ i < n the point (xi+1, yi+1) is constructible in one step from the setPi = {(x0, y0), (x1, y1), . . . , (xi, yi)}.

We deduce from Lemma 19 that for each 1 ≤ i < n both [Q(Pi)(xi+1) : Q(Pi)] and [Q(Pi)(yi+1) :Q(Pi)] equal 1 or 2. By the tower law [Q(Pn) : Q] is equal to

[Q(Pn−1)(xn,yn):Q(Pn−1)(xn)][Q(Pn−1)(xn):Q(Pn−1)]···[Q(P1)(x2,y2):Q(P1)(x2)][Q(P1)(xn):Q(P1)],

which is a product of 1s and 2s. Thus [Q(Pn) : Q] = 2r for some non-negative integer r. Again,by the tower law, we see that both [Q(x) : Q] and [Q(y) : Q] divide 2r, so are themselves apower of 2. �

Exercise. (i) Given three distinct points A,B and P , with P not on the line AB, show thatone can construct a line passing through P which is perpendicular to AB.

(ii) Given two distinct constructible points P and Q, show that one can construct a line passingthrough P which is perpendicular to PQ.

(iii) Combine (i) and (ii) to show that given distinct A,B and P , with P not on the line AB,one can construct a line through P which is parallel to AB.

(iv) Use (iii) to show that given any constructible points A and B, with the origin O not onthe line AB, one can construct the parallelogram with corners at O,A,B and B −A.

(v) Conclude that if the length l is constructible, so that it forms the distance between twoconstructible points A and B, then the point (l, 0) is also constructible.

Theorem 20. Angle trisection, squaring the circle and doubling the cube are all impossibleusing ruler and compass. More precisely:

(i) The constructible angle π/3, formed between the constructible lines which join (0, 0) to

(1, 0) and (0, 0) to (12 ,√

32 ), is not trisectable.

(ii) One cannot construct a length whose cube is twice the volume of the unit cube.(iii) One cannot construct a length whose square equals the area of the unit circle.

13

Proof. (i) Let θ = π/9. By way of contradiction, suppose that the angle in question istrisectable, then one of the lines y = (tan θ)x or y = (tan(2θ))x is constructible. Since anglebisection is possible, it follows in either case that the line y = (tan θ)x is constructible.Intersecting this line with the unit circle centred at the origin, we see that the point(cos θ, sin θ) is constructible. We claim that [Q(cos θ) : Q] = 3, which on comparison withTheorem 18 gives our desired contradiction.

To prove the claim, first note that by the double angle formulae, we have

1

2= cos(3θ) = cos θ cos(2θ)− sin θ sin(2θ)

= cos θ(cos2 θ − sin2 θ)− sin θ(2 sin θ cos θ)

= cos3 θ − 3 sin2 θ cos θ

= cos3 θ − 3 cos θ + 3 cos3 θ

= 4 cos3 θ − 3 cos θ.

Hence 2 cos θ is a zero of the cubic f = t3− 3− 1. Let us now show that this is irreducibleover Q. If not, it must have a linear factor over Q, hence a rational root a/b with a and bcoprime integers and b positive. Multiplying through by b3 we deduce that a3−3ab2−b3 =0. Thus b|a3 and a|b3. Since a and b are coprime, it follows from Euclid’s lemma thata|b and b|a, so that a/b = ±1. One can check that f(±1) 6= 0. Thus f is indeed monicirreducible over Q and must therefore be the minimal polynomial for 2 cos(θ) over Q. ByTheorem 14 we must have [Q(2 cos θ) : Q] = deg(f) = 3. Since 2 is a non-zero rational,Q(2 cos θ) = Q(cos θ), which yields the claim.

(ii) By the previous exercise, if the length l with l3 = 2 is constructible, then so is the point(l, 0) = ( 3

√2, 0). Since 3

√2 is irrational, the cubic f = t3 − 2 is irreducible over Q (it has

only one real root - think of the graph). It follows that f equals the minimal polynomialof 3√

2 over Q, and so by Theorem 14 we have [Q( 3√

2) : Q]. Hence by Theorem 18 thelength l cannot be constructible.

(iii) As above, if a length l with l2 = π is constructible, then (l, 0) = (√π, 0) is also con-

structible. Since π is transcendental over Q (Lindemann, 1882), we have [Q(π) : Q] =∞.But π ∈ Q(

√π), so Q(π) is a Q-subspace of Q(

√π). It follows that [Q(

√π) : Q] ≥ [Q(π) :

Q], so [Q(√π) : Q] =∞ and (l, 0) cannot be constructible by Theorem 18.

�

5. Some irreducibility criteria

It is apparent that it is convenient to have available a number of irreducibility criteria.

Definition (content). Let f ∈ Z[t]. The content of f is the highest common factor of itscoefficients. We call f primitive if it has content 1.

Lemma 21 (Gauss’ lemma). Let f ∈ Z[t] be primitive. Then f is irreducible over Q if andonly if f is irreducible over Z.

Proof. The ‘only if’ direction is clear, we therefore prove the ‘if’ direction. To this end, supposethat f = gh with g, h ∈ Q[t]. Clearing denominators, it follows that there exists a positiveinteger N1 ≥ 1 and g1, h1 ∈ Z[t] with deg(g1) = deg(g) and deg(h1) = deg(h) such thatN1f = g1h1.

Suppose that N1 > 1, so that N1 has a prime factor p. Working over the integral domainFp we have 0 ≡ g1h1 (mod p). Hence by the degree-product formula over Fp, either g1 ≡ 0 orh1 ≡ 0 (mod p). Re-labelling if necessary, we may assume the latter. Set N2 = N1/p, g2 = g1

and let h2 be the polynomial obtained from h1 by dividing all coefficients by p. Then N2f = g2h2

with deg(g2) = deg(g), deg(h2) = deg(h) and 1 ≤ N2 < N1. We can continue iterating in thismanner, reducing Ni by at least one on each iteration, until we obtain Ni = 1 for some i. It

14

then follows that f = gihi for some gi, hi ∈ Z[t] with deg(gi) = deg(g) and deg(hi) = deg(h). Itfollows that if f is reducible over Q then it is also reducible over Z. �

Example. Let us show that the polynomial t2 − 2 is irreducible over Q. By Gauss’ lemma, weneed only show that t2 − 2 is irreducible over Z. Suppose that

t2 − 2 = (a1t+ b1)(a2t+ b2)

with ai, bi ∈ Z. Then a1a2 = 1 and b1b2 = −2. Thus b1 = ±1 or b2 = ±1. Re-labelling indicesif necessary we may assume that b1 = ±1. Sine a1 = ±1, we see that one of t+ 1 or t− 1 is afactor of f . But then f(−1) = 0 or f(1) = 0, neither of which are true.

More sophisticated approaches are based on reduction modulo p, a tactic we have alreadyemployed in the proof of Gauss’ lemma. In this context, when f ∈ Z[t], we write [f ]mfor theimage of f under the canonical map from Z[t] to (Z/mZ)[t] which maps each coefficient in Zto its residue class mod m. One can check that the localisation map is a ring homomorphismZ[t]→ (Z/mZ)[t].

Theorem 22 (Localisation principle). Let f ∈ Z[t] and suppose there exists a prime p whichdoes not divide the leading coefficient of f and such that [f ]p is irreducible over Fp. Then f isirreducible over Q.

Proof. Suppose that f = gh for some g, h ∈ Z[t]. As localisation mod p is a homomorphismon Z[t], we have [f ]p = [g]p[h]p. Since p does not divide the leading coefficient of f and [f ]p isirreducible over Fp, we have (re-labelling g and h if necessary) that deg[g]p = deg[f ]p = deg(f).Therefore deg(g) ≥ deg(f). But since f 6= 0, the degree product formula gives that deg(f) ≥deg(g). Thus deg(g) = deg(f). Hence f is irreducible over Z, and therefore irreducible over Qby Gauss’ lemma. �

Theorem 23 (Eisenstein’s criterion). Let f ∈ Z[t] and suppose there exists a prime p dividingall coefficients of f except the leading coefficient, and such that p2 does not divide the constantcoefficient. Then f is irreducible over Q.

Proof. Suppose that f = gh for some integer polynomials of the form g = a0 + a1t+ . . .+ amtm

and h = b0 + b1t + . . . bntn. Since p|a0b0 and p2 - a0b0, we have that p divides exactly one of

a0 and b0. Without loss of generality, suppose that p - b0. Let i be minimal such that p - ai(clearly p - am since p - ambn). Suppose that i < deg(f). Then

p|aib0 + ai−1b1 + . . .+ a0bi.

Yet p|aj for all j < i, so that p|aib0. Hence by Euclid’s lemma either p|ai or p|b0, both of whichlead to contradictions. Thus i = deg(f). Hence deg(g) ≥ i = deg(f). It follows that f isirreducible over Z, and thus irreducible over Q by Gauss’ lemma. �

Proposition 24 (Change of variables). Let K be a field and a, b ∈ K with a 6= 0. Then f isirreducible over K iff f(at+ b) is irreducible over K.

Proof. Let q be a non-zero polynomial over K. Since the leading coefficient of q(at+ b) is equal

to the leading coefficient of q multiplied by adeg(q) , we see that deg(q) = deg(q(at+ b)). Thusf = gh with deg(g) and deg(h) both less than deg(f) implies that f(at+ b) = g(at+ b)h(at+ b)so that f(at+ b) is reducible.

For the reverse implication, we note that if f1 is defined to be the polynomial f(at+ b), thenf(t) = f1(a−1t− a−1b). Hence by the previous implication, if f1 is reducible then so is f . �

Examples.

(1) f = t7 + 3t2 + 9t+ 3 is irreducible over Q by Eisenstein’s criterion with p = 3.(2) f = 2

9 t5 + 5

3 t4 + t3 + 1

3 is irreducible over Q if and only if 9f = 2t5 + 15t4 + 9t3 + 3 isirreducible over Q. The latter holds by Eisenstein’s criterion with p = 3.

