SAG 2017: Principal Ideal Domains 8.2 # 1, 2, 3, 4, 5

Home , Principal ideal, Principal ideal domain, Quotient

Austin Alderete

Jun. 30th, 2017

For what follows, the term PID refers to a principal ideal domain. 8.2.1) Prove that in a PID two ideals (a) and (b) are comaximal if and only if a greatest common divisor of a and b is 1 (in which case we say that a and b are relatively prime or coprime).

Proof. Let R be a PID and let (a) and (b) be two ideal in R. (⇒) Suppose that (a) and (b) are comaximal. Then (a) + (b) = R. As (a) + (b) is the smallest ideal containing (a) and (b) it is the smallest ideal generated by the two elements, (a, b). Observe then that (1) = R = (a, b) = (a) + (b) and so 1 is a greatest common divisor of a and b. (⇐) Suppose instead that a and b are coprime. Then 1is a greatest common divisor of a and b which means that 1 is a generator for the smallest pricipal ideal containing a and b. But then (a) + (b) = (a, b) = (1) = R, and so (a) and (b) are comaximal.

8.2.2) Prove that any two nonzero elements of a PID have a least common multiple.

Proof. Let R be a PID and let a, b be nonzero elements of R. Recall that a least common multiple of a and b is an element e ∈ R such that a, b |e and if a, b |e0 then e |e0. We claim that the ideal (a)∩(b) is nonempty and its generator, call it e, is a least common multiple of a and b.

1 Recall that (a)(b) ⊆ (a) ∩ (b). As ab ∈ (a)(b) is nonzero, we have that (a) ∩ (b) is nonempty. As R is a PID, it has a generator e. Note that e ∈ (a) ∩ (b) implies e ∈ (a), (b) implies (e) ⊆ (a), (e) ⊆ (b) which is true if and only if a|e, b|e. Suppose now that we have e0 such that a, b divide e0. Then (e0) ⊆ (a) and (e0) ⊆ (b) and so e0 ∈ (a) ∩ (b) = (e) which implies e|e0.

8.2.3) Prove that a quotient of a PID by a prime ideal is again a PID.

Proof. Let R be a PID and let (p) be a prime ideal of R. We can assume that (p) is nonzero as if (p) is the zero ideal, its quotient is isomorphic to R itself. As every nonzero prime ideal in a PID is a maximal ideal, we have that (p) is maximal. Then the quotient ring R/(p) is a ﬁeld, which automatically implies the result. In fact, the result is sadly trivial here as ﬁelds don’t have any proper ideals of which to speak.

8.2.4) Let R be an integral domain. Prove that if the following two conditions hold then R is a PID: (i) any two nonzero elements a and b in R have a greatest common divisor which can be written in the form ra + sb for some r, s ∈ R, and

(ii) if a1, a2, a3, ... are nonzero elements of R such that ai+1|ai for all i, then there is a

positive integer N such that an is a unit times aN for all n ≥ N. Proof attempt: Reading through it, there are obviously many things wrong (mainly the assumption that I is countably generated and that one can form the gcd as I did). However, it inspires the correct proof that follows.

Proof. Let R be an integral domain and suppose that (i) and (ii) above hold. Let I be any ideal in R. Suppose we have a generating set for I (possibly I itself)

I = hb1, b2, ...i.

For each bi, bi+1 for i ≥ 1 we can form, from property (i), ci,i+1,...,j the greatest common

divisor of bi, bi+1, all the way to bj.. Then

I = hb1, c12, c123, c1234, ...i.

2 Observe that b1|c12 and ci,...,j|ci,...,j+1 by construction. Then from property (ii) it follows that we terminate the sequence at some point. That is,

I = hb1, c12, c123, c1234, ..., c123...N i

for some N. We can then ﬁnd a greatest common divisor of the entire ﬁnite set which implies that I is principal.

Actual proof:

Proof. Let R be an integral domain and suppose that (i) and (ii) above hold. Let I be any ideal in R. Pick an element b1 from I. If hb1i = I, we are done. If not, then pick an element c1 ∈ I − hb1i. Let b2 = gcd(b1, c1), b2 is in I by property (i). If hb2i = I then we are done. Suppose that we are never ‘done’. Continue on in this manner, to create the sequence

{b1, b2, b3, ...}.

Observe that as bi+1 = gcd(bi, ci) we have that bi+1|bi. Therefore we have the nested chain of ideals

(b1) ⊆ (b2) ⊆ ... ⊆ (bi) ⊆ (bi+1) ⊆ ...

By property (ii), there exists N such that for all n ≥ N, an = raN for r a unit in R. That −1 means, r being invertible, that r an = aN or that the chain eventually stabilizes (as an

both divides and is divided by aN for all n ≥ N):

(b1) ⊆ (b2) ⊆ ... ⊆ (bN ) = (bN+1) = ...

But this contradicts the assumption that we can always pick a ci and so this process must terminate at some point. That is, I is principal. √ 8.2.5) Let R be the quadratic integer ring Z[ −5]. Deﬁne the ideals I2 = (2, 1 + √ √ √ 0 −5),I3 = (3, 2 + −5), and I3 = (3, 2 − −5).

0 (a) Prove that I2,I3,I3 are nonprincipal ideals in R.

3 √ Proof. Let N be the norm N(a + b −5) = a2 + 5b2.

Suppose that I2 is principal. Then √ √ √ 2 = α(a + b −5) ; 1 + −5 = β(a + b −5)

for α ∈ R and a, b ∈ Z, and taking the norm of each

4 = N(α)[a2 + 5b2] ; 6 = N(β)[a2 + 5b2].

We have that N(α),N(β) ∈ Z and so

4 ≥ [a2 + 5b2] ; 6 ≥ [a2 + 5b2].

Noting that b2 ≥ 1 or b2 = 0, the left equation demands that b = 0 and so

4 = N(α)a2 ; 6 = N(β)a2

but note that 6 is square-free and so the right equation is never satisﬁed. From this

we conclude that I2 cannot be principal.

2 (b) Prove that the product of two nonprincipal ideals can be principal by showing that I2 is the principal ideal generated by 2.

Proof. It suﬃces to show that any product of two elements in I2 is a multiple of 2. In this manner, every ﬁnite sum of such products is also a multiple of two. We have √ √ √ 2(2) ; 2(1 + −5) ; (1 + −5)(1 + −5)

and as the ﬁrst two are obviously multiples of 2, we only need focus on the last. Computing, we ﬁnd √ √ √ √ √ (1 + −5)(1 + −5) = 1 + 2 −5 − 5 = −4 + 2 −5 = 2(−2 + −5)

and so we are done.