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.
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 , is an invertible fractional ideal of .
(⇒) This was proven earlier (propositions 2 and 5).
(⇐) It suffices to show . By definition . Let be any maximal ideal. Since M is finitely generated and is invertible, we have (by proposition 3 here):
If the A-module is contained in any maximal then , a contradiction. Hence . ♦
Complementing the above result, we have:
Let be a local domain. A fractional ideal M of A is invertible if and only if it is principal and non-zero. Thus .
Suppose M is invertible; write MN = A. Then for and . Since all and sum to 1, not all lie in . Thus is invertible for some i. Multiplying by a unit we may assume .
It remains to show . Clearly since , we have . Conversely if then since . ♦
Thus we have:
A non-zero fractional ideal M of A is invertible if and only if:
- it is finitely generated, and
- for each maximal , is a principal fractional ideal of .
Let with maximal ideal . We claim that is not invertible; it suffices to show that is not a principal ideal of
For that we apply Nakayama’s lemma to compute . Let ; we can check that so since . Since has 4 elements, we see that . Hence is not principal.
Let with . Is an invertible ideal?
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 if ; this follows from proposition 3 here: every ideal of is of the form for some ideal .
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 has height 1. Since is invertible, is principal by proposition 2. By exercise A.2 here, is a minimal non-zero prime so . ♦
A Dedekind domain A is normal.
Since normality is a local property, we may assume is local. And since is invertible, it is principal by proposition 2. Thus it suffices to prove the following.
- If is a noetherian local domain such that is principal, then A is normal.
The whole of the next section is devoted to such rings.
Discrete Valuation Rings
A discrete valuation ring (dvr) is a noetherian local domain such that is principal. If , we call a uniformizer of the dvr.
As we saw earlier, 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 . If , then is a dvr.
In particular, we can take any irreducible element π of a noetherian UFD A; then π is a prime element so is a dvr. Common examples include:
where k is a field.
Let A be a dvr with uniformizer . Then every non-zero ideal of A is uniquely of the form for some .
We have .
First we show that . Indeed if then we can write for . Then for each n so which is impossible since A is noetherian.
Now for each , let
, where .
We leave the remaining as an exercise.
- Prove that where .
- Hence show that any non-zero ideal of A is of the form 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.
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.
Let be a noetherian 1-dimensional local domain (i.e. A has exactly two prime ideals: 0 and ).
If A is normal then it is a dvr.
Pick so that is a noetherian ring with exactly one prime, i.e. it is a local artinian ring, so is nilpotent. Thus for some n > 0. We may assume .
Let , an ideal of A. Fix an and set . Then is an ideal of A. If , then is principal.
Otherwise . We claim that is integral over A. To prove this, pick a finite generating set for the ideal ; each can be written as an A-linear combination of so
where M is an matrix with entries in A. Hence where equality holds in . Multiplying by the adjugate matrix, we get
Since , we get ; expanding gives a monic polynomial relation in with coefficients in A.
Hence is integral over A; since A is normal, . Thus for all so . Since we have . But this means , a contradiction. ♦
Let A be a noetherian 1-dimensional domain. If A is normal, then it is a Dedekind domain.
Let be any ideal; it is finitely generated since A is noetherian. To prove it is invertible by proposition 1, it suffices to show is principal for each maximal ideal . But is a local 1-dimensional noetherian domain; since it is normal, by proposition 6 it is a dvr. Hence 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 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 is not smooth at the origin.