1
$\begingroup$

Let $A$ be a commutative noetherian ring (I do not mind to assume that $A$ is a UFD), and assume that $A$ is regular.

Recall that a commutative noetherian ring is called regular if all its localizations at maximal ideals are regular local rings.

Let $w$ be algebraic over $A$ (I do not mind to assume that $w$ is integral over $A$. I do NOT want to assume that $w$ is in the field of fractions of $A$), and assume that $A[w]$ is noetherian.

My question: When is $A[w]$ regular? (I suspect there exists a counterexample, even under the additional conditions: $A$ is a UFD, $w$ is integral over $A$).

Notice the following question, which deals with $w$ transcendental over $A$: Does the regularity of $A$ imply the regularity of $A[X]$?

$\endgroup$
4
  • $\begingroup$ Some simplifications that don't really answer your question: I think $A$ will automatically be a UFD (Auslander-Buchsbaum) and in particular normal. And certainly $A[w]$ will be Noetherian by the Hilbert basis theorem. $\endgroup$
    – Hoot
    Jun 11, 2015 at 1:35
  • $\begingroup$ Thanks for trying to help. A clarification: In the specific example I have in mind, $A$ is not local but a UFD. (If I am not wrong, generally: A regular ring is normal-- this can be found in Matsumura's book. A regular local ring is a UFD). I thought the Hilbert basis theorem works when $w$ is transcendental over $A$, not algebraic? $\endgroup$
    – user237522
    Jun 11, 2015 at 1:51
  • $\begingroup$ Thanks! Actually, I have once asked a similar question (in which $A$ is a polynomial ring in two indeterminates): math.stackexchange.com/questions/1287044/… $\endgroup$
    – user237522
    Jun 11, 2015 at 2:29
  • $\begingroup$ I don't understand your emphasis on $A[w]$ noetherian: AFAIK a regular ring is assumed noetherian (that is, $A$ is noetherian by definition), and I don't think there are many books in CA which miss this condition. $\endgroup$
    – user26857
    Jun 11, 2015 at 9:02

2 Answers 2

Reset to default
4
$\begingroup$

The nonregular ring $R=k[x,y]/\langle x^2 - y^3\rangle$, which is the coordinate ring of the cusp (not regular at the maximal ideal $\langle x,y\rangle$), is of this form. Indeed, you can set $A:=k[t^2]\cong k[t]$ which is a polynomial ring and then, $R=A[t^3]$. Here, $t^3$ is integral over $A$, but it is not in the field of fractions for reasons of degree.

Geometry: You are basically taking a smooth affine scheme $X$ and then you consider a closed subscheme $Z\subseteq \mathbb A^1_X$ whose projection to $X$ is finite. That closed subscheme may well be singular. However, this is not the case if that finite projection is actually étale, i.e. smooth of relative dimension zero - in this case $Z$ must be smooth. There are several equivalent definitions of being étale, one would be flat and unramified. These are now just properties of the ring extension $A\subseteq A[w]$.

Edit. When you can guarantee flatness and you want to use the above criterion, you have to check the following: For any prime ideal $\mathfrak q\subseteq A[w]$, the prime ideal $\mathfrak p := \mathfrak q\cap A$ (of $A$) generates the maximal ideal of the local ring $A[w]_{\mathfrak q}$. This is the condition for $A\hookrightarrow A[w]$ to be unramified.

$\endgroup$
5
  • 3
    $\begingroup$ very nice answer, especially the geometric interpretation. One can note that in this case the extension (or the morphism) is not flat. $\endgroup$
    – Parisien
    Jun 11, 2015 at 7:40
  • $\begingroup$ @Jesko Huttenhain: Thanks! What if I know that the extension $A \subset A[w]$ is flat? (in the special example I have in mind, $A[w]$ is actually a free $A$-module). $\endgroup$
    – user237522
    Jun 11, 2015 at 16:03
  • $\begingroup$ @user237522 In spite of the first comment above, the given example is such $A\subset A[w]$ is flat, in fact even free. (I've thought this is pretty clear from the way I've constructed this example in my answer.) $\endgroup$
    – user26857
    Jun 11, 2015 at 16:24
  • 2
    $\begingroup$ @user26857: That's what I thought...next time I will count on myself. $\endgroup$
    – user237522
    Jun 11, 2015 at 16:53
  • $\begingroup$ @user237522: I added an explanation of "unramified". I am not sure how easy that is to check, though. $\endgroup$
    – Jesko Hüttenhain
    Jun 11, 2015 at 17:20
1
$\begingroup$

If $A$ is regular, then $A[X]$ is regular. Now let $f$ be a monic polynomial in the square of a maximal ideal of $A[X]$. Then $A[X]/(f)=A[w]$ is not regular. (It's worthwhile to mention that $A[w]$ is however CM; see Bruns and Herzog, Proposition 2.2.11.)

Concrete example: $A=k[t]$, and $f(X)=X^2-t^3$. The maximal ideal $\mathfrak m=(t,X)$ has the property $f\in\mathfrak m^2$. (I've chosen this example on purpose, to show that the example given by the other answer is a particular case of the construction I proposed: just replace here $t$ by $t^2$.)

$\endgroup$
7
  • $\begingroup$ Thanks! In the special example I have in mind (in which $A$ is a UFD and $A[w]$ is a domain), $A[w]$ is even a complete intersection ring (regular $\subset$ complete intersection $\subset$ Gorenstein $\subset$ Cohen-Macaulay). However, I am not able to show it is regular. Actually it is good enough for me to show that $A[w]$ is integrally closed. $\endgroup$
    – user237522
    Jun 11, 2015 at 16:18
  • $\begingroup$ @user237522 Please rephrase your comment in such a way I can understand exactly what's the hypothesis and what you want to show. $\endgroup$
    – user26857
    Jun 11, 2015 at 16:21
  • $\begingroup$ Thank you very much! Now, $A$ is a regular UFD, $w$ is not in the field of fractions of $A$, $w$ is integral over $A$, so $A[w]$ is a finitely generated $A$-module (then by a corollary on page 58 of one of Bourbaki's books, $A[w]$ is a projective $A$-module. But my $A$ is $K[x,y]$, so $A[w]$ is a f.g. free $A$-module, by Quillen-Suslin). I wish to show that $A[w]$ is integrally closed (if I am not wrong, for a domain being normal and integrally closed is the same. If one shows $A[w]$ is regular, then it is normal; however, showing normality is good enough for me). $\endgroup$
    – user237522
    Jun 11, 2015 at 16:43
  • 1
    $\begingroup$ @user237522 "Now, A is a regular UFD, w is not in the field of fractions of A, w is integral over A, so A[w] is a finitely generated A-module. I wish to show that A[w] is integrally closed." As far as I can see you wanted to generalize this to A a regular UFD. I suggest you to ask the question for A = K[X,Y] as a new question, as now we are far from the original one. (I don't understand why you ask a general question if have in mind a clear one.) $\endgroup$
    – user26857
    Jun 11, 2015 at 18:49
  • 1
    $\begingroup$ I admit it was not clear if my question was about a regular UFD $A$, or $A=K[x,y]$. The reason for me asking a more general question than the specific example I have in mind, is that only recently I have started to be interested in regularity, separability, normality etc., so I thought that maybe I am missing some known facts that can be applied to my case. After asking some relevant questions and getting counterexamples, it seems that my specific case needs special attention. $\endgroup$
    – user237522
    Jun 11, 2015 at 20:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .