1. Galois Theory

1.1. A homomorphism of fields F → F ′ is simply a homomorphism ofrings. Such a homomorphism is always injective, because its kernel isa proper ideal (it doesnt contain 1), which must therefore be zero.

A field E containing a field F is called an extension field of F . Wewrite [E : F ] = dimF E for the degree of E over F . We say that E isfinite over F if [E : F ] <∞.

For E/F extension and α ∈ E. Consider F [x] → E, f 7→ f(α).There are two possibilities:

Case 1: ker = (0). We say that α is transcendental over F . We havean isomorphism F (x)→ F (α).

Case 2: ker = (f). Here f is a monic irreducible polynomial. Wesay that α is algebraic over F and f is a minimal polynomial of α overF . We have an isomorphism F [x]/(f)→ F (α).

Proposition 1.1. E/F is finite iff E is algebraic and finitely generated(as a field) over F .

Let E,E ′ be fields containing F . An F -homomorphism is a homo-morphism ϕ : E → E ′ such that ϕF = id.

Proposition 1.2. Let F (α) be a simple field extension of a field F ,and let ϕ0 : F → Ω be a homomorphism of fields.

(a) If α is transcendental over F , then the map ϕ 7→ ϕ(α) defines aone-to-one correspondence between extensions ϕ : F (α)→ Ω of ϕ0 andelements of Ω transcendental over ϕ0(F ).

(b) If α is algebraic over F , with minimum polynomial f ∈ F [x],then the map ϕ 7→ ϕ(α) defines a one-to-one correspondence betweenextensions ϕ : F (α) → Ω of ϕ0 and the roots of ϕ0(f) over Ω. Inparticular, the number of such maps is the number of distinct roots ofϕ0(f) in Ω.

1.2. Let f ∈ F [x]. We say E/F splits f if f(x) = Π(x−αi) for αi ∈ E.If in addition, E is generated by the roots of f , then we say that E isa splitting field of f .

Proposition 1.3. Every polynomial f ∈ F [x] has a splitting field Efand [Ef : E] 6 (deg f)!.

Proposition 1.4. Let f ∈ F [x], E/F generated by roots of f . Ω/Fsplits f . Then

(1) There exists F -homomorphism E → Ω.1

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(2) ]F−hom : E → Ω 6 [E : F ] and equals [E : F ] if f has onlysimple roots in Ω.

Corollary 1.5. Let f ∈ F [x]. The splitting field Ef is unique up toisomorphism.

Corollary 1.6. Let E/F be a splitting field of f . Then |Aut(E/F )| 6[E : F ] and equals [E : F ] if f is separable over F (i.e., none of itsirreducible factor in F [x] has multiple roots) Here Aut(E/F ) is thegroup of F -automorphism E → E.

1.3. E/F is called separable if minimal polynomial of any element of Eis separable (i.e., no multiple roots). E/F is called normal if minimalpolynomial of any element of E splits in E[x].

Example 1.7. Q[ 3√

2] is separable, but not normal over Q.Fp(T ) is normal, but not separable over Fp(T p).

Lemma 1.8. (Dedekind) Let F be a field and G be a group. If σ1, · · · , σnare distinct characters from G to F . Then σi are linearly indepdentover F .

Proof. Otherwise,∑aiσi(g) = 0 with ]i; ai 6= 0 as small as pos-

sible. So∑aiσi(h)σi(g) = 0 for all g, h ∈ G. Thus

∑ai(σi(h) −

σ1(h))σi(g) = 0. However, ]i; ai(σi(h) − σ1(h)) 6= 0 < ]i; ai 6= 0.Contradiction.

Theorem 1.9. (Artin) Let G be a finite group of automorphisms of Eand F = EG the fixed field of G. Then [E : F ] 6 |G|.