(3) Consider f = t4 +4t3 +6t2 +4t+8. By the binomial theorem f = (t+1)4 +7 = f1(t+1)where f1 = t7 + 7 is irreducible over Q by Eisenstein’s criterion with p = 7. Thus bychange of variables f is also irreducible over Q.

15

(4) Consider f = t2 + 9t+ 1. Localising over F3 gives [f ]3 = t2 + 1. This is irreducible overF3 since a2 6= −1 for any a ∈ F3. Thus by the localisation principle, f is irreducibleover Q.

(5) Consider f = 16t4 + 15t3 + 7. Localising over F5 gives [f ]5 = t4 + 2. This has no zerosin F5, so has no linear factors. Working over F5 suppose that

t4 + 2 = (t2 + a1t+ b1)(t2 + a2t+ b2).

Then

a1 + a2 = 0, a1a2 + b1 + b2 = 0, a1b2 + a2b1 = 0, b1b2 = 2 (mod 5).

Since a2 = −a1 and b1b2 = 2, we can multiply a1b2 + a2b1 = 0 through by b1 to obtainthat 2a1 − a1b

21 = 0. Since F5 is an integral domain, we can factorise to deduce that

either a1 = 0 or b21 = 2. The squares in F5 are 0, 1 and 4, so we must have that a1 = 0.This gives that a2 = 0 and b1 + b2 = 0, so b1 = −b2. But then −b21 = 2, so that b21 = 3.Again 3 is not a square in F5, so we have a contradiction.

Thus [f ]5 is irreducible and 5 does not divide the leading coefficient of f , so by thelocalisation principle f is irreducible over Q. Warning: Over F3 we have

[f ]3 = t4 + 1 = (t2 + t− 1)(t3 − t− 1),

so we must choose the prime p in our localisation carefully.

Proposition 25. The field A consisting of complex numbers algebraic over Q is an infinitealgebraic extension of Q.

Proof. We verified that A is a field in Theorem 16. By definition the extension A/Q is algebraic.This is not a finite extension since the degree of the minimal polynomial of an element of A isunbounded: the polynomial tn − 2 is irreducible over Q by Eisenstein with p = 2, so that anycomplex root α of this polynomial is an element of A with [Q(α) : Q] = n. Thus for any n ∈ Nwe have [A : Q] ≥ n. �

6. Splitting fields and normal extensions

From the fundamental theorem of algebra we know that every non-constant polynomial overQ can be written as a product of linear factors over C. The motivation for studying splittingfields is that if we are studying a specific polynomial f ∈ Q[t], then f has such a factorisationover a much smaller field than C, in fact over a finite extension of Q.

Definition (Splits / splitting field). Let L/K be a field extension and let f be a polynomialover K. We say that f splits over L if it can be written as a product of linear factors over L.Equivalently, if λ is the leading coefficient of f , then f splits over L if one has

f = λ(t− α1) · · · (t− αn)

for some α1, . . . , αn ∈ L.We say that L/K is a splitting extension for f if

(i) f splits over L.(ii) If f splits over a field M which is intermediate between L and K (in the sense that

L ⊃M ⊃ K), then L = M .

Equivalently, L is a minimal extension of K over which f splits. We also say L is a splittingfield for f over K in this case.

Notice that if f ∈ K[t] splits over some extension L of K, then the intersection

Σ =⋂{M : M is a subfield of L, K ⊂M and f splits over M}

is itself a field containing K and over which f splits. Clearly this intersection is a minimal suchextension of K (why? ). It follows that if f splits over some extension of K then there exists asplitting field for f over K, an observation we record as a proposition.

16

Proposition 26. Let f be a polynomial over K and suppose that there is an extension L/Ksuch that f splits over L. Then there exists a splitting field for f over K.

We would like to determine exactly what a splitting field for f over K looks like. Clearly if fhas the factorisation λ(t−α1) · · · (t−αn) then f splits over the field K(α1, . . . , αn). Moreover,it is clear that this is a minimal field containing K and the roots α1, . . . , αn (by definition of thefield generated by K and α1, . . . , αn). The only way that K(α1, . . . , αn) could not be a splittingfield for f over K is if f has another decomposition into linear factors f = λ(t−β1) . . . (t−βn),with K(β1, . . . , βn) a proper subfield of K(α1, . . . , αn). For this to be true the sets {α1, . . . , αn}and {β1, . . . , βn} cannot be equal, but this contradicts unique factorisation in L[t] with L =K(α1, . . . , αn), since either side of the identity

(t− α1) · · · (t− αn) = (t− β1) · · · (t− βn)

would constitute genuinely distinct decompositions into irreducible factors. We have thereforeproved the following.

Proposition 27 (Form of a splitting field). Let f ∈ K[t]. Suppose that Σ is a splitting fieldfor f over K. Then Σ takes the form K(α1, . . . , αn) where α1, . . . , αn are the roots of f in Σ.

Corollary 28. Let f ∈ K[t]. If Σ/K is a splitting extension for f , then Σ/K is a finiteextension (hence also algebraic).

Proof. From Theorem 17, the extension Σ/K is finite if and only if Σ takes the formK(α1, . . . , αn)with each αi algebraic over K. By the proposition Σ certainly takes this form. It also followsfrom Theorem 17 that a finite extension is algebraic. �

So far, to construct a splitting field for f over K we have required the existence of a largerfield over which f splits. Our aim is now to show that such fields always exist.

Theorem 29 (Existence of splitting fields). Let f be a non-zero polynomial over K. Then thereexists an extension L/K such that f splits over L. Thus by Proposition 26, f has a splittingfield over K.

Proof. The result is trivial if deg(f) ≤ 1, so let us assume that the degree of f exceeds 1. Itfollows that f has an irreducible factor g over K, so f = gh for some h ∈ K[t]. Consider thequotient ring K[t]/(g) and the element α = t+ (g). We know from Proposition 8 and Theorem13 that K[t]/(g) is a simple algebraic extension of K equal to K(α) and with α a zero of g.Thus over K(α) we have g = (t−α)g1 for some g1 ∈ K(α)[t]. By induction, g1 splits over someextension M of K(α).

The polynomial h has all its coefficients in M since K ⊂ M . As deg(h) < deg(f), we canagain apply the induction hypothesis to conclude that there exists an extension L of M suchthat h splits over L. It follows that f splits over L. �

We have therefore shown that if f ∈ K[t] then there exists a splitting field extension Σ/K forf . We also know that Σ must take the form K(α1, . . . , αn) where the αi are the roots of f in Σ.There is a lot of algebraic rigidity in this structure, so that if K(β1, . . . , βn) is another splittingfield for f over K, surely the roots β1, . . . , βn are in some sense ‘the same’ as α1, . . . , αn.One would hope that we could ‘pair roots off’, so that there exists a field isomorphism τ :K(α1, . . . , αn) → K(β1, . . . , βn) with (τ(α1), . . . , τ(αn)) a permutation of (β1, . . . , βn). Thisis not such a trivial task. In essence we are proving the uniqueness of splitting fields up toisomorphism.

To prove this it helps to clarify what we mean by saying that the extensions L/K and L′/K ′

are isomorphic.

Definition. Let i1 : K1 → L1 and i2 : K2 → L2 be field extensions (i.e. field monomorphisms).We say these extensions are isomorphic if all of the following hold:

(i) There exists an isomorphism σ : K1 → K2.(ii) There exists an isomorphism τ : L1 → L2.

17

(iii) We have the identity τ ◦ i1 = i2 ◦ σ.

The last condition can be restated in the language of category theory by saying that thefollowing diagram commutes:

K1σ−−−−→ K2yi1 yi2

L1τ−−−−→ L2

To show the uniqueness of splitting extensions, our task is therefore to prove that if i1 : K →Σ1 and i2 : K → Σ2 are splitting extensions for the same polynomial f ∈ K[t], then there existsan isomorphism τ : Σ1 → Σ2 such that the following diagram commutes

Kid−−−−→ Kyi1 yi2

Σ1τ−−−−→ Σ2

In other words τ ◦ i1 = i2. It actually helps to prove something more general: that if i1 :K1 → Σ1 is a splitting extension for f ∈ K1[t] and if i2 : K2 → Σ2 is a splitting extension for15

σ(f) ∈ K2[t], then these extensions are isomorphic. In other words, there exist isomorphismsσ : K1 → K2 and τ : Σ1 → Σ2 such that the following diagram commutes

K1σ−−−−→ K2yi1 yi2

Σ1τ−−−−→ Σ2

(To say this diagram commutes means that τ ◦ i1 = i2 ◦ σ.)We begin with an important special case of this result.

Lemma 30. Suppose that K1(α)/K1 and K2(β)/K2 are simple algebraic extensions, that K1

is isomorphic to K2 and that the minimal polynomial of α over K1 corresponds to the minimalpolynomial of β over K2 (up to isomorphism). Then the extensions are isomorphic. Moreover,given any isomorphism σ : K1 → K2 we can extend this to an isomorphism τ : K1(α)→ K2(β)which maps α to β.

Proof. Let σ : K1 → K2 be an isomorphism. First note that since σ is a ring isomorphismfrom K1 to K2, it induces a ring isomorphism from K1[t] to K2[t], obtained by applying σ tocoefficients. Also note that if I is an ideal of K1[t] then σ(I) is an ideal of K2[t] and we havean isomorphism K1[t]/I ∼= K2[t]/σ(I) via the map

f + I 7→ σ(f) + σ(I).

Let mα denote the minimal polynomial of α over K1. Then we are assuming that mβ = σ(mα)is the minimal polynomial of β over K2. One can check that σ applied to the ideal (mα) coincideswith the ideal (mβ). Hence from our previous remark we see that K1[t]/(mα) is isomorphic toK2[t]/(mβ) via the map

f + (mα) 7→ σ(f) + (mβ).

Using this and Theorem 13, we obtain a series of isomorphisms which mapK1(α)→ K1[t]/(mα)→K2[t]/(mβ)→ K2(β) given by

f(α) 7→ f + (mα) 7→ σ(f) + (mβ) 7→ σ(f)(β).

