Math 254A. Integral Bases

This handout covers much of the material on integral bases from the book, but in (I hope) a more organized way.

Recall that we are assuming the “usual picture,” in which A is an integrally closed noetherian ring, K is its fraction field, L is a finite extension of K (not assumed separable), and B is the integral closure of A in L.

Here we make the further assumption that A is finite over ℤ or of finite type over a field. Recall that an A -algebra B is of finite type if it is finitely generated as an A-algebra (i.e., as a ring over the image of A → B).

Lemma 1. There exists a basis for L over K which is contained in B.

Proof. Let ω_{1}, …, ω_{n} be any basis for L over K. As was shown in class earlier (a claim on Monday, August 28), for each i there is a nonzero a_{i} ∈ A such that a_{i}ω_{i} ∈ B. Then a_{1}ω_{1}, …, a_{n}ω_{n} is a basis for L over K which is contained in B. □

We continue with a slight extension of Proposition 2.10.

Proposition 2. Under the above assumptions on A, B is finite over A.

Proof. If A is of finite type over a field, then this is “finiteness of the integral closure”— see Eisenbud Cor. 13.13.

Assume now that A is finite over ℤ. If char A ≠ 0 , then since A is entire we have char A = p is prime, and then A is of finite type over the field 𝔽_{p}. This is covered by the preceding case.

Therefore we may assume that char A = 0 . Then char K = 0 , so in particular L/K is separable and the proposition can be proved by the book’s method of using the discriminant.

Indeed, let ω_{1}, …, ω_{n} be a basis for L over K which is contained in B.

Let d be the discriminant d(ω_{1}, …, ω_{n}) . Since d ≠ 0 , it follows that B ≅ dB (as modules over A). Also, by Lemma 2.9 in the book, dB ⊆ ∑ Aω_{i}. Since A is noetherian and ∑ Aω_{i} is a finitely generated A-module, it follows that dB is also finitely generated (as an A-module), and therefore so is B. □

The remainder of the handout will handle the more general situation of certain A-submodules of L.

Definition 3. An A-submodule of L is full if it is not contained in a proper K-linear subspace of L.

Lemma 4. Let M be a nonzero, finitely generated B-submodule of L. Then M is also finitely generated as an A-module, and is full.

Proof. Let M be as above. Since M is finitely generated as a B-module, and B is finite over A, it follows that M is finitely generated as an A-module.

Next let m ∈ M be a nonzero element, and let ω_{1}, …, ω_{n} be a basis for L over K which is contained in B. Then {mω_{1}, …, mω_{n}} is a basis for L over K which is contained in mB, which in turn is contained in M . Thus M is full. □

Throughout the rest of this handout, M is a full, finitely generated A-submodule of L.

Definition 5. An integral basis for M over A is a collection ω_{1}, …, ω_{n} ∈ M such that each m ∈ M there are unique a_{1}, …, a_{n} ∈ A such that m = a_{1}ω_{1} + ⋯ + a_{n}ω_{n}.

Proposition 6. The following are equivalent.

(i). M has an integral basis over A; (ii). M is a free A-module of rank [L : K] . (iii). M is a free A-module of finite rank; (iv). M is a free A-module.

Proof. (i) ⟺ (iii) is immediate from the definition of free module. Also (ii) ⟹ (iii) ⟹ (iv) is trivial. Finally, to show (iv) ⟹ (ii), assume that M is a free A-module. Let {ω_{i}}_{i∈I} be any basis for M over A. This basis is linearly independent over A, hence over K, so it must be finite, and in fact |I| ≤ n. Finally, since M is full, this basis must span L, so we must have |I| = n. □

Proposition 7. If A is principal (i.e., A is a PID, since A was assumed to be entire), then M has an integral basis.

Proof. Since M is contained in the field L, it is torsion-free (as an A-module). Therefore, by the theory of finitely generated modules over a PID, M is free. □

Therefore, if A is principal, then M (and in particular B) has an integral basis, and we can define the discriminant of M to be the discriminant of an integral basis of M . It is an element of K, well defined up to multiplication by an element of (A^{∗})^{2} (by Lemma 0 from class on September 1). It is denoted d(M) . If M ⊆ B then d(M) ∈ A.

The special case in which A = ℤ is the main case considered in number theory. In this case, (ℤ^{∗})^{2} = (±1)^{2} = {1}, so the discriminant of a module M as above is a well-defined element of ℤ.

In this case we also have a generalization of Proposition 2.12 to the type of modules considered here. This will be needed for working with proper subrings of B.

Proposition 8. Let K be a number field, and let M ⊆ M′ be full, finitely generated subgroups of the additive group of K. Then the index (M′ : M) is finite and satisfies

d(M) = (M′ : M)^{2}d(M′) .

Proof. As in the book, by Lemma 0 on September 1, this is a matter of noting that (M′ : M) is the absolute value of the determinant of the base change matrix in passing from an integral basis of M to an integral basis of M′. This, in turn, follows either from Gaussian elimination of integer matrices, or from multivariable calculus on K ⊗_{ℚ} ℝ ≅ ℝ^{n}. □