Commutative Algebra 45

Invertibility is Local

In this article, we again let A be an integral domain and K its field of fractions. We continue our discussion of invertible fractional ideals of A.

Proposition 1.

A fractional ideal M of A is invertible if and only if the following hold.

  • M is a finitely generated A-module.
  • For any maximal ideal \mathfrak m \subset A, M_{\mathfrak m} is an invertible fractional ideal of A_{\mathfrak m}.

Proof

(⇒) This was proven earlier (propositions 2 and 5).

(⇐) It suffices to show (A:M) M = A. By definition (A:M) M\subseteq A. Let \mathfrak m \subset A be any maximal ideal. Since M is finitely generated and M_{\mathfrak m} is invertible, we have (by proposition 3 here):

A_{\mathfrak m} = (A_{\mathfrak m} : M_{\mathfrak m})M_{\mathfrak m} = (A:M)_{\mathfrak m} M_{\mathfrak m} = [(A:M)M]_{\mathfrak m}.

If the A-module (A:M) M is contained in any maximal \mathfrak m then [(A:M)M]_{\mathfrak m} \subseteq \mathfrak m A_{\mathfrak m}, a contradiction. Hence (A:M)M = A. ♦

Complementing the above result, we have:

Proposition 2.

Let (A,\mathfrak m) be a local domain. A fractional ideal M of A is invertible if and only if it is principal and non-zero. Thus \mathrm{Pic} A = 0.

Proof

Suppose M is invertible; write MNA. Then 1 = x_1 y_1 + \ldots + x_n y_n for x_i \in M and y_i \in N. Since all x_i y_i \in A and sum to 1, not all x_i y_i lie in \mathfrak m. Thus x_i y_i \in A is invertible for some i. Multiplying x_i by a unit we may assume x_i y_i = 1.

It remains to show M = Ax_i. Clearly since x_i \in M, we have M\supseteq A x_i. Conversely if m\in M then m = x_i (my_i) \in Ax_i since MN\subseteq A. ♦

Thus we have:

Corollary 1. 

A non-zero fractional ideal M of A is invertible if and only if:

  • it is finitely generated, and
  • for each maximal \mathfrak m \subset A, M_{\mathfrak m} is a principal fractional ideal of A_{\mathfrak m}.

Example

Let A = \mathbb Z[2\sqrt 2] = \{a + 2b\sqrt 2 : a,b \in \mathbb Z\} with maximal ideal \mathfrak m = \{2a + 2b\sqrt 2: a,b\in \mathbb Z\}. We claim that \mathfrak m is not invertible; it suffices to show that \mathfrak m A_{\mathfrak m} is not a principal ideal of A_{\mathfrak m}.

For that we apply Nakayama’s lemma to compute \dim_{A/\mathfrak m} \mathfrak m /\mathfrak m^2. Let t = 2\sqrt 2; we can check that \mathfrak m = 2A + tA so \mathfrak m^2 = 4A + 2tA = 2\mathfrak m since t^2 = 8. Since \mathfrak m/\mathfrak m^2 has 4 elements, we see that \dim_{A/\mathfrak m} \mathfrak m /\mathfrak m^2 = 2. Hence \mathfrak m A_{\mathfrak m} is not principal.

Exercise

Let A = \mathbb C[X, Y]/(Y^2 - X^3) with \mathfrak m = (X, Y). Is \mathfrak m an invertible ideal?

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Dedekind Domains

Definition.

Dedekind domain A is an integral domain in which every fractional ideal is invertible (or equivalently, every ideal is invertible).

We obtain some immediate facts about Dedekind domains.

  • Since invertible ideals are finitely generated, a Dedekind domain is noetherian.
  • If A is a Dedekind domain so is its localization S^{-1}A if 0\not\in S; this follows from proposition 3 here: every ideal of S^{-1}A is of the form \mathfrak a(S^{-1}A) for some ideal \mathfrak a \subseteq A.

Proposition 3.

In a Dedekind domain A, every non-zero prime ideal is maximal. Thus its Krull dimension is at most 1.

Hence Spec A is as follows:

one-dimensional_domains

Proof

If A is a field we are done. Otherwise, we will show that every maximal ideal \mathfrak m\subset A has height 1. Since \mathfrak m is invertible, \mathfrak m A_{\mathfrak m}\subset A_{\mathfrak m} is principal by proposition 2. By exercise A.2 here, \mathfrak m A_{\mathfrak m} is a minimal non-zero prime so \mathrm{ht} \mathfrak m = 1. ♦

Proposition 4.

A Dedekind domain A is normal.

Proof

Since normality is a local property, we may assume (A, \mathfrak m) is local. And since \mathfrak m is invertible, it is principal by proposition 2. Thus it suffices to prove the following.

  • If (A,\mathfrak m) is a noetherian local domain such that \mathfrak m is principal, then A is normal.

The whole of the next section is devoted to such rings.

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Discrete Valuation Rings

Definition.

discrete valuation ring (dvr) is a noetherian local domain (A,\mathfrak m) such that \mathfrak m \ne 0 is principal. If \mathfrak m = (\pi), we call \pi a uniformizer of the dvr.

Note

As we saw earlier, (\pi) has height 1, so the spectrum of A is easy to describe:

spec_of_dvr

Here is an easy way to construct such rings.

Lemma.