Proof. Assume that G = 1 = σ1, · · · , σn and α1, · · · , αm ∈ E,basis over F . If m > n, then the equations

∑i σi(αj)xj = 0 for all

1 6 j 6 m has a nontrivial solution (x1, · · · , xm) = (a1, · · · , am) inE. We may also assume that this is a solution with fewest possiblenonzero elements. WLOG, x1 6= 0 and x1 ∈ F after rescalaring. Thenfor any σ ∈ G, (σ(x1), · · · , σ(xm)) and (σ(x1) − x1, · · · , σ(xm) − xm)are also solutions. By minimality assumption, σ(xi) = xi for all i andσ ∈ G. So xi ∈ F for all i. Thus

∑xiai = 0. Contradiction.

Corollary 1.10. Let G be a finite group of automorphisms of E. ThenG = Aut(E/EG).

Proof. G ⊂ Aut(E/EG). So |G| 6 |Aut(E/EG)| 6 [E : EG] 6 |G|.Theorem 1.11. For a finite extension E/F , the following statementsare equivalent:

(a) E is the splitting field of a separable polynomial f ∈ F [x].(b) F = EAut(E/F ).(c) F = EG for some finite group G of automorphisms of E.(d) E is normal and separable over F .In this case, we call that E is Galois over F and write Gal(E/F ) for

Aut(E/F ).

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Proof. (a)⇒ (b). Let G = Aut(E/F ). Then EG ⊃ F . E is splittingfield of f over EG. So [E : EG] = |Aut(E/EG)| = |G| = |Aut(E/F )| =[E : F ] and EG = F .

(b)⇒ (c). Obvious.(c) ⇒ (d). Let α ∈ E and f minimal polynomial. Let α1, · · · , αm

be orbit of α under G. Set g(x) = Π(x− αi) ∈ F [x]. Then f | g sinceg(α) = 0. Also g | f since f(αi) = 0. Thus f = g splits into distinctfactors in E.

(d) ⇒ (a). E = F [α1, · · · , αn]. Let fi be minimal polynomial of αiover F . Then fi splits in E. Hence f = Πfi splits in E. This is alsoseparable since E/F is separable.

Corollary 1.12. Every finite separable extension is contained in afinite Galois extension.

Proof. Let E = F [α1, · · · , αn]. Then the splitting field of f = Πficontains E and is Galois over F .

Corollary 1.13. Let E ⊃ M ⊃ F . If E is Galois over F , then it isGalois over M .

Proof. Assume that E = Ef with f ∈ F [x]. Then E is also thesplitting field of f ∈M [x].

Theorem 1.14. (Fundamental theorem) Let E be a finite Galois ex-tension of F and G = Gal(E/F ). Then H 7→ EH , M 7→ Gal(E/M)gives bijection between subgroup of G and intermediate extension F ⊂M ⊂ E. Moreover,

(a) H1 ⊃ H2 iff EH1 ⊂ EH2.(b) (H1 : H2) = [EH2 : EH1 ].(c) σHσ−1 ↔ σ(M).(d) H normal in G iff EH normal over F . In this case, EH is Galois

over F and Gal(EH/F ) ∼= G/H.

Proof. H = Gal(E/EH) is proved. Also E is Galois over M . SoM = EGal(E/M).

(a) H1 ⊃ H2 ⇒ EH1 ⊂ EH2 ⇒ H1 = Gal(E/EH1) ⊃ Gal(E/EH2) =H2.

(b) |Hi| = [E : EHi ].

(c) ∀τ ∈ G,α ∈ E, τα = α ⇐⇒ στσ−1(σα) = σα. So Gal(E/σM) =σGal(E/M)σ−1.

(d) If H normal in G. Then σEH = EσHσ−1= EH . Now σ 7→

σ |EH gives G → Aut(EH/F ) with kernel H. So induces G/H →Aut(EH/F ). Since (EH)G/H = F , EH is Galois over F and G/H ∼=Gal(EH/F ).

If M = F [α1, · · · , αm] normal over F , then σ(αi) is a root of theminimal polynomial of αi over F , so in M . Thus σ(M) = M for allσ ∈ G. Hence σHσ−1 = H.

