Solutions to Homework 7 35. (Dummit-Foote 8.2 #6) (a) Let S be the ...

Solutions to Homework 735. (Dummit-Foote 8.2 #6) (a) Let S be the set of all ideals of R that are not principal, and let{C α } α∈I be a chain in S. The chain C α has as upper bound ⋃ α∈I C α. This union is clearly an ideal.To see that it is non-principal, note that any generator d would have to live in C α for some α,implying C α = (d), a contradiction. As every chain in S has an upper bound, a maximal elementof 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 I a , we must haveI a = (α) for some α ∈ R, by maximality of I.As bI a ⊆ I, we have b ∈ J, and it follows from the definition of ideal that I ⊆ J. It is now clearthat I I b ⊆ J. Again by maximality of I, J is principal, say J = (β). The product of principalideals is principal, so I a J = (αβ). But by the definition of J, we have I a J ⊆ I.(c) Let x ∈ I ⊆ I a = (α). Then x = sα for some s ∈ R, and it is easy to see from the definitionof J that s ∈ J. Thus x ∈ I a J, but x was arbitrary, so it follows that I ⊆ I a J = (αβ). By part(b), we now know that I = (αβ), which contradicts the fact that I is non-principal. Therefore Smust be empty, i.e. R is a PID.36. (Dummit-Foote 8.2 #8) Let I be an ideal in D −1 R. LetI ′ = {r ∈ R | there exists d ∈ D such that r/d ∈ I }.It is straightforward to verify that I ′ is an ideal of R. As R is a PID, we must have I = (α) forsome α ∈ R. Then the same α will generate D −1 I ′ = 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 isirreducible by Proposition 18, the ideal (1+i) is prime and hence Z[i]/(1+i) is a field. To count itscardinality note that 2 = (1 + i)(1 − i) and 2i = (1 + i) 2 , and that i = i + 1 − 1. Thus in Z[i]/(1 + i),we have ¯2 = 0 = ¯2i, and ī = ¯1. From this it is easy to see that {0, 1} make up a complete list ofcoset 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 = π¯π in Z[i]. Now π and ¯π are distinct irreducibles, and their gcd (Z[i] is a Euclideandomain) is equal to 1. Thus the ideals they generate are coprime, henceZ[i]/(p) ∼ = Z[i]/(π) × Z[i]/(¯π)by the Chinese Remainder theorem. By the same reasoning as in part (b), the ring Z[i]/(p) hasorder p 2 . The only non-trivial factorization of p 2 is p · p, hence both factors on the left-hand side1

of the above equation have order p.38. (Dummit-Foote 9.2 #4) Suppose p 1 , . . . , p n were a complete list of primes in F [x]. Letq = p 1 · · · p n + 1. Then p i ∤ 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 thatdeg(p) = 0. Now X is also a multiple of p, so there exists some b + cX ∈ R[X] such thatp(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), whichsays that there exist s, t ∈ R[X] such thatsa + tX = 1.Equating coefficients again immediately yields sa = 1, and hence a is a unit in R.2