One can check that the resulting isomorphism τ : K1(α) → K2(β) satisfies τ |K1 = σ andτ(α) = β. �

We apply this lemma iteratively to obtain the following theorem.

15Here σ(f) denotes the polynomial obtained from f by applying σ to the coefficients of f .

18

Theorem 31 (Uniqueness of splitting fields). Let σ : K1 → K2 be a field isomorphism and letf ∈ K1[t]. Suppose that Σ1/K1 is a splitting extension for f and that L2/K2 is an extensionover which σ(f) splits. Then there exists a monomorphism τ : Σ1 → L2 which restricts to σon K1. Moreover, if L2 = Σ2 is a splitting field for σ(f) over K2, then Σ1/K1 and Σ2/K2 areisomorphic extensions via

K1σ−−−−→ K2y y

Σ1τ−−−−→ Σ2

Proof. We proceed by induction on n = deg(f). Since Σ1 is a splitting field for f over K1, wehave that Σ1 = K(α1, . . . , αn) where f = λ(t−α1) · · · (t−αn) and λ is the leading coefficient off . Let m1 denote the minimal polynomial of α1 over K1. Then m1|f , so σ(m1)|σ(f) and σ(m1)must therefore split over L2. Let β1 be a zero of σ(m1) in L2. Since σ is an isomorphism, we seethat σ(m1) is monic irreducible over K2. Thus σ(m1) is the minimal polynomial for β1 over K2.We now apply Lemma 30 to deduce that there exists an isomorphism σ1 : K1(α1) → K2(β1)which restricts to σ on K1. Let K ′1 = K1(α1) and K ′2 = K2(α) and let g ∈ K ′1[t] be the quotientof f on division by (t− α1), so that f = (t− α1)g. Notice that Σ1 is a splitting field for g overK ′1 and σ1(g) splits over L2. We can therefore apply the induction hypothesis to conclude thatthere exists an isomorphism τ : Σ1 → L2 which restricts to σ1 on K ′1, and thus restricts to σon K1.

Finally, suppose that L2 = Σ2 is a splitting field for σ(f) over K2. We claim that τ : Σ1 → Σ2

is then an isomorphism. From the first part of the theorem, it suffices to establish surjectivity,i.e. that τ(Σ1) ⊃ Σ2. This follow from the fact that σ(f) splits over τ(Σ1). �

Combining the results from Proposition 26 to Theorem 31 we see that given a polynomialf ∈ K[t] there exists a splitting field extension Σ/K for f , and this extension is unique up toisomorphism.

Example (The splitting field of t4− 2 over Q). Let α = 4√

2. Over C we can factorise t4− 2 as

(t− α)(t+ α)(t− iα)(t+ iα).

Thus Σ = Q(α,−α, iα,−iα) = Q(α, iα) is a splitting field for t4− 2 over Q. Clearly Q(α, iα) ⊂Q(α, i). However, since

i =iα× α3

2∈ Q(α, iα),

we in fact have Q(α, iα) = Q(α, i).For later purposes, let us calculate the degree of some intermediate fields between Q and

Q(α, i). The polynomial t4 − 2 is irreducible over Q (Eisenstein with p = 2). Thus [Q(α) :Q] = deg(t4 − 2) = 4. The polynomial t2 + 1 is irreducible over Q (it has no roots in Q), thus[Q(i) : Q] = 2. If [Q(α)(i) : Q(α)] = 1 then i ∈ Q(α) ⊂ R, which is a contradiction. Thus[Q(α, i) : Q(α)] = 2 (it is at most 2 since [Q(i) : Q] = 2. It follows from the tower law that[Q(α, i) : Q] = 8. Again using the tower law, we deduce that [Q(α, i) : Q(i)] = 4. We summariseour deliberations in the following diagram, where the number next to an arrow indicates the

19

degree of the extension.C

Q(α, i)

∞∧

Q(i)

4>

Q(α)

2

<

Q

8

∧

4

>

2

<

(6.1)

Definition (Normal extension). We say the extension L/K is normal if every irreduciblepolynomial f ∈ K[t] with a zero in L also splits over L.

Examples. The extension C/Q is normal by the fundamental theorem of algebra. The exten-sion R/Q is not normal since t3 − 2 is irreducible over Q with a zero in R, but does not splitover R (it has two complex roots).

Theorem 32. An extension L/K is normal and finite if and only if L is the splitting field forsome polynomial over K.

Proof. Suppose L/K is normal and finite. By the form of finite extensions (Theorem 17), wesee that L takes the form K(α1, · · · , αn) for some αi all algebraic over K. Let mi denote theminimal polynomial of αi over K. Then f = m1 · · ·mn ∈ K[t] and f splits over L by normality.Moreover if L/M/K and f splits over M then (by the fundamental theorem of algebra) αi ∈Mfor all i, hence M ⊃ K(α1, . . . , αn) = L. It follows that L is indeed a splitting field for f overK.

Conversely, suppose that L is a splitting field for f ∈ K[t]. The L = K(α1, . . . , αn) wherethe αi are the roots of f . Hence L is finitely generated with each generator algebraic over K,thus L/K is a finite extension by Theorem 17. It remains to prove normality. Let g ∈ K[t] beirreducible over K with g(α) = 0 for some α ∈ L. We wish to show that g splits over L. ByTheorem 29, there exists an extension M of L over which g splits. We’re done if we can showthat all the roots of g in M also lie in L. To this end, let β ∈M with g(β) = 0. Since α and βboth have the same minimal polynomial over K, Lemma 30 shows there exists an isomorphismσ : K(α)→ K(β) which restricts to the identity on K and maps α to β. Since L is a splittingfield for f over K, we see that L(α) is a splitting field for f over K(α) and L(β) is a splittingfield for f = σ(f) over K(β). Hence by uniqueness of splitting fields (Theorem 31), there existsan isomorphism τ : L(α) → L(β) which restricts to σ on K(α). It follows that the extensionsL(α)/K(α) and L(β)/K(β) are isomorphic. In particular [L(α) : K(α)] = [L(β) : K(β)]. Bythe tower law, we therefore have that

[L(β) : L] =[L(β) : K(β)]

[L : K(β)]=

[L(β) : K(β)]

[L : K]/[K(β) : K]=

[L(α) : K(α)]

[L : K]/[K(α) : K]= [L(α) : L] = 1.

Thus β ∈ L as required. �

Corollary 33. Let L/M/K be a tower of field extensions and suppose that L/K is a finitenormal extension. Then L/M is also a finite normal extension.

Proof. From Theorem 32 we have that L/K is a splitting extension for some f ∈ K[t]. Sincef ∈M [t] also, L/M is also a splitting extension for f . �

7. The Galois Group of an Extension

So far we have studied finite extensions L/K through their vector space structure alone. Thisis not subtle enough - all vector spaces of dimension n over K are isomorphic. One of the most

20

important insights of Galois theory is that the structure of an extension is determined by itssymmetries.

What are symmetries in this case? They are bijections σ : L→ L which leave all polynomialequations over K invariant and respect the field structure of L. We say σ : L → L is anautomorphism of L if it is a bijective isomorphism from a field to itself. The set of allautomorphisms of L forms a group Aut(L) under composition.

Let K be a subfield of L. A K-automorphism of L is an element σ of Aut(L) which fixesK in the sense that σ|K = idK . The set of K-automorphisms of L form a subgroup of Aut(L),denoted Gal(L/K) and called the Galois group of L over K.

Example (Gal(C/R)). Let σ ∈ Gal(C/R). Then 0 = σ(0) = σ(i2 + 1) = σ(i)2 + 1. Thusσ(i) = ±1. If σ(i) = i, then for all complex numbers z = a + ib with a, b ∈ R we haveσ(z) = a + ib = z. Thus σ = idC and clearly this is an element of Gal(C/R). If on the otherhand we have σ(i) = −i, then σ(a + ib) = a − ib, so that σ is complex conjugation, which isagain clearly an element of Gal(C/R). Therefore Gal(C/R) ∼= Z/2Z.

Example (Gal(Q(21/3)/Q)). Let α = 3√

2. As in the previous example, if σ ∈ G := Gal(Q(21/3)/Q)then one can show that σ(α) must be a real root of t3− 2. Since α is the only such real root, itfollows that σ(α) = α. As 1, α, α2 form a basis for this extension over Q, we deduce that thereis only one element of G, namely the identity automorphism. Thus G is the trivial group.

Remark 2. For a finitely generated algebraic extension16 L = K(α1, . . . , αn) of K, the elementsσ ∈ Gal(L/K) are completely determined by their values on α1, . . . , αn. This is because thenon-negative powers of αi span K(αi) as a vector space over K, hence by the proof of the tower

law, the monomials αi11 · · ·αinn span K(α1, . . . , αn).

Proposition 34. Let f ∈ K[t] and σ ∈ Gal(L/K). Then for all α ∈ L we have f(α) = 0 ifand only if f(σ(α)) = 0. In particular, f is the minimal polynomial for α over K if and only iff is the minimal polynomial for σ(α) over K.

Proof. Exercise. �

The above result says that Gal(L/K) permutes those roots of f ∈ K[t] which lie in L. Whenf is irreducible over K and L = Σ is a splitting field for f over K, then Gal(L/K) in factpermutes all the roots of f transitively.

Proposition 35. Let f ∈ K[t] be irreducible and Σ be a splitting field for f over K. ThenGal(Σ/K) permutes the roots of f in Σ transitively. In other words, for α, β ∈ L with f(α) =f(β) = 0 there exists τ ∈ Gal(Σ/K) such that σ(α) = β.

Proof. By Lemma 30 we can extend the identity idK to an isomorphism σ : K(α) → K(β)which maps α to β. By uniqueness of splitting fields (Theorem 31) we can then extend σ to anisomorphism τ : Σ→ Σ. Then τ ∈ Gal(Σ/K) and τ(α) = β. �

Example. Let α = 4√

2. We saw in the last section that Σ = Q(α, i) = Q(α,−α, iα,−iα)is a splitting field for t4 − 2 over K = Q. By the above remark and Proposition 34, eachσ ∈ Gal(Σ/K) is completely determined by how it permutes the roots α,−α, iα,−iα. Wetherefore deduce that Gal(Σ/K) is isomorphic to a subgroup of S4. By Proposition 35 thismust be a transitive subgroup of S4.

