MATH 250B: COMMUTATIVE ALGEBRA HOMEWORK #8

JOHN VOIGHT

Problem 1. Let K be a Galois extension of the ℚ with group G. Let B the integral closure of ℤ in K, and let α ∈ B be such that K = ℚ(α). Let f(X) be the irreducible polynomial for α over ℚ. Let p be a prime number, and assume that f remains irreducible modulo p over ℤ/pℤ. What can you say about the Galois group G?

Solution. Let 𝔭 be a prime above p in B. Then the decomposition group G_{𝔭} is a subgroup of G; by Proposition 2.5, the extension B/𝔭 over ℤ/pℤ is Galois with group cyclic of order n = [K : ℚ] = deg f. By proposition 2.8, G_{𝔭} is isomorphic to this Galois group, so ℤ/nℤ ⊂ G. Since both groups are finite of order n, G ≅ ℤ/nℤ, i.e. G is cyclic.

Problem 2. Let A be an entire ring and K its quotient field. Let t be transcendental over K. If A is integrally closed, show that A[t] is integrally closed.

Solution. Let x ∈ K(t) be integral over A[t]. Since K[t] is a UFD, it is integrally closed, so x ∈ K[t] (Proposition 1.7). Let x satisfy the integral equation

f(X) = X^{n} + a_{n−1}X^{n−1} + ⋯ + a_{0}

with coefficients a_{i} ∈ A[t]. Write x = x(t) = c_{n}t^{n} + ⋯ + c_{0} with c_{i} ∈ K. Substituting these into f, looking at the coefficients of each t^{i} we see that the coefficients c_{i} are themselves integral over A, hence c_{i} ∈ A, and x ∈ A[t].

Problem 4. Let L be a finite extension of ℚ and let O_{L} be the ring of algebraic integers in L. Let σ_{1}, …, σ_{n} be the distinct embeddings of L into the complex numbers. Embed O_{L} into a Euclidean space by the map

α ↦ (σ_{1}α, …, σ_{n}α).

Show that in any bounded region of this Euclidean space, there is only a finite number of elements of O_{L}. [Hint: The coefficients in an integral equation for α are elementary symmetric functions of the conjugates of α and thus are bounded integers.] Use Exercise 5 of Chapter III to conclude that O_{L} is a free ℤ-module of dimension ≤ n. In fact, show that the dimension is n, a basis of O_{L} over ℤ also being a basis of L over ℚ.

Solution. Suppose α lies in the bounded region. Then the numbers σ_{i}α are all bounded; in particular, the coefficients of an integral equation for α (being elementary symmetric functions of the conjugates of α) are bounded integers. Since the degree of an integral equation for α is also bounded (by the degree [L : ℚ]), there are only finitely many such integral equations, hence only finitely many α.

Date: April 10, 2003.

VII: 1, 2, 4, 6, 7, 9, 10.

By Exercise III.5, O_{L} is a free ℤ-module of rank ≤ n. Let α_{1}, …, α_{n} be a basis for L over ℚ. For each α_{i}, there exists an integer c_{i} ∈ ℤ_{>0} such that c_{i}α_{i} ∈ O_{L}; therefore c_{i}α_{i} ∈ O_{L} are ℤ-linearly independent, so O_{L} contains a free ℤ-submodule of rank n, hence O_{L} is itself free of rank n.

Problem 7. Let O be an entire ring which is noetherian, integrally closed, and such that every nonzero prime ideal is maximal. Define a fractional ideal 𝔞 to be a nonzero 𝒪-submodule of the quotient field K such that there exists c ∈ O, c ≠ 0 for which c𝔞 ⊂ O. Prove that fractional ideals form a group under multiplication:

(a) Given an ideal 𝔞 ≠ 0 in O, there exists a product of prime ideals 𝔭_{1} … 𝔭_{r} ⊂ 𝔞.

(b) Every maximal ideal 𝔭 is invertible, i.e. if we let 𝔭^{−1} be the set of elements x ∈ K such that x𝔭 ⊂ O, then 𝔭^{−1}𝔭 = O.

(c) Every nonzero ideal is invertible, by a fractional ideal. [Use the noetherian property that if this is not true, there exists a maximal noninvertible ideal 𝔞 and get a contradiction.]

Solution. For (a), consider the set of ideals which do not contain a product of primes. If the set is nonempty, since O is noetherian, there exists a maximal such ideal 𝔞. We cannot have 𝔞 prime, therefore there exists x, y ∈ O such that xy ∈ 𝔞 but x, y ∉ 𝔞. Since 𝔞 is maximal, the ideals 𝔞 + (x) and 𝔞 + (y) contain a product of prime ideals: but

(𝔞 + (x))(𝔞 + (y)) ⊂ 𝔞

so 𝔞 contains a product of primes, a contradiction.

For (b), it is clear that 𝔭^{−1}𝔭 is an ideal of O, so since 𝔭 is maximal, we must have either 𝔭^{−1}𝔭 = 𝔭, or 𝔭^{−1}𝔭 = O. Let a ∈ 𝔭 be nonzero: then by (a) there exists a product of primes 𝔭_{1} … 𝔭_{r} ⊂ (a) ⊂ 𝔭. We may assume that 𝔭_{1} … 𝔭_{r} is a minimal such product (r taken as small as possible). Since every prime is maximal, we know 𝔭 = 𝔭_{i} for some i, so we may assume 𝔭 = 𝔭_{1}. Then 𝔭_{2} … 𝔭_{r} ⊄ (a), so there exists b ∈ 𝔭_{2} … 𝔭_{r} such that b ∉ (a). But b𝔭 ⊂ (a), so ba^{−1}𝔭 ⊂ O, so ba^{−1} ∈ 𝔭^{−1}. Since b ∉ aO, ba^{−1} ∉ O, so 𝔭^{−1} ≠ O. This implies 𝔭^{−1}𝔭 = O.

For (c), suppose that the set of nonzero noninvertible ideals is nonempty. Then since O is noetherian, there exists a maximal such noninvertible ideal 𝔞. If 𝔞 is maximal, then by (b) it is invertible; otherwise, it is contained (properly) in a maximal ideal 𝔭, and

𝔞 ⊂ 𝔞𝔭^{−1} ⊂ 𝔭𝔭^{−1} = O.

We cannot have 𝔞 = 𝔞𝔭^{−1}, so 𝔞𝔭^{−1} is invertible. But then

O = (𝔞𝔭^{−1})𝔞𝔭

implies that 𝔞 is invertible, a contradiction.

To conclude, note that every fractional ideal 𝔞 has c𝔞 ⊂ 𝔒 for some c ≠ 0: then (1/c)(c𝔞)^{−1} is an inverse for 𝔞.

Problem 9. Let A be an entire ring, integrally closed. Let B be entire, integral over A. Let 𝔮_{1}, 𝔮_{2} be prime ideals of B with 𝔮_{1} ⊃ 𝔮_{2} but 𝔮_{1} ≠ 𝔮_{2}. Let 𝔭_{i} = 𝔮_{i} ∩ A. Show that 𝔭_{1} ≠ 𝔭_{2}.

Solution. Suppose 𝔭_{1} = 𝔭_{2} = 𝔭. Then the ring A_{𝔭} is a local ring with maximal ideal 𝔭A_{𝔭}, and B_{𝔭} is an integral extensions of A_{𝔭} with 𝔮_{i}B_{𝔭} ∩ A_{𝔭} = 𝔭A_{𝔭}, so since the latter is maximal, so too is each 𝔮_{i}. In particular, if 𝔮_{1} ⊂ 𝔮_{2}, then 𝔮_{1} = 𝔮_{2}, a contradiction.