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Proposition 1.15. Let E,L be field extensions of F contained in somecommon field. If E/F is Galois, then EL/L and E/E ∩ L are Galois,and the map

Gal(EL/L)→ Gal(E/E ∩ L), σ 7→ σ |Eis an isomorphism. In this case

[EL : F ] =[E : F ][L : F ]

[E ∩ L : F ].

Proof. E is a splitting field of separable f ∈ F [x]. So EL (resp. E)is splitting field of f over L (resp. E∩L). For σ ∈ Gal(EL/L), σ mapsroots of f to roots of f . So σ(E) = E. Now σ 7→ σ |E is defined.

If σ ∈ Gal(EL/L) fixes the elements in E, then it also fixes theelements in EL. So σ = 1. Thus injective. If α ∈ E is fixed byGal(EL/L), then α ∈ E∩L. So image is Gal(E/E∩L) by fundamentaltheorem.

Proposition 1.16. Let E,L be field extensions of F contained in somecommon field. If E,L are Galois over F , then EL,E ∩ L are Galoisover F , and

Gal(EL/F )→ Gal(E/F )×Gal(L/F ), σ 7→ (σ |E, σ |L)

is an isomorphism of Gal(EL/F ) onto the subgroup

H = (σ1, σ2);σ1 |E∩L= σ2 |E∩Lof Gal(E/F )×Gal(L/F ).

Proof. E is a splitting field of separable f ∈ F [x] and L is a splittingfield of separable g ∈ F [x]. Then EL is a splitting field of separablefg ∈ F [x]. As in the previous Prop., Gal(EL/F ) → Gal(E/F ) ×Gal(L/F ), σ 7→ (σ |E, σ |L) is defined, injective with image containedin H.

Now for α ∈ E ∩ L with minimal polynomial h ∈ F [x]. Then h isseparable, and has distinct deg(h) roots in E (resp. L and EL), sinceE,L,EL are Galois over F . Thus all these roots are in E ∩ L. SoE ∩ L/F is normal and separable, hence Galois.

The remaining part follows from comparing the size of Gal(EL/F )with |H|.

1.4. For separable f ∈ F [x], let Ef be the splitting field. Assumethat f has only simple roots in Ef , f(x) = Πn

i=1(x − αi) ∈ Ef [x]. SetGf = Gal(Ef/F ). Then Gf ⊂ Sn.

Proposition 1.17. (1) Let ∆(f) = πi<j(αi − αj). Then ∆(f) ∈ F iffGf ⊂ An.

(2) The discriminant D(f) = ∆(f)2 is always in F .

Proposition 1.18. f is irreducible iff Gf acts transitively on the setof roots f .

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Proof. ⇐. If g | f and α is a root of g. Then for any root β of f ,β = σα for some σ ∈ Gf . Thus g(σα) = σg(α) = 0. So β is a root ofg and f = g.⇒. If α, β are roots of f . Then F (α) ∼= F (β) ∼= F [x]/(f). There

exists F -homomorphism φ : F (α) → Ef , α 7→ β. Since Ef is splittingfield, φ extends to F -homomorphism σ : Ef → Ef . Now σ(α) = β.

Example 1.19. If f(x) irreducible and separable of degree 3. ThenGf = A3 or S3, according to D(f) is square in F or not.

Fact. The discriminant of x3 + px+ q is −4p3 − 27q2.

2. application

2.1. Finite fields. E/Fp of degree n. Then E has pn elements. E× isa cyclic group of pn−1 elements. Thus xp

n−1 = 1 for all nonzero x ∈ E.So any element of E is a root of f(x) = xp

n − x. Since f ′(x) = −1, fhas pn distinct roots and E is the splitting field of f and is Galois overFp. We write it as Fpn .

Frobenius map Fr : x 7→ xp is injective, hence surjective on E andEFr = Fp. So by fundamental theorem, Gal(E/Fp) =< Fr > is acyclic group of order n.

