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!


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.


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