The function K 7→ Gal(L/K) maps a subfield K of L to a subgroup Gal(L/K) of Aut(L).Can we reverse this map? Given a subgroup G ≤ Aut(L) define

LG = {x ∈ L : (∀σ ∈ G)[σ(x) = x]} .One can check that LG is a subfield of L, called the fixed field of G.

We now state a number of properties of these maps. These are simple deductions which weleave as exercises for the reader (although we proved them in the lectures).

16Equivalently, an extension of finite degree.

21

The maps K 7→ Gal(L/K) and G 7→ LG reverse inclusions, so that if K,M are subfields of Lthen

K ⊂M =⇒ Gal(L/K) ⊃ Gal(L/M). (7.1)

Similarly, if H,G are subgroups of Aut(L) then

H ⊂ G =⇒ LH ⊃ LG. (7.2)

For any subfield K of L we have

K ⊂ LGal(L/K). (7.3)

For any subgroup G of Aut(L) we have

G ⊂ Gal(L/LG). (7.4)

We always have the identities

Gal(L/K) = Gal(L/LGal(L/K)) and LG = LGal(L/LG). (7.5)

Definition (Galois extension). An extension L/K is called Galois if LGal(L/K) = K. In words:no element of L \K is fixed by the K-automorphisms of L.

Examples.

• Recall that Gal(C/R) consists of the identity and complex conjugation. Since the onlycomplex numbers fixed by conjugation are the reals, we see that C/R is Galois.• Let α = 2

√3. Recall that Gal(Q(α)/Q) is the trivial group consisting solely of the

identity. Therefore every element of Q(α) is fixed by the Q-automorphisms of Q(α), sothis extension is far from being Galois.

We aim to prove the following alternative (numerical) characterisation of when an extensionis Galois.

L/K is Galois ⇐⇒ [L : K] = |Gal(L/K)|. (7.6)

In fact, we eventually show that [L : K] is always at least as large as |Gal(L/K)|, so Galoisextensions are those whose Galois group is as large as possible.

We prove (7.6) using a series of arguments with the flavour of abstract linear algebra. Firsta definition.

Definition. Let L be a field and S a non-empty set. Then the set of functions f : S → Lhas the structure of an L-vector space when functions f are regarded as tuples

(f(x)

)x∈S with

addition and scalar multiplication defined in the obvious manner.

Lemma 36. [Dedekind’s lemma] The monomorphisms σ : K → L are linearly independent overL.

Proof. Suppose otherwise. Then there exist monomorphisms σ1, . . . , σn : K → L and λ1, . . . , λn ∈L not all zero such that

0 = λ1σ1 + . . .+ λnσn. (7.7)

Let n be minimal with respect to the existence of such a dependence. Then all λi are non-zero.Since σn−1 and σn are distinct, there exists y ∈ L with σn−1(y) 6= σn(y). Evaluating (7.7) atyx with x ∈ L arbitrary, we deduce that

0 = λ1σ1(y)σ1 + . . .+ λnσn(y)σn. (7.8)

Multiplying (7.7) through by σn(y) and subtracting the resulting identity from (7.8), we con-clude that

0 = λ1

(σ1(y)− σn(y)

)σ1 + . . .+ λn−1

(σn−1(y)− σn(y)

)σn.

Since λn−1

(σn−1(y)− σn(y)

)6= 0, we have contradicted the minimality of n. �

Theorem 37. Let L be a field and G a subgroup of Aut(L). Then [L : LG] = |G|.22

Proof. Let B be a basis for L over LG. Consider the (possibly infinite) matrix with entries inL defined by

M =(g(x)

)g∈Gx∈B

.

Here rows are indexed by elements of G and columns are indexed by elements of our basis B.

Claim. (i) The rows of M are linearly independent over L.(ii) The columns of M are linearly independent over L.

Since the row rank of M equals the column rank, the claim yields the theorem when combinedwith the observations:

row rank of M ≤ (number of entries in the rows of M) = |X| = [L : LG],

column rank of M ≤ (number of entries in the columns of M) = |G|.

Let us therefore prove the claim. Suppose that there exists an L-linear dependence amongstthe rows of M , so that there exists g1, . . . , gn ∈ G and λ1, . . . , λn ∈ L such that for all x ∈ Bwe have

λ1g1(x) + . . .+ λngn(x) = 0. (7.9)

Note that any LG-linear map T : L → L is completely determined by its values on the basisB. Therefore the map λ1g1 + . . . λngn : L → L is identically zero. By Dedekind’s lemma, theautomorphisms g : L→ L are linearly independent over L, so we must have λ1 = . . . = λn = 0.Claim (i) therefore follows.

For claim (ii), let us suppose that there is an L-linear dependence amongst the columns ofM . Choose x1, . . . , xn ∈ B minimal for which there exist λ1, . . . , λn ∈ L not all zero with∑n

i=1 λig(xi) = 0 for all g ∈ G. By minimality, all λi must be non-zero, so multiplying this

linear relation through by λ−11 , we may assume that for all g ∈ G we have

g(x1) + λ2g(x2) + . . .+ λng(xn) = 0. (7.10)

Applying an arbitrary h ∈ G to either side of this expression, we deduce that for all g ∈ G wehave

hg(x1) + h(λ2)hg(x2) + . . .+ h(λn)hg(xn) = 0. (7.11)

It is well known from group theory that multiplying a group by any of its elements permutesthe group, so that hG = G. Therefore (7.11) tells us that for any g ∈ G we have

g(x1) + h(λ2)g(x2) + . . .+ h(λn)g(xn) = 0. (7.12)

Subtracting (7.10) from (7.12), we find that for any g ∈ G we have(h(λ2)− λ2

)g(x2) + . . .+

(h(λn)− λn

)g(xn) = 0. (7.13)

This gives a linear dependence amongst the columns of M corresponding to x2, . . . , xn, contra-dicting the minimality of n unless h(λi) − λi = 0 for all i. Since h ∈ G is arbitrary we musthave λi ∈ LG for all i. Taking g = idL in (7.10) therefore implies that

x1 + λ2x2 + . . .+ λnxn = 0,

and this contradicts the linear independence of B over LG. �

Corollary 38. (i) We have the inequality |Gal(L/K)| ≤ [L : K], with equality if and only ifL/K is Galois.

(ii) If G is a finite subgroup of Aut(L), then Gal(L/LG) = G.

Proof. (i) By Theorem 37 we have

[L : LGal(L/K)] = |Gal(L/K)|.

Since K ⊂ LGal(L/K) we also have [L : LGal(L/K)] ≤ [L : K], with equality iff K =

LGal(L/K), i.e. iff L/K is Galois.23

(ii) We know that G ⊂ Gal(L/LG). Also by Theorem 37 we have

|G| = [L : LG] = [L : LGal(L/LG)] = |Gal(L/LG)|.

Hence if G is finite, then G = Gal(L/LG).�

Example. Let L = Q(√

2,√

3) and K = Q. Any element of Aut(L) fixes Q (since σ(1) = 1 forall σ ∈ Aut(L) and L has characteristic zero). Thus Gal(L/K) = Aut(L).

One can check17 that√

3 /∈ Q(√

2), hence√

3 has minimal polynomial t2 − 3 over Q(√

2).Using the proof of the tower law, we see that 1,

√2,√

3,√

6 then constitute a basis for Q(√

2,√

3)over Q. It follows that each element of Aut(L) is completely determined by its values on

√2 and√

3. Since each σ ∈ Gal(L/K) permutes roots of polynomials over K, we see that σ(√

2) = ±√

2and σ(

√3) = ±

√3. There are therefore four possibilities for σ:

σ0(a+ b√

2 + c√

3 + d√

6) = a+ b√

2 + c√

3 + d√

6,

σ1(a+ b√

2 + c√

3 + d√

6) = a− b√

2 + c√

3− d√

6,

σ2(a+ b√

2 + c√

3 + d√

6) = a+ b√

2− c√

3− d√

6,

σ3(a+ b√

2 + c√

3 + d√

6) = a− b√

2− c√

3 + d√

6,

where a, b, c, d ∈ Q. One can check that each of these maps is a genuine automorphism of L.Moreover, each satisfies σ2

i = id. Therefore Gal(L/K) ∼= C2 × C2, the Klein 4-group (here Cndenotes the cyclic group of order n).

Let us consider fixed fields of subgroups:

LAut(L) = Q,

L{id,σ1} = Q(√

3),

L{id,σ2} = Q(√

2),

L{id,σ3} = Q(√

6),

L{id} = Q(√

2,√

3).

Notice that |Gal(L/K)| = 4 = [L : K], so L/K is Galois. Later, we show that since L/K isfinite and Galois, all intermediate fields between L and K correspond to fixed fields of subgroups.Thus we have given a complete description of subfields of Q(

√2,√

3).

8. More on normal extensions

We will eventually prove that an extension is Galois if and only if it is both normal andseparable. Before defining separable extensions, let us give an alternate criterion for normalextensions. This criterion is part of a general philosophy of classifying extensions in terms ofextensions of embeddings (injective homomorphisms) into algebraically closed fields. We beginwith two key facts.

Theorem 39. Let K be a field. Then there exists an algebraically closed field containing K asa subfield.

Proof. Essentially an “infinite induction” version of the proof of Theorem 29. �

Theorem 40. Let L/K be an algebraic extension and σ : K → M be an embedding of K intoan algebraically closed field M . Then there exists an extension of σ to an embedding of L in M .

17Mimic the proof that√

2 /∈ Q.

