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 MN = A. 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.

A 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.

A 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 a noetherian 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 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 $n\ge 0$ .

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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7 Responses to Commutative Algebra 45

Vanya says:

July 23, 2020 at 10:52 am

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

- limsup says:
  
  July 25, 2020 at 4:24 pm
  
  Nakayama’s lemma says: if $B = A_{\mathfrak m}$ is local and $\mathfrak n = \mathfrak m A_{\mathfrak m}$ is its unique maximal ideal, then the minimum number of generators of $\mathfrak n$ is the dimension $\dim_{B/\mathfrak n} \mathfrak n / \mathfrak n^2$ .
  
  Then you also need the fact that $\mathfrak n / \mathfrak n^2 \cong \mathfrak m / \mathfrak m^2$ . This was covered earlier (I think).
  
  - limsup says:
    
    July 25, 2020 at 4:26 pm
    
    Read two paragraphs after exercise B in https://mathstrek.blog/2020/04/22/commutative-algebra-34/
Vanya says:

July 23, 2020 at 11:03 am

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.

- limsup says:
  
  July 25, 2020 at 4:29 pm
  
  Actually $\mathfrak m$ is not principal here. Use the logic in the example to conclude.
  
Vanya says:

July 23, 2020 at 1:13 pm

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?

- limsup says:
  
  July 25, 2020 at 4:50 pm
  
  I think you’re right. Noetherian needs to be specified.
  
  Thanks – corrected.