Proposition 2.1. Let F be a field and G a finite subgroup of F×. ThenG is cyclic.

Proof. G is a finite abelian group. By the classification of finiteabelian group, there exists r ∈ N such that G contains an element oforder r, r | |G| and xr = 1 for all x ∈ G. Thus every element of G is aroot of xr − 1, thus |G| 6 r.

Corollary 2.2. (1) For any d | n, Fpn contains exactly one field withpd elements.

(2) Fpm ⊂ Fpn iff m | n.(3) xp

n−x is the product of all monic irreducible polynomial over Fpwhose degree divides n.

Proof. (1) Since Gal(Fpn/Fp) is cyclic, there is exactly one subgroupof order n/d.

(2) ⇒ Gal(Fpm/Fp) is a quotient group of Gal(Fpn/Fp). So m | n.(3) First, the factors of xp

n−x are distinct because it has no commonfactor with its derivative.

If g is an irreducible factor of xpn − x and α is a root of g in Fpn ,

then Fp(α) ⊂ Fpn is a subfield of with pdeg g elements. Thus deg g | n.If f is an irreducible monic polynomial of degree m with m | n, then

f has a root α in a field of degree m over Fp, thus in Fpn . Henceαp

n − α = 0 and f | xpn − x.

Corollary 2.3. The algebraic closure F of Fp is F = ∪Fpn!.

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Theorem 2.4. (Fundamental theorem of algebra) C is algebraicallyclosed.

Proof. C is the splitting field of x2 + 1 ∈ R[x]. Let f ∈ R[x] andE the splitting field of f(x)(x2 + 1). Then E is Galois over R. LetG = Gal(E/R) and H a Sylow 2-subgroup. Set M = EH . Then[M : R] = (G : H) is odd. We show that

(1) M = R.Let α ∈ M and g the minimal polynomial. Then [R(α) : R] is odd

and deg(g) is odd. Since g(−∞) = −∞ and g(∞) = ∞, g has a rootin R. Thus deg(g) = 1 and α ∈ R.

(2) E = C.Set G′ = Gal(E/C). Then by (1), G′ is a 2-group. If not 1, then

there is a normal subgroup N of index 2. Thus EN has degree 2 overC. But all the square roots of C are in C. So impossible.

2.2. Constructible numbers. Recall that α ∈ R is constructible if itis in a subfield of the form Q[

√a1, · · · ,

√ar], where ai ∈ Q[

√a1, · · · ,

√ai−1].

Then in particular, [Q(α) : Q] is a power of 2.

Theorem 2.5. If E ⊂ R is Galois over Q of degree 2s, then α isconstructible for all α ∈ E.

Remark 2.6. It fails if E is not Galois over Q. For example, f(x) =x4 − 4x + 2 has Gf = S4. If all the roots of f are constructible, thenso is Ef . Let H be a Sylow 2-subgroup of S4, then [EH : Q] is odd andany element in EH\Q is not constructible.

Proof. Let G = Gal(E/Q). Recall that any 2-group has nontrivialcenter and hence a subgroup in the center of order 2. Now a sequenceof groups

1 = G0 ⊂ G1 ⊂ · · · ⊂ Gs = G

with |Gi+1/Gi| = 2. By fundamental theorem, a sequece of fields

E = E0 ⊃ E1 ⊃ · · · ⊃ Es = Q

with [Ei : Ei+1] = 2. Thus Ei = Ei+1[√ai] for some ai ∈ Ei+1.

Corollary 2.7. A regular p-gon with p prime is constructible iff p is aFermat prime, i.e., p is of the form 22r

+ 1.

Proof. [Q(e2πi/p) : Q(cos(2π/p))] is a power of 2. Now Q(e2πi/p)is Galois over Q with Galois group (Z/pZ)× of order p − 1. Thuscos(2πi/p) is constructible iff p− 1 is a power of 2. Easy to check thatsuch p is of the form 22r

+ 1.