24

Proof. (Proof given only for finite extensions) We will prove this by induction on [L : K]. WriteL = K(α1, α2, . . . , αn). We may assume without loss of generality that αn /∈ K(α1, α2, . . . , αn−1).Put L1 = K(α1, α2, . . . , αn−1). Then [L1 : K] < [L : K]. By induction hypothesis, there is anextension of σ to an embedding of L1 in M . We denote this extension also by σ. Let M1 bethe image of L1 under σ. Then σ : L1 → M1 is an isomorphism and L = L1(α). Let m(x)be the minimal polynomial of α over L1 and put n(x) = σ(m(x)). Then n(x) ∈ M1[X] is anirreducible polynomial. Let β ∈M be a root of n(x). By Lemma 30, there exists an extensionof σ to an isomorphism from L = L1(α) to M1(β). This completes the proof.

�

Remark 3. A close look at the proof shows that the number of such extensions of σ is lessthan or equal to [L : K].

Definition. Let K be a subfield of L. We say that an embedding σ of L into a field M (withM containing K) is a K-embedding if σ fixes all elements of K.

We can now restate the property of an extension being normal.

Theorem 41. Let L/K be an algebraic extension and M be an algebraically closed field con-taining L. Then the following are equivalent.

(1) L/K is a normal extension.(2) Any K-embedding of L in M is an automorphism of L.

Proof. Assume that L/K is a normal extension. Let σ be a K-embedding of L in M . We needto show that σ is an automorphism of L. Let α ∈ L. It suffices to show that σ(α) ∈ L. Byassumption, mα,K splits in L. Since σ(α) is a root of mα,K , this shows that σ(α) ∈ L.

Conversely, suppose that any K-embedding of L in M is an automorphism of L. We need toshow that L/K is a normal extension. To do this, it is enough to show that every irreduciblepolynomial over K which has a root in L also splits over L. Let p(x) be such a polynomial witha root α ∈ L. It suffices to show that all the other roots of p(x) belong to L. Let β ∈ M beany other root of p(x). Let σ : K(α)→ K(β) be a K-embedding; this exists by Lemma 30. ByTheorem 40, this extends to a K-embedding of L in M taking α to β. By assumption, this isan automorphism of L. Hence β ∈ L, as required.

�

We end this section with a few other important properties of normal extensions.

Definition (Compositum). Let K1,K2 be two fields that are contained in some field L. Thenwe denote by K1K2 the smallest subfield of L containing both K1 and K2, and call it thecompositum of K1 and K2 in L.

Remark 4. The isomorphism class of K1K2 does not depend on the choice of L. In practisewe will often fix some algebraically closed field containing all the fields under consideration andtake compositums in it.

Example. Let K be a subfield of L and let α1, α2, . . . , αm, β1, β2, . . . , βn be elements of L.Then K(α1, α2, . . . , αm)K(β1, β2, . . . , βn) = K(α1, α2, . . . , αm, β1, β2, . . . , βn).

Theorem 42. Let E/K and F/K be algebraic extensions and suppose that E,F are bothcontained in some field L.

(1) Suppose that E/K is normal. Then EF/F is normal.(2) Suppose that E/K and F/K are both normal. Then EF/K and (E ∩ F )/K are also

normal.

Proof. Fix an algebraically closed field M containing L. To prove the first assertion, let σ be aF -embedding of EF in M . Then σ(EF ) = σ(E)σ(F ) = σ(E)F = EF ; hence EF/F is normal.

Next, suppose that E/K and F/K are both normal. Let σ be a K-embedding of EF in M .Then σ(EF ) = σ(E)σ(F ) = EF ; hence EF/K is normal.

25

Finally, if σ is a K-embedding of E ∩ F in M , we first extend σ to a K-embedding of EF inM and then observe that σ(E ∩ F ) = E ∩ F . �

Example. Let α = 4√

2. Consider the diagram

C

Q(α, i)

∞∧

Q(i)

4>

Q(α)

2

<

Q

8

∧

4

>

2

<

Which of the extensions are normal?

9. Separable extensions

In this section we define separable extensions and prove various properties about them.

Theorem 43. Let E/F be a finite extension and let σ : F → M be an embedding of F intoan algebraically closed field. Then there exists an integer [E : F ]s which depends only on theextension E/F (and is independent of σ) such that σ has exactly [E : F ]s extensions to anembedding of E in M .

Proof. Let Sσ be the set of extensions of σ to E. Let τ : F →M be another embedding. Thenτ ◦ σ−1 is an isomorphism from σ(F ) to τ(F ). We may assume without loss of generality thatM is algebraic over σ(F ). Let λ : M →M be an extension of τ ◦ σ−1. Then σ∗ 7→ λ ◦ σ∗ givesa bijection from Sσ to Sτ .

�

The integer [E : F ]s is called the separable degree of the extension E/F . It has several niceproperties.

Theorem 44. (1) Let E/F/K be a tower of finite extensions. Then [E : K]s = [E : F ]s[F :K]s.

(2) Let E/K be a finite extension. Then [E : K]s ≤ [E : K].

Proof. Let σ : K 7→ L be an embedding of K in an algebraically closed field L. Let {σi}i∈Ibe the family of distinct extensions of σ to F , and for each i let τij be the family of distinctextensions of σi to E. Then every extension of σ to E is one of the τij . Clearly the set of τijhas [E : F ]s[F : K]s elements. This gives us [E : K]s = [E : F ]s[F : K]s.

To show that [E : K]s ≤ [E : K], we write E/K as a tower of simple extensions E =K(α1, α2, . . . αn)/K(α1, α2, . . . αn−1)/ . . . /K(α1)/K. By multiplicativity of the separable de-gree, it suffices to show that [M(α) : M ]s ≤ [M(α) : M ] for each field M . Let N be analgebraically closed field containing M . We need to show that the number of M -embeddingsof M(α) in N is less than or equal to d = [M(α) : M ]. Let p(x) be the minimal polynomial ofα over M . Then p(x) has degree d. Moreover, each M -embedding σ of M(α) in N takes α tosome root of p(x) and where α goes completely determines σ. So the number of such σ equalsthe number of distinct roots of p(x) which is bounded by d. �

Definition. A finite extension E/F is called separable if [E/F ]s = [E/F ].

Theorem 45. Let E/F/K be a tower of finite extensions. Then E/K is separable if and onlyif E/F and F/K are both separable.

26

Proof. Follows immediately from Theorem 44. �

Definition. Let K be a field and p(x) ∈ K[x] be an irreducible polynomial over K. Then wesay that p(x) is separable if it has no multiple zeroes in its splitting field, or equivalently, if ithas no multiple zeroes in an algebraically closed field containing K.

Definition. Let K be a field and α be an element algebraic over K. Then we say that α isseparable over K if the minimal polynomial of α over K is separable.

Remark 5. It is clear from the above definition that if L/K is an algebraic extension with αseparable over K, then α is also separable over L.

Lemma 46. Let K be a field and α be an element algebraic over K. Then α is separable overK if and only if K(α)/K is separable.

Proof. Let m(x) denote the minimal polynomial of α over K. From the proof of Theorem 44, itfollows that [K(α) : K]s equals the number of distinct roots of m(x) in some algebraically closedfield. So K(α)/K is separable if and only if the number of distinct roots of m(x) equals thedegree of m(x), i.e., if and only if m(x) has no multiple roots, i.e., if and only α is separable. �

Thus, the concepts of separability for elements, polynomials, and field extensions are allessentially the same. The next Theorem completes this picture.

Theorem 47. Let L/K be a finite extension of fields. Then, the following are equivalent.

(1) L/K is separable.(2) All elements of L are separable over K.(3) There exist elements α1, . . . , αn in L which are separable over K and such that L =

K(α1, . . . , αn).

Proof. Proof that (1) implies (2): Suppose L/K is separable. Let α ∈ L. Then, by Theorem 45,K(α)/K is separable. By Lemma 46, α is separable over K.

Proof that (2) implies (3): Trivial.Proof that (3) implies (1): Follows from Theorem 45, Lemma 46 and Remark 5. �

The concept of separable extensions extends easily to infinite algebraic extensions.

Definition. An algebraic extension E/F is called separable if every finite sub-extension isseparable, i.e. M/F is separable for any intermediate field E/M/F with M/F finite.

Remark 6. It is now an easy exercise that Theorem 45 and (1)⇔ (2) of Theorem 47 continueto hold for infinite separable extensions.

Theorem 48. Let E/K and F/K be finite extensions and suppose that E,F are both containedin some field L.

(1) Suppose that E/K is separable. Then EF/F is separable.(2) Suppose that E/K and F/K are both separable. Then EF/K and (E ∩ F )/K are also

separable.

Proof. Proof of (1): Suppose that E = K(α1, . . . , αn). Then EF = F (α1, . . . , αn). We knowthat each αi is separable over K. So, by Remark 5, each αi is separable over F . This completesthe proof.

Proof of (2): If E/K and F/K are both separable then EF/F is separable. Hence EF/Kis separable by Theorem 45. The proof that (E ∩ F )/K is separable also follows immediatelyfrom Theorem 45. �

Remark 7. The above Theorem continues to hold true for infinite algebraic extensions (proveit!)

27

So how do we check if an extension is actually separable? By Theorem 47, it is enoughto check if certain elements are separable, and for that it is enough to check if their minimalpolynomials are separable. Thus it all boils down to the question of determining when anirreducible polynomial can have multiple roots in an algebraically closed field. For this, we needthe concept of formal derivative.

Definition. Let K be a field and

f(x) = anxn + an−1x

n−1 + . . .+ a1x+ a0 ∈ K[x].

Then we define the formal derivative of f , denoted Df to be the polynomial in K[x] given by

Df(x) = nanxn−1 + (n− 1)an−1x

n−2 + . . .+ a1.

It is easy to check that the usual rules of differentiation hold in this context: D(f + g) =Df + Dg and D(fg) = (Df)g + f(Dg) for all f, g ∈ K[x]. The usefulness of this concept isbecause of the following Lemma.

Lemma 49. A nonzero polynomial f over a field K has a multiple zero in an algebraicallyclosed field containing K if and only if f and Df have a common factor (in K[x]) of degree≥ 1.

