3201: COMMUTATIVE ALGEBRAFALL 2010. EXERCISE SHEET 4

TO BE HANDED IN ON NOVEMBER 17TH

Problems marked with a star (∗) will be assessed and count towards the finalmark. Problems marked with an (O) are optional, you are strongly advised to trythese. They will be marked for your own convenience, but the mark will not counttowards your final mark. Problems marked with a (C) are challenge questions.

Exercise 1 (*). Let α : R→ S be a ring homomorphism.i) Suppose that R is a field, and S is not the trivial ring. Show that α must be

injective.ii) Suppose that S is an ID. Show that Kerα is a prime ideal of R.

iii) If R is a PID and S is an ID, show that either α is injective, or Imα is a field.iv) Let S be an ID. By considering the ring homomorphism e0 : S[x]→ S given

by p(x) 7→ p(0) (i.e. evaluation at 0), show that if R = S[x] is a PID, then Sis a field.

Solution.i) For any α : R→ S ring homomorphism, KerαER is an ideal. But R is a field,so it is simple, and thus it must be either Kerα = R or Kerα = 0. But for anyring morphism one has α(1) = 1, and since S is not trivial, 1 6= 0, thus 1 /∈ Kerα,so it cannot be Kerα = R, hence it must be Kerα = 0, and thus α is injective.ii) Let a, b ∈ R such that ab ∈ Kerα, then 0 = α(ab) = α(a)α(b), but since S isan ID, then it must be either α(a) = 0 (and thus a ∈ Kerα) or α(b) = 0 (yieldingb ∈ Kerα). Hence, Kerα is prime.

An alternative proof is the following: by the first isomorphism theorem, one hasR/Kerα ∼= Imα ≤ S. As S is an ID, Imα is also an ID, and we know that aquotient R/I is an ID if and only if I is prime, so Kerα must be prime.iii) Assume α is not injective. By the first isomorphism theorem R/Kerα ∼=Imα ≤ S. As S is an ID, by ii) Kerα must be prime. We know that R is a PID, soit must be Kerα = (a) for some a ∈ R, now, we know that the ideal (a) is primeif and only if a is prime, since every prime is irreducible, we get a irreducible, andthus the ideal (a) is maximal among principal ideals. But R is a PID, so all idealsare principal, implying that (a) is actually maximal, and thus Imα ∼= R/(a) is afield.iv) For any s ∈ S, the constant polynomial s ∈ S[x] satisfies e0(s) = s, thus thehomomorphism e0 is surjective, i.e. Im e0 = S. Also, the polynomial x ∈ S[x]satisfies e0(x) = 0, hence x ∈ Ker e0 and e0 is not injective. If S[x] is a PID,applying iii) we obtain that S = Im e0 is a field. �

Remark: Note that this exercise provides a proof that S[x] is a PID if and only ifS is a field which is different from the one obtained in Coursework 3, Ex 4.

Exercise 2 (*). LetR be a UFD, and a ∈ R∗\U(R) an element that can be writtenas a = pα1

1 · · · pαtt for distinct, non-associated primes pi ∈ R, where αi ≥ 1 for all

i = 1, . . . , t.1

2

i) If d|a, show that then d ∼ pβ11 · · · pβtt for some βi such that 0 ≤ βi ≤ αi.

ii) Show that there are (1+α1) · · · (1+αt) distinct principal ideals (d) such that(a) ⊆ (d).

iii) Deduce that R satisfies ACC on principal ideals.

Solution.i) If d|a, then there is some c ∈ R such that a = dc, write d = uq1 · · · qs, c =vq′1 · · · q′r as product of irreducibles, where u, v are units to take care of the casethat either c or d were units, one has

a = pα11 · · · p

αtt = cd = uvq1 · · · qsq′1 · · · q′r,

by uniqueness of the factorization, each of the qj must be (an associate of) one ofthe pi. By reordering and putting equal terms together one gets d ∼ pβ11 · · · p

βtt ,

where 0 ≤ βi ≤ αi.ii) Each principal ideal (d) containing (a) must be generated by an element d suchthat d|a. By i), d must be an associate of some pβ11 · · · p

βtt with 0 ≤ βi ≤ αi. Thus

one has 1 + αi different choices for βi, making a total of (1 + α1) · · · (1 + αt)different choices (recall that associate elements generate the same principal ideal).iii) Let (a1) ⊆ (a2) ⊆ · · · be a chain of principal ideals in R. If all the ideals are0, the chain trivially stabilizes, so we can assume that there some of the ideals (aj)is nonzero, i.e. aj 6= 0; then for any n ≥ j one has (aj) ⊆ (an). By ii), there isonly a finite number of principal ideals containing aj , thus the chain must stabilizeat some point. �

Exercise 3 (*). Let R be a UFD and a, b ∈ R∗ \ U(R). If a = upα11 · · · p

αtt and

b = vpβ11 · · · pβtt , where u, v ∈ U(R) are units, pi ∈ R are primes, and αi, βi ≥ 0.

i) Let γi := min{αi, βi}. Show that gcd(a, b) = pγ11 · · · pγtt .

ii) Let δi := max{αi, βi}. Show that the element e = pδ11 · · · pδtt satisfies the

following properties:(a) a|e, b|e.(b) If f ∈ R is such that a|f and b|f , then e|f .