2.3. Primitive element. Finite extension E/F is simple if E = F [α]for some α ∈ E. Such α is primitive element.

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Theorem 2.8. (1) Finite extension E/F is simple iff there are onlyfinitely many intermediate extensions.

(2) If E is separable over F , then it is a simple extension.

Proof. If F is a finite field, then E is also finite and E× is cyclic. So(1) and (2) follows. Now assume that F is infinite.

(1) ⇒. Assume that E = F [α] and f is the minimal polynomialof α. For any intermediate field M , let fM ∈ M [x] be the minimalpolynomial of α. Then fM | f is a monic polynomial in E[x]. Soin particular, there are only finitely many possible fM . Now for fM ,let M ′ be the subfield of M generated by the coefficient of fM . ThenE = M [α] = M ′[α] and [E : M [α]] = deg fM = [E : M ′[α]]. SoM = M ′.⇐. Assume that E = F [α, β]. The case with more generators can

be proved by induction. Since |F | is infinite, there exists c 6= c′ ∈ Fsuch that F (α+ cβ) = F (α+ c′β). Then (c′− c)β ∈ F (α+ cβ) and β,hence α also in F (α + cβ). So α + cβ is a primitive element.

(2) E is contained in a Galois extension E ′ of F . By fundamentaltheorem, there are only finitely many intermediate extension betweenF and E ′. Hence by (1), E/F is simple.

2.4. Cyclotomic extension.

Proposition 2.9. Let n ∈ N and F a field of char 0 or p with p - n.Let E be the splitting field of xn − 1. Then

(1) There exists a primitive n-th root ξ of 1 in E, i.e., the order ofξ is n.

(2) E = F [ξ].(3) E is Galois over F . The map Gal(E/F ) → (Z/nZ)×, σ 7→ i,

where σ(ξ) = ξi, is injective.

Proof. (1)&(2) gcd(xn − 1, nxn−1) = 1. So only simple roots. Allthe n-th roots of 1 in E form a subgroup of order n, so cyclic and thereexists primitive element.

(3) Any primitive root is of the form ξi for some i ∈ (Z/nZ)×.

Proposition 2.10. Let F = Q. Then the cyclotomic polynomialΨn(x) = Πξ primitive (x − ξ) is in Z[x] and is irreducible over Q. Theabove map Gal(E/F )→ (Z/nZ)× is an isomorphism.

Proof. For any n-th root ω of unity, let d be the order of ω. Thenω is a primitive d-th root of unity. Now we have a bijection betweenthe set of n-th root of unity, and the pair (d, ω), where d | n and ω isa primitive d-th root of unity. So

xn − 1 = Πd|nΨd(x).

In particular, by Gauss Lemma, Ψd(x) ∈ Z[x] for all d | n.

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Now assume that Ψn(x) = f(x)g(x), where f(x) irreducible andf(x), g(x) are monic polynomials in Z[x]. Assume that ξ is a root off . We’ll show that for any d ∈ Z with gcd(d, n) = 1, ξd is also a rootof f . It is suffices to prove the case where d is a prime p.

If ξd is not a root of f , then it is a root of g. Thus f(x), g(xp) has acommon root and g(xp) = f(x)h(x). Hence g(x)p ≡ g(xp) = f(x)h(x)mod p as elements in Fp[x]. Thus f , g has common roots. Thereforexn − 1 ∈ Fp[x] has multiple roots, which is impossible.

Now |Gal(E/F )| = [E : F ] = deg Ψn = ϕ(n) = |(Z/nZ)×|. SoGal(E/F )→ (Z/nZ)× is an isomorphism.

2.5. Cyclic and Kummer extesion.

Theorem 2.11. (Hilbert’s 90) Let E/F be cyclic extension with Galoisgroup G =< σ >. For β ∈ E, define the norm NE/F (β) = Πσiβ. ThenNE/F (β) = 1 iff β = α/σα for some α ∈ E×.