Proof. Fix an algebraically closed field M containing K. Suppose f(x) = (x−α)2g(x) for someα ∈ M , g ∈ M [x]. Then Df = 2(x − α)g(x) + (x − α)2Dg, so α is a zero of both f and Df .Thus the minimal polynomial of α over K divides both f and Df .

Conversely, suppose that over M , f(x) = λ∏ni=1(x − αi), where the various αi ∈ M are all

distinct. Then

Df(x) = λ((x− α2)(x− α3) . . . (x− αn) + (x− α1)(x− α3) . . . (x− αn) + . . .

. . .+ (x− α1)(x− α2) . . . (x− αn−1)).

It follows that

Df(αi) = (αi − α1) . . . (αi − αi−1)(αi − αi+1) . . . (αi − αn) 6= 0.

So f and Df have no common roots in M . It follows that f and Df cannot have a commonfactor (in K[x]) of degree ≥ 1 (since any root of this common factor in M would be a commonroots of f and Df in M). �

This leads to the following Theorem.

Theorem 50. Over a field of characteristic zero, all irreducible polynomials are separable. Overa field of characteristic p, an irreducible polynomial is inseparable if and only if it has the form

arxrp + ar−1x

(r−1)p + . . .+ a1xp + a0, r ≥ 1

Proof. Let K be a field and f ∈ K[x] be irreducible. Then f is not separable if and only if f ,Df have a common factor. But f is irreducible and Df is of lower degree than f . Hence thisis only possible if Df = 0. It follows that if f = anx

n + . . .+ a0, then the quantities rar are allequal to 0. This is clearly impossible if the characteristic of K is 0. And if the characteristic isp, then this happens provided ar = 0 for all r that are not multiples of p. �

Corollary 51. All algebraic extensions of a field of characteristic 0 are separable.

In fact there are certain fields of characteristic p with the property that all algebraic extensionsare separable, e.g., any finite field. It is not completely obvious how to cook up a non-separableextension. The key point is that one has to find an irreducible polynomial of the form

arxrp + ar−1x

(r−1)p + . . .+ a1xp + a0.

Example. Let Fp = Z/pZ denote the finite field of p elements. Let K = Fp(u) be the field ofrational functions over Fp in the indeterminate u, and let f = xp− u ∈ K[x]. It can be checkedthat f is irreducible, yet all its roots (in the splitting field) are equal. Thus, if L = K(τ) whereτ is a root of f , then L/K is an extension of degree p that is not separable. (Incidentally, it isnormal.)

28

10. The main theorems of Galois Theory

Let L/K be an algebraic extension. Recall that the set of K-automorphisms of L form agroup, denoted Gal(L/K). Recall that L/K is called Galois if the fixed field of Gal(L/K) equals

K, i.e. LGal(L/K) = K. Recall also that if L/K is a finite extension, then it is Galois if andonly if |Gal(L/K)| = [L : K].

Our first theorem is as follows.

Theorem 52. Let L/K be a finite extension. Then L/K is Galois if and only if it is normaland separable.

Proof. Let n = [L : K]. Let M be an algebraically closed field containing L. Suppose thatL/K is normal and separable. We know from previous work that |Gal(L/K)| ≤ n. Since L/Kis separable, there are exactly n distinct K-embeddings of L in M . Since L/K is normal eachof these embeddings is an automorphism of L. Thus |Gal(L/K)| ≥ n. Hence |Gal(L/K)| = n.This implies that L/K is Galois.

Conversely, suppose that L/K is either not separable or separable but not normal. In each ofthese cases we need to show that L/K is not Galois. So first assume that L/K is not separable.Then there are strictly less than n K-embeddings of L in M . Therefore there are strictly lessthan n K-automorphisms of L, i.e. |Gal(L/K)| < n. So L/K is not Galois. Next, assume thatL/K is separable but not normal. Then there exists a K-embedding of L in M that is not aK-automorphism of L. Since the total number of K-embeddings of L in M equals n, this impliesthat the total number of K-automorphism of L is strictly less than n, i.e., |Gal(L/K)| < n. SoL/K is not Galois. �

Remark 8. The above Theorem continues to hold for infinite algebraic extensions. However,we will not prove this.

Lemma 53. Let L/M be a finite extension. Let τ be an automorphism of L. Then Gal(L/τ(M)) =τGal(L/M)τ−1.

(Note that we are viewing both sides as subgroups of the group of all automorphisms of L;multiplication is just composition of automorphisms)

Proof. Suppose σ ∈ Gal(L/M) and x ∈ M . Then τστ−1(τ(x)) = τ(σ(x)) = τ(x). So τστ−1 ∈Gal(L/τ(M)). Thus, τGal(L/M)τ−1 ⊂ Gal(L/τ(M)). A similar argument proves the otherdirection. �

Theorem 54 (Fundamental theorem of Galois theory). Let L/K be a finite Galois extensionwith G = Gal(L/K). For any subgroup H of G, let LH denote the subfield of elements of Lthat are fixed by all elements of H.

(1) The map M 7→ Gal(L/M) is an inclusion-reversing bijection between intermediate fieldsM (i.e., L/M/K) and subgroups H of G. The inverse of this map is H 7→ LH .

(2) For any intermediate field M , L/M is Galois and [M : K] = |G|/|H| where H =Gal(L/M). Further, M/K is Galois if and only if H is a normal subgroup of G, inwhich case Gal(M/K) is isomorphic to G/H.

Proof. First of all, note that for all intermediate fields M , L/M is Galois. This follows fromthe fact (already proved) that these extensions are normal and separable.

To prove the first part, it suffices to show that for all intermediate fields M , LGal(L/M) = M ,and for all subgroups H of G, Gal(L/LH) = H. The first assertion follows from the fact thatL/M is Galois, and the second follows from Corollary 38.

For part (2), we have already shown that L/M is Galois. Put H = Gal(L/M). Note that[M : K] = [L : K]/[L : M ] = |G|/|H|. So we are left to prove that M/K is Galois if and only ifH is a normal subgroup of G, and to show that if so, then Gal(M/K) is isomorphic to G/H.

Suppose M/K is Galois. Let τ ∈ G. Then τ induces a K-automorphism of M . So τ(M) = M .So, by the previous Lemma, H = τHτ−1. This shows that H is normal in G. Conversely if

29

M/K is not Galois, then it is not normal, so we can pick σ ∈ G such that σ(M) 6= M , henceby the previous part, H 6= Gal(L/σ(M)). Hence by the Lemma above, H 6= σHσ−1; so H isnot normal.

Finally, if M/K is Galois, we need to show that Gal(M/K) = G/H. But the map σ 7→ σ|Mis a homomorphism from G to Gal(M/K). It is surjective by Theorem 40 and it’s kernel isclearly H. �

Next suppose that L/M/K is a tower of finite extensions with L/K Galois. We have seenthat if M/K is Galois, then Gal(M/K) = Gal(L/K)/Gal(L/M). What about the case whenM/K is not Galois? We would still have a Galois group Gal(M/K); can we identify it?

Recall: If H is a subgroup of G, then we define the normalizer of H as follows:

NG(H) = {g ∈ G : gHg−1 = H}.

In particular NG(H) is the maximal subgroup of G in which H is normal.

Proposition 55. Let L/M/K be a tower of extensions with L/K Galois. Let Gal(L/K) = Gand Gal(L/M) = H. Then Gal(M/K) is isomorphic to NG(H)/H.

Proof. For any τ ∈ NG(H), consider τ |M . We claim that τ |M is an automorphism of M .To prove this, it is enough to show that τ(M) = M . But this follows from the fact thatGal(L/τ(M)) = τGal(L/M)τ−1 = Gal(L/M).

So τ 7→ τ |M is a homomorphism from NG(H) to Gal(M/K). The kernel of this homomor-phism consists of those elements in NG(H) that fix M ; thus the kernel is H∩NG(H) = H. More-over the homomorphism is surjective since any element σ of Gal(M/K) can be first extendedto an automorphism σ′ of L, and this automorphism must belong to NG(H) by Lemma 53. �

Next, we prove a very useful result about compositums of Galois extensions.

Theorem 56. Let E/K and F/K be finite extensions and suppose that E,F are both containedin some field L.

(1) Suppose that E/K is Galois. Then EF/F is Galois. Moreover the map σ 7→ σ|E givesan isomorphism from Gal(EF/F ) to Gal(E/E ∩ F ).

(2) Suppose that E/K and F/K are both Galois. Then EF/K and (E ∩ F )/K are alsoGalois. Moreover, the map σ 7→ (σ|E , σ|F ) gives an isomorphism from Gal(EF/(E∩F ))to Gal(E/E ∩ F )×Gal(F/E ∩ F ).

Proof. Suppose that E/K is Galois. Then the fact that EF/F is Galois follows from thecorresponding statements about normal and separable extensions. Now, consider the mapσ 7→ σ|E . It is clearly a homomorphism from Gal(EF/F ) to Gal(E/E ∩ F ). It is injectivebecause if σ|E is trivial then σ fixes all elements of E and of F and hence σ is identity on EF .It is surjective because if not, let the image of Gal(EF/F ) be a subgroup H of Gal(E/E ∩ F ).The fixed field ofH in E consists of all those elements of E that are fixed by all F -automorphismsof EF , which is precisely E ∩ F . So we are done by the main theorem.

Suppose that E/K and F/K are both Galois. The facts that EF/K and (E ∩ F )/K arealso Galois follows from the corresponding statements about normal and separable extensions.Now, consider the map σ 7→ (σ|E , σ|F ), viewed as a homomorphism from Gal(EF/(E ∩ F ))to Gal(E/E ∩ F ) × Gal(F/E ∩ F ). This is injective because if σ|E and σ|F are both trivial,then σ fixes all of EF . This is surjective because (by the first part!) the image contains{1} ×Gal(F/E ∩ F ) and Gal(E/E ∩ F )× {1}. �

Corollary 57. Let E/K and F/K be finite extensions with E/K Galois. Then [EF : F ] divides[E : K].

Example. Consider the splitting fields of x4 +5x2 +6 and x6−8, over Q. What are the variousGalois groups?