Remark: The element e is called a least common multiple of a and b, and isdenoted by lcm(a, b).

iii) Deduce that gcd(a, b) lcm(a, b) = ab.

Solution.i) Take d = pγ11 · · · p

γtt . By 2 i), we know that d|a and d|b. Let e ∈ R be such that

e|a and e|b. Again by 2 i), since e|a one gets e = pr11 · · · prtt with 0 ≤ ri ≤ αi. But

since we also have e|b it must be 0 ≤ ri ≤ βi as well, thus ri ≤ min(αi, βi) = γi,and hence, applying 2 i) once more, e|d, proving that d is a greatest common divisorof a and b.ii) Since αi ≤ δi for all i, from 2 i) it follows that a|e. Similarly, as βi ≤ δi for alli one gets b|e. Let f ∈ R be such that a|f , b|f . Then f = am = pα1

1 · · · pαtt m. By

uniqueness of factorizations, f must contain the factor pi with multiplicity at leastαi. Similarly, as b|f , then for each i f must contain the factor pi with multiplicityat least βi, hence pi appears in the factorization of f at least max(αi, βi) = δitimes; thus, all the prime factors of e appear in the factorization of f with greatermultiplicity, and thus e|f .iii) Note that for any pair of real numbers numbers x, y one has max(x, y) +min(x, y) = x + y. Using this one gets γi + δi = min(αi, βi) + max(αi, βi) =

3

αi + βi and hence

gcd(a, b) lcm(a, b) = pγ11 · · · pγtt p

δ11 · · · p

δtt

= pγ1+δ11 · · · pγt+δtt

= pα1+β11 · · · pαt+βt

t

= ab

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Exercise 4 (*). Let R be a UFD, and a, b ∈ R. Prove that the ideal (a) ∩ (b) isprincipal.

Solution. By 3, we know that there exists e = lcm(a, b). Now, by definition ofleast common multiple one has a|e, thus (e) ⊆ (a), and b|e, thus (e) ⊆ (b), andtherefore (e) ⊆ (a) ∩ (b). Conversely, let f ∈ (a) ∩ (b), then f ∈ (a), and so a|f ,and f ∈ (b), yielding b|f . By definition of least common multiple, one gets e|f ,and thus f ∈ (e), so we get (a)∩ (b) ⊆ (e), and consequently (a)∩ (b) = (e). �

Exercise 5 (C). Show that a ring R satisfies the ACC on all ideals (i.e. is Noether-ian) if and only if every ideal is finitely generated, i.e. if for any ideal I ER thereexist a1, . . . , an ∈ R such that I = (a1) + · · ·+ (an).

Solution.⇐ Assume that every ideal is finitely generated. Let

I1 ⊆ I2 ⊆ · · ·be an ascending chain of ideals of R. Then the set I =

⋃n In is also an ideal of R,

so it must be finitely generated, i.e. there are ai ∈ R such that I = (a1)+· · ·+(an),but then, each of these ai ∈ I =

⋃In, so there is some ki such that ai ∈ Iki . By

taking N = max(k1, . . . , kn) one has ai ∈ IN for all i, thus IN = I , and one has

I = IN ⊆ IN+1 ⊆ · · · ⊆ I,therefore In = I = In for all n ≥ N , and thus the chain stabilizes, and hence R isNoetherian.⇒ Assume that R is Noetherian. Suppose there is some I E R ideal of R which

is not finitely generated. If I = 0, then I = (0), so I must be nonzero. Picka1 ∈ I . Since I is not finitely generated, it must be I1 = (a1) ( I , so there issome element a2 ∈ I such that a2 /∈ (a1). Define I2 = (a1, a2), then I2 is alsofinitely generated, so there must exist a3 ∈ I such that a3 /∈ I2. Inductively, eachIn = (a1, . . . , an) is finitely generated, so it cannot be equal to I , so there mustexist an+1 ∈ I such that an+1 /∈ In and we can form In+1 = In + (an+1). In thisway, we build a chain of ideals

I1 ( I2 ( · · · ( In ( In+1 ( · · ·that never stabilizes, contradicting the Noetherian property for R.

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