Let $$R=\frac{K[[x,y,z]]}{\left<xz,yz\right>}\text{ and } M=\frac{R}{\left<z+\left<xz,yz\right>\right>}.$$ Compute the projective dimension of $M$ as an $R$-module.

My attempt to arrive at a solution:

I haven't completely solved the problem and I'm not really sure, but I think Auslander-Buchsbaum formula can be applied to this problem to compute projective dimension:

$R$ is isomorphic to $K[[x,y]]\oplus K[[z]]$ and $M$ as an $R$-module is isomorphic to $K[[x,y]]$.

$R$ is Noetherian, because both $K[[x,y]]$ and $K[[z]]$ are Noetherian. R is local with the maximal ideal $m=<x,y> \oplus <z>$. It is maximal because the quotient is a field and it is the unique maximal because $x,y$ and $z$ being non-units must lie in maximal ideals and I guess this will tell us something about why $m$ is maximal.

Now $\rm{depth_R(M)}=2$ because x,y is a $M$-regular sequence in $m$ $\rm{depth(R)}=3$ because z,x,y is a $R$-regular sequence in $m$

Therefore, by Auslander-Buchsbaum formula $$\rm{pd_R(M)}+\rm{depth_R(M)}=\rm{depth(R)}$$

which gives $\rm{pd_R(M)}=1$.

$\rm{pd_R(M)}<\infty$ because of Hilbert's syzygy theorem or something.

Sorry for being intuitive and non-rigorous. I shared my attempt only to show that I have given the problem some thought.

  • $\begingroup$ Having done your intuitive work, now try to be rigorous. First, are you sure $R$ is isomorphic to the direct sum you write down? Also, Hilbert's syzygy or something is a bit too much. Please read the theorem and see whether it applies. $\endgroup$ – Mohan Jun 12 '16 at 13:14
  • $\begingroup$ A direct product $A\times B$ of rings is rarely a local ring, @user246836... $\endgroup$ – Mariano Suárez-Álvarez Jun 12 '16 at 16:43
  • $\begingroup$ $R$ is not regular (not even CM), so you have no reason to think that $M$ has finite projective dimension. However, if you assume it is, then get immediately a contradiction (from A-B formula), so pd M is infinite. $\endgroup$ – user26857 Jun 12 '16 at 16:45
  • $\begingroup$ :( This was an exam problem. I naively ruled out $\rm{pd_R(M)=\infty}$. Now the exam's over, I'm going to give it some thought how to get a contradiction from A-B formula. @user26857 $\endgroup$ – user246836 Jun 12 '16 at 17:00
  • $\begingroup$ @Mohan: Well, I was trying to rewrite things down more rigorously until user246836 proved me wrong. I didn't have enough time in the exam to write things down rigorously and I didn't know the statement for Hilbert's syzygy theorem. So, I just tried my luck. I'm sorry for the wrong proof, but I think it's actually a good problem for learning to actually compute things rather than just learning only the theory. $\endgroup$ – user246836 Jun 12 '16 at 17:08

You should complete the following steps:

  • $x+z$ is regular on $R$.
  • $R/(x+z)$ has depth $0$.
  • Deduce that $R$ has depth $1$.

Since you have computed the depth of $M$ to be $2$, Auslander-Buchsbaum cannot hold, i.e. the projective dimension of $M$ cannot be finite.


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