30

11. Galois groups of polynomials

Let K be a field and f ∈ K[x] be an irreducible separable polynomial. We define the Galoisgroup of f to be the Galois group of the splitting field of f over K. (Note that the splittingfield and Galois group does not change if we replace f(x) by f(ax+ b) for a, b ∈ K, a 6= 0.)

Let now f be as above of degree n. Let L be the splitting field. Then f has n distinct rootsin L. Let G be the Galois group of f . Then any element of G permutes the roots of f . Thus,we get a homomorphism from G to Sn. This homomorphism is injective because if σ ∈ G fixesall the roots of f , then it fixes all elements of L (since L is generated over K by the roots).Furthermore the image under this homomorphism must be a transitive subgroup of Sn becausegiven any two roots, there is an element of G that takes one root to the other. We note theabove observations in the following Theorem.

Theorem 58. Let K be a field and f ∈ K[x] be an irreducible separable polynomial of degreen. Then the Galois group of f over K is isomorphic to a transitive subgroup of Sn.

In particular the size of the Galois group must be a multiple of n and must also divide n!.Let us do some case studies of polynomials of various degrees.Polynomials of degree 2Let f(x) = x2 + ax+ b with a, b ∈ K. When is it irreducible? It is irreducible if and only if

f has no roots in K. If char(K) 6= 2 this happens if and only if a2 − 4b /∈ K2. So for simplicitylet us assume that char(K) 6= 2 and a2 − 4b is not a square in K. In that case f is also alwaysseparable.

Then if α is a root of f(x) then K(α) contains the other other root β as well, since β = −a−α.So K(α) is the splitting field of f over K. Since [K(α) : K] = 2, the Galois group must be ofsize 2. It follows that the Galois group of f is Z/2Z.

Conversely, if char(K) 6= 2, then any any extension L/K of degree 2 is Galois, because wecan take the minimal polynomial of any element α ∈ L −K and apply the above analysis. Infact we can always choose α so that the minimal polynomial is of the form x2 − a; this followsfrom “completing the square.”

Polynomials of degree 3Let K be a field. For simplicity we assume that char(K) 6= 2, 3. Let f(x) = x3 + ax+ b with

a, b ∈ K. Any polynomial of degree 3 can be brought into this form by making a substitutionof the form x 7→ rx + s, with r, s ∈ K, r 6= 0. Suppose that f has no root in K. Then f isirreducible because any factorization of f must have a factor of degree 1. Also f is separable,using previous results.

Let α be any root of f . So [K(α) : K] = 3. Let L be the splitting field of f . Then L/K(α)/K.Put G = Gal(L/K). Then G is a subgroup of S3. Since 3 divides |G| = [L : K] and |G| divides6, it follows that |G| equals either 3 or 6. In the former case L = K(α) (so K(α)/K is Galois)and G = A3. In the latter case, [L : K(α)] = 2 (so K(α)/K is not Galois) and G = S3.

How can we tell, just looking at the coefficients of f , which of the above cases occur?Suppose that in the splitting field, we have

f(x) = (x− α1)(x− α2)(x− α3).

Comparing coefficients, we note that α1 +α2 +α3 = 0, α1α2 +α2α3 +α1α3 = a, α1α2α3 = −b.Put

δ = (α1 − α2)(α2 − α3)(α3 − α1)

and ∆ = δ2. Clearly δ and ∆ are elements of L. But in fact, ∆ ∈ K. This is because anypermutation of the roots αi leaves ∆ unchanged. Hence ∆ is fixed by all elements of G, andhence ∆ ∈ K.

Using the above relations, in fact we can write down explicitly ∆ = −4a3− 27b2. ∆ is calledthe discriminant of f .

31

Now we note that δ ∈ K if and only if ∆ is a square in K. By the main theorem of Galoistheory, δ ∈ K happens if and only if all elements of G fix δ. And now comes the key point: Ifσ ∈ A3, then σ(δ) = δ and if σ ∈ S3 − A3, then σ(δ) = −δ. So whether or not ∆ is a square isequivalent to whether or not the Galois group is A3! We summarize thusly:

Theorem 59. Let f(x) ∈ K[x] be an irreducible cubic polynomial where char(K) 6= 2, 3. Thenf is separable. Let α be any root of f , L be the splitting field of f over K and G = Gal(L/K).Then:

(1) L = K(α) if and only if G ' Z/3Z if and only if ∆ is a square in K.(2) L 6= K(α) if and only if [L : K(α)] = 2 if and only if G ' S3 if and only if ∆ is not a

square in K. In that case L contains a unique subfield of degree 2 over K; this subfieldis K(

√∆).

Example. Let f(x) = x3 − x + 1. Let us investigate the Galois group of f over Q. It is easyto check that f has no integer roots. So, f is irreducible by Gauss’ Lemma. Its discriminant∆ = −23 which is not a square. So the Galois group of f is S3.

Example. Let f(x) = x3 − 3x+ 1. Let us investigate the Galois group of f over Q. It is easyto check that f has no integer roots. So, f is irreducible by Gauss’ Lemma. Its discriminant∆ = 81 which is a square. So the Galois group of f is Z/3Z. In particular, this means that ifα is a root of f , then the other two roots can be written as polynomials in α with coefficientsin Q.

A degree 4 exampleLet α = 4

√2. Consider the diagram

L = Q(α, i)

Q(i)

4>

Q(α)

2<

Q

8

∧

4

>

2

<(11.1)

We have seen that L/Q is Galois. What is the Galois group? What are its various subgroupsand corresponding subfields?

12. Finite fields

In this short section, we will use the tools of Galois theory to study finite fields. Recall thatfor any prime p, Fp = Z/pZ denotes the field with p elements.

Lemma 60. Let K be a finite field. Then char(K) = p where p is a prime. Moreover, thenumber of elements in K is a power of p.

Proof. Since K has finitely many elements, 1 + 1 + . . .+ 1︸ ︷︷ ︸m

= 0 for some integer m. So m = 0 in

K. So p = 0 for some prime p dividing m. It follows that K is an extension of Fp. So |K| = pn

where n is the degree of the field extension. �

Lemma 61. Let M be a field and let H be a finite subgroup of M×. Then H is cyclic.

Proof. By structure theory, H is isomorphic to a product of cyclic groups as follows:

H ' Cr1 × Cr2 × . . . Crtwith r1|r2|r3| . . . |rt.

So every element of H is a root of xrt − 1. So r1r2r3 . . . rt = |H| ≤ rt. So, t = 1, and H iscyclic.

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Theorem 62. Let p be a prime and q = pn be a positive power of p. Then,

(1) There exists a field of order q. This field is unique up to isomorphism and will be denotedhenceforth by Fq.

(2) There is a unique copy of Fq inside any algebraically closed field containing Fp.(3) All elements of Fq satisfy the polynomial xq = x. In particular Fq/Fp is a splitting

extension.(4) F×q is a cyclic group of order q − 1.(5) Fq/Fp is a Galois extension with the Galois group isomorphic to the cyclic group Z/nZ.

The Galois group is generated by the Frobenius map x 7→ xp.(6) Fq contains a subfield of order pd if and only if d divides n. In that case, the subfield is

unique.

Proof. Put K = Fp. Let L be the splitting field of f(x) = xq − x over K. By Lemma 49, f hasq distinct zeroes in L. Let R be the set of zeroes. By assumption, R is a finite set of size q.

Let φ : L 7→ L denote the map x 7→ xp. It is easy to check that φ is an automorphism ofL. φ is known as the Frobenius map. Note that R is also the set of elements in L fixed by theautomorphism φn. It follows that R is a subfield of L. But since R generates L, it follows thatR = L.

Thus we have shown that L is a field of order q and that all elements of L satisfy thepolynomial xq = x. In particular L/K is a splitting extension.

We now show that every field of order q is of the above form. Indeed, let L be a field of orderq. Then L× is a group of order q− 1. So xq = x for all x ∈ L. So f splits over L. But f cannotsplit over a smaller field because any splitting field of f must contain at least q elements. So Lis the splitting field for f . This determines L uniquely up to isomorphism and also shows thatthere is a unique copy of L in any algebraically closed field containing Fp.

Thus we have proved the first three assertions of the theorem. it remains to prove the lastthree.

Consider again F×q . It is a group of order q − 1. By the lemma earlier it is cyclic.The fact that Fq/Fp is a Galois extension follows from the fact that is a splitting field. It

is clear that the Galois group G = Gal(Fq/Fp) has size n and that the Frobenius element φbelongs to G. Now φn = 1. But if φi = 1 for some i < n then all elements of L must satisfy

xpi−1 = 0 which means L cannot have more than pi elements. So G is a cyclic group generated

by φ.Finally there is a unique subgroup of Z/nZ of any any size that divides n. This proves the

last assertion.�

13. Solubility of polynomials by radicals

Definition. An extension L/K is said to be radical if we can write L = K(α1, α2, . . . , αn) wherefor each i = 1, . . . , n there exists an integer ri such that αr11 ∈ K and αrii ∈ K(α1, α2, . . . , αi−1).

Remark 9. We may assume that each ri is a prime number. This does not change the definition.

Definition. Let f be a polynomial over a field K of characteristic 0. We say that f is solubleby radicals if the splitting field of f is contained in a radical extension of K.

Our goal is to characterize which polynomials are soluble by radicals. We will do this usinggroup theory.

Definition. A finite group G is said to be soluble if there is a series of subgroups

{1} = G0 ⊂ G1 ⊂ . . . ⊂ Gn = G

such that Gi is a normal subgroup of Gi+1 and Gi+1/Gi is abelian.

Some facts:

(1) All abelian groups are soluble.33

(2) The smallest non-soluble group has size 60; it is the alternating group A5.

Our main theorem is the following.

Theorem 63. A polynomial f over a field of characteristic 0 is soluble by radicals if and onlyif its Galois group is a soluble group.

In particular, all polynomials of degree ≤ 4 are soluble by radicals.The proof will be quite long. We begin with recalling some facts from group theory that we

will need.