Proof. ⇒ N(β) = N(α)/N(σα) = 1.⇐Assume that |G| = n. Then by Dedekind’s theorem,

∑n−1i=0 (Πi−1

j=0σj(β))σi

is not identically zero on E. Hence there exists a ∈ E such thatα :=

∑n−1i=0 (Πi−1

j=0σj(β))σi(a) 6= 0. Now βσ(α) = α.

Theorem 2.12. Let F be a field containing a primitive n-th root of 1.Then

(1) For any nonzero a ∈ F , the splitting field E for xn−a is a cyclic(Galois) extension.

(2) If E is a cyclic extension of F of degree n, then E is the splittingfield of xn − a for some a ∈ F .

Proof. (1) gcd(xn− a, nxn−1) = 1. So xn− a is separable and E is aGalois extension. Let α be a root. Then all the roots are of the formξα for some n-th root ξ of unity. Now Gal(E/F )→ µn, σ 7→ σ(α)/α isinjective since E = F [α]. Here µn is the group of all n-th root of unity.

(2) N(ξ) = ξn = 1. By Hilbert’s 90, ξ = α/σα for some α ∈ E.Since G is cyclic, any subgroup of G is normal and F (α) is Galois overF . The map G → Gal(F (α)/F ), σi 7→ σi |F (α) is injective. HenceE = F (α). Moreover, σ(αn) = (ξα)n = αn and αn ∈ EG = F .

Theorem 2.13. Let F be a field containing a primitive n-th root ξ of1. Let E/F be a finite extension. Then E/F is a Kummer extension(i.e., with abelian Galois group) whose Galois group has an exponentdividing n (i.e., any element in G is of order dividing n) iff E is asplitting field of (xn − a1) · · · (xn − ar) for some a1, · · · , ar ∈ F .

Proof. ⇐. We have that E = F (α1, · · · , αr), where αi is a root ofxn − ai. Then for any σ ∈ G = Gal(E/F ), σ(αi) = ξui(σ)αi. Hence forτ ∈ Gal(E/F ), we have that ui(σ)+ui(τ) = ui(τ)+ui(σ) and στ = τσ.Thus G is abelian. Also for any σ ∈ G, σn = 1.

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⇒. Since G is a finite abelian group, it is a direct product of cyclicgroups C1, · · · , Cr. Set C ′i = C1× · · · × Ci× · · ·Cr ⊂ G and Ei = EC′i .Then Ei/F is a cyclic extension. Hence Ei = F (αi) for some αi withαni = ai ∈ F . By fundamental theorem of Galois theory, E = E∩C

′i

is the unique field extension that contains Ei for all i. Hence E =F (α1, · · · , αr).

2.6. Galois’s solvability theorem. Let F be a field of char 0. Apolynomial f ∈ F [x] is called solvable if any root of f can be obtainedby ±,×, / and n

√·.

A group G is solvable if there is a sequence 1 = G0 ⊂ · · · ⊂ Gs =G, where Gi is normal in Gi+1 and Gi+1/Gi is cyclic for all i. Anysubquotient of a solvable group is again solvable.

Theorem 2.14. f ∈ F [x] is solvable iff Gf is solvable.

Proof. ⇐ Let F ′ = F [ξ], where ξ is a primitive (deg f)!-th rootof 1. Set G′f = Gal(F ′f/F

′). Then the natural map G′f → Gf iswell-defined and injective. Hence G′f is solvable. Now there exists asequence 1 = G0 ⊂ G1 ⊂ · · ·Gs = G′f , here Gi is normal in Gi+1 and

Gi+1/Gi is cyclic. Let Fi = (F ′f )Gi . Then F ⊂ F ′ = Fs ⊂ · · ·F0 = F ′f

and Fi/Fi+1 is a cyclic extension.⇒ There exists F = F0 ⊂ · · ·Fm, where Fi+1 = Fi(αi), α

nii ∈ Fi

and Fm splits f . Set n = n1 · · ·nm. Let Ω′ be a field containing Fmand a primitive n-th root of unity ξ. Then Ω′ is contained in a Galoisextension of F . We denote this field by Ω. Set G = Gal(Ω/F ) and H =Gal(Ω/Fm[ξ]). Set H ′ = ∩σ∈GσHσ−1. Then H ′ is normal in G and isthe maximal subgroup in σHσ−1 for all σ. Thus ΩH′ = Πσ∈GΩσHσi =Πσ∈Gσ(Fm[ξ]) is Galois over F and is the minimal subfield containingσΩH for all σ ∈ G.

