Solutions to Homework 7
35. (Dummit-Foote 8.2 #6) (a) Let S be the set of all ideals
S of R that are not principal, and let
{Cα }α∈I be a chain in S. The chain Cα has as upper bound α∈I Cα . This union is clearly an ideal.
To see that it is non-principal, note that any generator d would have to live in Cα for some α,
implying Cα = (d), a contradiction. As every chain in S has an upper bound, a maximal element
of S exists by Zorn’s Lemma.
(b) Let I be a maximal element of the set S defined in part (a). By hypothesis, I is not prime,
hence we can find a, b ∈ R such that ab ∈ I but neither a nor b is in I. As I ( Ia , we must have
Ia = (α) for some α ∈ R, by maximality of I.
As bIa ⊆ I, we have b ∈ J, and it follows from the definition of ideal that I ⊆ J. It is now clear
that I ( Ib ⊆ J. Again by maximality of I, J is principal, say J = (β). The product of principal
ideals is principal, so Ia J = (αβ). But by the definition of J, we have Ia J ⊆ I.
(c) Let x ∈ I ⊆ Ia = (α). Then x = sα for some s ∈ R, and it is easy to see from the definition
of J that s ∈ J. Thus x ∈ Ia J, but x was arbitrary, so it follows that I ⊆ Ia J = (αβ). By part
(b), we now know that I = (αβ), which contradicts the fact that I is non-principal. Therefore S
must be empty, i.e. R is a PID.
36. (Dummit-Foote 8.2 #8) Let I be an ideal in D−1 R. Let
I 0 = {r ∈ R | there exists d ∈ D such that r/d ∈ I }.
It is straightforward to verify that I 0 is an ideal of R. As R is a PID, we must have I = (α) for
some α ∈ R. Then the same α will generate D−1 I 0 = I in D−1 R.
37. (Dummit-Foote 8.3 #6) (a) The ring Z[i] is a PID, hence prime ideals are maximal. As 1 + i is
irreducible by Proposition 18, the ideal (1 + i) is prime and hence Z[i]/(1 + i) is a field. To count its
cardinality note that 2 = (1 + i)(1 − i) and 2i = (1 + i)2 , and that i = i + 1 − 1. Thus in Z[i]/(1 + i),
¯ and ī = 1̄. From this it is easy to see that {0, 1} make up a complete list of
we have 2̄ = 0 = 2i,
coset representatives, hence Z[i]/(1 + i) has order 2.
(b) By hypothesis, (q) is prime and thus Z[i]/(q) is a field by the same reasoning as in part (a).
To count its order, note that the same
{a + bi |0 ≤ a ≤ q − 1, 0 ≤ b ≤ q − 1 }
is a complete list of coset representatives, and has order q 2 .
(c) Let p = ππ̄ in Z[i]. Now π and π̄ are distinct irreducibles, and their gcd (Z[i] is a Euclidean
domain) is equal to 1. Thus the ideals they generate are coprime, hence
Z[i]/(p) ∼
= Z[i]/(π) × Z[i]/(π̄)
by the Chinese Remainder theorem. By the same reasoning as in part (b), the ring Z[i]/(p) has
order p2 . The only non-trivial factorization of p2 is p · p, hence both factors on the left-hand side
1
of the above equation have order p.
38. (Dummit-Foote 9.2 #4) Suppose p1 , . . . , pn were a complete list of primes in F [x]. Let
q = p1 · · · pn + 1. Then pi - q for all i, which implies that q is a unit because F [x] is a UFD.
But deg(q) > 0, a contradiction.
39. Clearly if a ∈ R× then I = (a, X) = (1), so I is principal.
For the converse, assume I = (p) is principal. Then a is a multiple of p, which implies that
deg(p) = 0. Now X is also a multiple of p, so there exists some b + cX ∈ R[X] such that
p(b + cX) = pb + pcX = X.
Equating the coefficients of X, we see that pc = 1, i.e. that p is a unit in R. So (a, X) = (1), which
says that there exist s, t ∈ R[X] such that
sa + tX = 1.
Equating coefficients again immediately yields sa = 1, and hence a is a unit in R.
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