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}.


(⇒) 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.


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}.


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.


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


Dedekind Domains


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:



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.


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.


Discrete Valuation Rings


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.


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


Here is an easy way to construct such rings.


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


Trivial. ♦

In particular, we can take any irreducible element π of a noetherian 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.


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.


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.


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.


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. ♦


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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7 Responses to Commutative Algebra 45

  1. Vanya says:

    Where was Nakayama’s lemma used in the example after Corollary 1?

  2. Vanya says:

    In the exercise after Example, I obtained \mathfrak{m}^2 = X \mathfrak{m} so that only $Y$ is the basis of the vector space \mathfrak{m}/ \mathfrak{m}^2 over A/\mathfrak{m} \cong \mathbf{C}. How does one conclude from this that \mathfrak{m} is principal? Thank you.

  3. Vanya says:

    In the first lemma in the section “discrete valuation rings”, should A be noetherian too in addition to being a domain, or does it follow that A_{\mathfrak{p}} is noetherian?

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