Let A be an integral domain with a prime element \pi \in A. If \mathfrak p = (\pi), then A_{\mathfrak p} is a dvr.

Proof

Trivial. ♦

In particular, we can take any irreducible element π of a UFD A; then π is a prime element so A_{(\mathfrak \pi)} is a dvr. Common examples include:

\mathbb Z_{(2)} = \{\frac a b \in \mathbb Q : b \text{ odd}\},\quad k[X]_{(X)} = \{ \frac {f(X)}{g(X)} : g(0) \ne 0\},

where k is a field.

Proposition 5.

Let A be a dvr with uniformizer \pi. Then every non-zero ideal of A is uniquely of the form (\pi^n) for some n\ge 0.

Proof

We have A \supsetneq (\pi) \supsetneq (\pi^2) \supsetneq \ldots.

First we show that \cap_n (\pi^n) = 0. Indeed if x \in \cap_n (\pi^n), x\ne 0 then we can write x = \pi^n u_n for u_1, u_2, \ldots \in A. Then u_n = \pi u_{n+1} for each n so (u_1) \subsetneq (u_2) \subsetneq \ldots which is impossible since A is noetherian.

Now for each x\in A-\{0\}, let

\nu(x) = \max(n\ge 0 : x \in (\pi^n)), where (\pi^0) = A.

We leave the remaining as an exercise.

  • Prove that (x) = (\pi^k) where k = \nu(x).
  • Hence show that any non-zero ideal of A is of the form (\pi^k) for some k. ♦

Thus a dvr is a special type of UFD with exactly one prime element (its uniformizer). This makes prime factoring in the ring rather trivial: every non-zero element is a unit times a power of the uniformizer.

Corollary 2.

A dvr is a PID, hence a UFD so it is a normal domain.

To recap, we have shown that Dedekind domains are noetherian, of Krull dimension at most 1, and normal. Next we will see that the converse is true.

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Conditions for DVR

We have already seen:

dvr ⟹ PID ⟹ UFD ⟹ normal.

The following is a converse statement.

Proposition 6.

Let (A,\mathfrak m) be a noetherian 1-dimensional local domain (i.e. A has exactly two prime ideals: 0 and \mathfrak m).

If A is normal then it is a dvr.

Proof

Pick x\in \mathfrak m- \{0\} so that A/(x) is a noetherian ring with exactly one prime, i.e. it is a local artinian ring, so \mathfrak m/(x) \subset A/(x) is nilpotent. Thus \mathfrak m^n \subseteq (x) \subseteq \mathfrak m for some n > 0. We may assume \mathfrak m^{n-1} \not\subseteq (x).

Let \mathfrak a = \{a\in A: a\mathfrak m\subseteq (x)\}, an ideal of A. Fix an a\in \mathfrak a and set \alpha := \frac a x \in\mathrm{Frac} A. Then \alpha \mathfrak m = \frac a x \mathfrak m \subseteq A is an ideal of A. If \alpha \mathfrak m = A, then \mathfrak m = \frac 1 {\alpha} A is principal.

Otherwise \alpha \mathfrak m \subseteq \mathfrak m. We claim that \alpha is integral over A. To prove this, pick a finite generating set y_1, \ldots, y_n for the ideal \mathfrak m; each \alpha y_i can be written as an A-linear combination of y_1, \ldots, y_n so

\alpha (y_1, \ldots, y_n)^t = M (y_1, \ldots, y_n)^t

where M is an n \times n matrix with entries in A. Hence (\alpha I - M)(y_1, \ldots, y_n)^t = 0 where equality holds in \mathrm{Frac} A. Multiplying by the adjugate matrix, we get

\det(\alpha I - M) \cdot (y_1, \ldots, y_n)^t = 0.

Since \mathfrak m \ne 0, we get \det(\alpha I - M) = 0; expanding gives a monic polynomial relation in \alpha with coefficients in A.

Hence \alpha \in \mathrm{Frac} A is integral over A; since A is normal, \alpha \in A. Thus \frac a x \in A for all a\in \mathfrak a so \mathfrak a \subseteq (x). Since x \in \mathfrak a we have \mathfrak a = (x). But this means \mathfrak m^{n-1} \subseteq \mathfrak a = (x), a contradiction. ♦

Corollary 3.

Let A be a noetherian 1-dimensional domain. If A is normal, then it is a Dedekind domain.

Proof

Let \mathfrak a \subseteq A be any ideal; it is finitely generated since A is noetherian. To prove it is invertible by proposition 1, it suffices to show \mathfrak aA_{\mathfrak m} is principal for each maximal ideal \mathfrak m. But A_{\mathfrak m} is a local 1-dimensional noetherian domain; since it is normal, by proposition 6 it is a dvr. Hence \mathfrak aA_{\mathfrak m} is principal. ♦

Summary.

Let A be an integral domain. Every non-zero ideal of A is invertible if and only if it is noetherian, normal and of Krull dimension at most 1.

If A = k[V] is the coordinate ring of a variety V, geometrically the above conditions translate as follows:

  • Krull dimension = 1 ⟺ V is a curve;
  • A is a domain ⟺ V is irreducible;
  • A is normal ⟺ V is “smooth”.

Hence philosophically, a Dedekind domain is “like a smooth curve”.

We put the term “smooth” in quotes because we based the concept on geometric intuition rather than rigour. E.g. in our example above, the curve Y^2 = X^3 is not smooth at the origin.

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