Now ΩH′ is generated by ξ, σ(αi) over F . Consider F ⊂ F [ξ] ⊂F [ξ, α1] ⊂ · · · ⊂ ΩH′ . Here each field is obtained from its predecessorby adding a n-th root of some element, hence cyclic extension. SoGal(ΩH′/F ) is solvable. Since Gal(Ff/F ) is a quotient of Gal(ΩH′/F ),it is also solvable.

2.7. Symmetric polynomial. A polynomial f(x1, · · · , xn) ∈ F [x1, · · · , xn]is symmetric if f(x1, · · · , xn) = f(xσ(1), · · · , xσ(n)) for σ ∈ Sn. Thepolynomials

p1 =∑i

xi,

p2 =∑i<j

xixj,

· · · ,pn = Πixi

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are called the elementary symmetric polynomials.

Theorem 2.15. (1) Every symmetric polynomial is a polynomial ofelementary symmetric polynomials, i.e., if f ∈ F [x1, · · · , xn] is sym-metric, then f ∈ F [p1, · · · , pn].

(2) The field F (x1, · · · , xn) is the splitting field of F (p1, · · · , pn) forthe polynomial Πi(X − xi) = Xn − p1X

n−1 + · · · + (−1)npn and theGalois group is Sn.

Proof. (1) For example, x31 + 4x2

1x2 + 4x1x22 + x3

2 = p31 + p1p2.

We define the multi-degree on polynomials as follows. The multi-degree for xk11 · · ·xkn

n is (k1, · · · , kn) ∈ Nn. We use lexicographic orderon Nn. For any polynomial f , let xk11 · · · xkn

n be the highest monomialoccurring in f with nonzero coefficient and we call (k1, · · · , kn) themulti-degree of f .

Now for any symmetric polynomial f , its multi-degree is of theform (k1, · · · , kn) for k1 > k2 > · · · kn. In other words, there ex-ists (d1, · · · , dn) ∈ Nn such that k1 = d1 + · · · + dn, k2 = d2 + · · · +dn, · · · , kn = dn. Notice that pd11 · · · pdn

n has multi-degree (k1, · · · , kn).Thus f − cpd11 · · · pdn

n is again symmetric and has multi-degree strictlyless than (k1, · · · , kn), here c is the coefficient of xk11 · · ·xkn

n in f . Nowpart (1) follows from induction on multi-degree.

(2) Let f = g/h with g, h ∈ F [x1, · · · , xn] be a symmetric function.Set H = Πσ∈Snσ(h). Then H and Hf are symmetric polynomials andhence in F [p1, · · · , pn]. Thus f = Hf/H ∈ F (p1, · · · , pn). ThereforeF (x1, · · · , xn)Sn = F (p1, · · · , pn).

Corollary 2.16. We define the general polynomial of degree n to be

f(X) = Xn − t1Xn−1 + · · ·+ (−1)ntn ∈ F (t1, · · · , tn)[X].

Then the Galois group of f(X) over F (t1, · · · , tn) is Sn.

Proof. The monomials pd11 · · · pdnn for (d1, · · · , dn) ∈ Nn are linearly

independent since they have different multi-degree as polynomials ofx1, · · · , xn. Hence p1, · · · , pn are algebraically independent. Thus themap F [t1, · · · , tn] → F [p1, · · · , pn], ti 7→ pi is bijective. So it extendsto an isomorphism F (t1, · · · , tn) → F (p1, · · · , pn). Now the corollaryfollows from part (2) of the previous theorem.

References