3
$\begingroup$

I found this tricky problem in trying to understand some properties of local rings at non-smooth points of embedded curves. But this would be a very long story. So I make it short and I try to go directly to the question.

Fix a field $k$, of characteristic $0$ and consider the ring of power series in $n+1$ variables, $B=k[[x, y_1, \ldots, y_n]]$. Let $a> 1$ be an integer and let $I$ be the ideal of $B$ given by the equations $x=y_n^{a}$ and $y_i^a= g_i(x,y_i)$ for some $g_i\in k[[x, y_i]]$ for $i=1, \ldots, n-1$. Suppose that $B/I$ is a regular (local) ring (I'm not totally sure that this is my case, but one has to start somewhere).

Let now $A=k[x]/(x^m)$ be my favorite Artin local ring. Is it true that the module $$M=\frac{k[[x, y_1,\ldots, y_n]]}{(x^m, x=y_n^{a}, y_i^a= g_i(x,y_i), i=1,\ldots, n-1)}$$ has finite (projective) homological dimension over $A[[y_1, \ldots, y_n]]$, i.e. it admits a finite projective resolution?

This question can also be a reference request: if you know where I can find something useful I would also be very happy!

$\endgroup$
1
$\begingroup$

If $B/I$ is regular local, then $I$ is a prime ideal in $B$. Notice that $x^m \not \in I$ (if $x \in I$, then the equations defining $I$ would imply $y_1, \ldots, y_n \in I$, so then $I$ would be the maximal ideal, which it is not, being $n$-generated). Thus $x^m$ is regular on $B/I$ (and on $B$), so taking a finite free $B$-resolution of $B/I$, and tensoring with $B/(x^m)$, produces a finite free $B/(x^m) = A[[y_1, \ldots, y_n]]$-resolution of your module $M = B/I \otimes_B B/(x^m)$.

By the way, the condition that $B/I$ is regular seems to be just that each $g_i$ contains $y_i$ as a linear term (up to a unit of $k$). This is certainly sufficient, and in order for the generators of $I$, along with one more element, to form a regular sop of $B$, it is also necessary.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.