Proposition 64. Let G be a finite group, H a subgroup, and N a normal subgroup.

(1) If G is soluble, so is H.(2) If G is soluble, so is G/N .(3) If N and G/N are both soluble, then so is G.

Proposition 65. Any group of prime power order is soluble.

Proposition 66. If n ≥ 5, then neither Sn nor An is soluble.

Proposition 67. Sylow’s theorems.

Let us now go back to the proof of Theorem 63. We will in fact prove a slightly differenttheorem which will imply this.

Theorem 68. Let K be a field of characteristic 0 and L/K be a Galois extension with Galoisgroup G. Then G is soluble if and only if there is an extension M/L such that M/K is a radicalextension.

We begin with a few lemmas.

Lemma 69. Let K be a field of characteristic 0 and L/K be a radical extension. Then thereexists an extension N/L such that N/K is normal and radical.

Proof. Let α1, α2, . . . , αn be as in the definition of radical extensions. Let mi be the minimalpolynomial of αi over K and let N be the splitting field of

∏ni=1mi over K. Clearly N/L

is an extension and N/K is normal. Moreover, if {σi}1≤i≤r denotes the various elements ofGal(N/K), then N = K({σi(αj)}i,j). This shows that N/K is radical (why?) �

Lemma 70. Let K be a field of characteristic 0 and p be a prime. Let L be a splitting field forthe polynomial xp − 1 over K. Then Gal(L/K) is abelian.

Proof. Special case of homework problem (in fact Gal(L/K) is a cyclic group). �

Lemma 71. Let K be a field of characteristic 0 and n an integer such that the polynomialxn − 1 splits in K. Let L be a splitting field over K for the polynomial xn − a for some a ∈ K.Then Gal(L/K) is abelian.

Proof. Let α ∈ L be a root of the polynomial xn− a. Then L = K(α). So if σ1, σ2 are elementsof Gal(L/K), then σ1(α) = ω1α, σ2(α) = ω2α for some ω1, ω2 in K that are roots of xn − 1.Hence

σ1σ2(α) = ω2ω1α = ω1ω2α = σ2σ1(α).

Thus σ1 and σ2 commute. This proves that Gal(L/K) is abelian.�

We will now prove one half of the theorem. Namely we will prove

Theorem 72. Let K be a field of characteristic 0 and L/K be a Galois extension with Galoisgroup G. Suppose that there is an extension M/L such that M/K is a radical extension. ThenG is soluble.

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Proof. By Lemma 69, we may assume that M/K is normal (hence Galois). It suffices to provethat Gal(M/K) is soluble. Let us write M = K(α1, α2, . . . , αn) where for each i = 1, . . . , nthere exists a prime number ri such that αrii ∈ K(α1, α2, . . . , αi−1). We will prove the theoremby induction on n.

If α1 ∈ K, we can drop this generator and so are done by induction. Suppose α1 /∈ K; letαp1 = a ∈ K. Let M0 be the splitting field of xp−a over K. Then M0 = K(ε, α1) where ε ∈ L isa generator of the group of pth roots of unity. So we have a tower of extensions M/M0/K(ε)/Kwhere by the previous lemmas, M0/K(ε) and K(ε)/K are both Galois with abelian Galoisgroups. It follows that Gal(M0/K) is soluble. Moreover, since M = M0(α2, . . . , αn), it followsby induction that Gal(M/M0) is soluble. Since Gal(M/K)/Gal(M/M0) = Gal(M0/K), itfollows that Gal(M/K) is soluble, as desired.

�

Next, we will prove the other half of Theorem 68. That is, we will prove:

Theorem 73. Let K be a field of characteristic 0 and L/K be a Galois extension with Galoisgroup G such that G is soluble. Then there is an extension M/L such that M/K is a radicalextension.

The key step to the proof is the following lemma.

Lemma 74. Let K be a field of characteristic 0 and p a prime such that the polynomial xp− 1splits in K. Let L/K be a Galois extension with Galois group Z/pZ. Then L = K(β) where βis a zero of an irreducible polynomial xp − a for some a ∈ K.

Proof. Let ω 6= 1 be a zero of xp−1 inK. Let σ be a generator of the cyclic groupG = Gal(L/K).Then 1, σ, σ2, .. ,σp−1 are distinct automorphisms of L. So by Lemma 36, it follows that theyare linearly independent over L. In particular 1 + ωσ + . . . + ωp−1σp−1 6= 0. So there existsα ∈ L such that

β = α+ ωσ(α) + . . .+ ωp−1σp−1(α) 6= 0.

We now that σ(β) = ω−1β. So β /∈ K. But σ(βp) = ω−pβ = β. So θ = βp ∈ K. SoL/K(β)/K. Since [L : K] is a prime, it follows that L = K(β).

�

Now, we prove Theorem 73.

Proof. We will induct on the size of G. If |G| = 1, the theorem is trivial. So we assume thatnot to be the case. Let H be a maximal proper normal subgroup. Then G/H is abelian andsimple and so is a cyclic group of prime order p.

Let L0 be the field obtained from L by adjoining the roots of xp − 1 and K0 be the fieldobtained from K by adjoining the roots of xp − 1. It suffices to find an extension M/L0 suchthat M/K0 is radical.

Note that L0 is normal over K0 and G0 = Gal(L0/K0) is a subgroup of G. If this is a propersubgroup, then by the induction hypothesis L0/K0 is radical, so we are done.

Else G0 is isomorphic to G. Let H0 ⊂ G0 be the subgroup corresponding to H. Then thefixed field R0 of H0 is a Galois extension of K0 with Galois group of order p. So R0/K0 is aradical extension. Also L0/R0 has Galois group of size less than G. So there exists an extensionM/L0 such that M/R0 is radical. So M/K0 is radical, as desired. �

Corollary 75. Any polynomial of degree ≤ 4 is soluble.

What about degree 5 or above?

Proposition 76. Let p be a prime and f be an irreducible polynomial of degree p over Q.Suppose that f has exactly 2 non-real zeroes. Then the Galois group of f is Sp. In particular,if p ≥ 5, then f is not solvable by radicals.

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Proof. (This proof assumes that the complex numbers are algebraically closed; this will beproved in the next section). Let L ⊂ C be a splitting field for f and G be the Galois group,viewed as a subgroup of Sp via its action on the roots. Complex conjugation permutes the twonon-real roots, so f contains a transposition. On the other hand since any root of f gives anextension of degree p, it follows that p | [L : Q] | |G|. But p2 cannot divide |G| since p2 - p!. Sothe Sylow p-subgroup of G is a cyclic group of order p and hence G contains a p-cycle. The proofnow follows from the group theory result that a transposition and a p-cycle generate Sp. �

Example. The polynomial x5 − 6x+ 3 has Galois group S5.

14. Miscellaneous topics

14.1. The field of complex numbers is algebraically closed. Let C = R(i). We will usethe tools of Galois theory to prove that C is algebraically closed. Suppose not. Then there is anirreducible polynomial over C whose degree is greater than 1. Thus, C has a finite non-trivialextension K. By enlarging K if necessary, we may assume that K/R is normal, and henceGalois (since R has characteristic 0).

Thus, assuming that C is not algebraically closed, we have a constructed a non-trivial ex-tension K of C such that K/R is Galois. We will show that this leads to a contradiction. LetG = Gal(K/R) and let H be a 2 Sylow subgroup of G. Let F be the fixed field of H. Then[F : R] is odd. Suppose F 6= R. Taking any α ∈ F − R, we see that f = mα,R is an irreduciblepolynomial over R of odd degree. But f(x) → −∞ as x → −∞, and f(x) →∞ as x →∞; sof must have a real root. So f must be of degree 1 which contradicts α /∈ R. This shows thatF = R. Hence G = H is a 2 group.

Hence G1 = Gal(K/C) is also a 2 group. Since we are assuming that G1 is non-trivial, it hasa subgroup G2 of index 2. Let M be the fixed field of G2. Then [M : C] = 2. Thus M = C(α)where α2 = a for some a ∈ C. In other words, there is an element a ∈ C whose square-root isnot in C. But writing a = x+ iy we see that the square roots of a are ±(c+ id) where

c =

√x+

√x2 + y2

2, d =

√−x+

√x2 + y2

2.

Thus the square roots of a are indeed in C. This contradiction shows that C is algebraicallyclosed.

14.2. The primitive element theorem. Let L/K be a finite extension. Thus, we can writeL = K(α1, α2, . . . , αn) with αi ∈ L. Now the question is, can we take n = 1? That is, can wewrite L = K(α) for some α ∈ L? If such an α exists, it is called a primitive element for theextension. We will show that under certain assumptions, a primitive element always exists.

Theorem 77. Let L/K be a finite extension. Then there exists α ∈ L such that L = K(α) ifand only if there are only finitely many fields F such that L/F/K. Moreover, supposing thatL/K is separable, then such an α always exists.

Proof. If K is finite, this is easy and follows from the fact that L× is cyclic. So assume thatK is infinite and suppose there are only finitely many intermediate fields F as above. To provethe result, it is enough (by induction) to show that for all α, β ∈ L, there exists c ∈ K suchthat K(α, β) = K(α + cβ). Indeed, as c varies over elements of K, we must have c1, c2 suchthat K(α+ c1β) = K(α+ c2β). Put L0 = K(α+ c1β). Then L0 contains (c1 − c2)β and hencecontains β. It follows that L0 also contains α. Thus L0 = L.

For the second part, we can let M/L be an extension such that M/K is normal and separable,and hence Galois. Then the intermediate fields between M and K correspond to subgroups ofthe Galois group, and hence there are only finitely many intermediate fields. It follows thatthere are finitely many fields between L and K. This completes the proof.

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How to actually find primitive elements? We can use the following fact: Suppose K(α, β) 6=K(α+ cβ). Let L be any Galois extension containing K(α, β). Then there exists an element ofthe Galois group which does not fix both α and β but does fix α+ cβ.

References

[Gar86] D. J. H. Garling, A course in Galois theory, Cambridge University Press, Cambridge, 1986.

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