Let $R=k[v,x,y,z]/I$, with $I=\langle v^2,z^2,xy,vx+xz,vy+yz,vx+y^2,vy-x^2\rangle$,and let $f:R^2 \rightarrow R^2$ denote the map given by the matrix $$M=\begin{pmatrix} v & y \\ x & z \end{pmatrix}$$ I guess that there is a module $N$ such that $\operatorname{coker}f \cong N\oplus k$, but I don't know how to prove it. Any comment is welcome. Thanks a lot!
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$\begingroup$ I don't think this is true. According to Macaulay2, the Betti numbers of the cokernel of your matrix are $2,2,4,11,32,95,...$, while those of the residue fields are $1,4,13,40,121,364,...$. $\endgroup$– Graham LeuschkeCommented Jun 25, 2013 at 14:39
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$\begingroup$ Thank you, Graham. Your idea is good. But I know little of Macaulay2. $\endgroup$– TmobiusXCommented Jun 26, 2013 at 7:06
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$\begingroup$ The point is that the Betti numbers (ranks of free modules in the resolution) of $\mathrm{coker} M$ would have to be greater than or equal to those of $k$, if $k$ were a direct summand. They aren't, so it isn't. $\endgroup$– Graham LeuschkeCommented Jun 26, 2013 at 11:18
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$\begingroup$ Graham: I haven't tried to compute the Betti numbers, but I did compute the cokernel, and it appears to have a direct summand of $k$. Are you sure of your calculation? $\endgroup$– Steven LandsburgCommented Jun 26, 2013 at 13:16
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$\begingroup$ As sure as one can ever be with computer algebra -- I checked it on Singular and got the same answer. $\endgroup$– Graham LeuschkeCommented Jun 26, 2013 at 15:26
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2 Answers
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Edited to add: Well, now I feel embarrassed to have gotten an answer accepted which is absolute garbage, so I think I should offer an actual answer in addition to the indirect proof in a comment I made above (i.e. the Betti numbers of the cokernel $A$ of your matrix are 2,2,4,11,32,95,..., while those of the residue field $k$ are 1,4,13,40,121,364,..., so $k$ can't be a direct summand of $A$).
Here's another that doesn't rely on computer algebra software. It does rely on $A$ being a graded module over the (naturally) graded ring $R$. Suppose $A \cong N \oplus k$. It's easy to check that $A$ has hilbert function $(2,6,1)$. Since $A$ is generated in a single degree, the copy of $k$ must also be generated in that degree, so $N$ has hilbert function $(1,6,1)$. In particular $N$ must be cyclic, $N \cong R/J$ for some $J$. But $R$ has hilbert function $(1,4,3)$, so can't have a quotient with hilbert function $(1,6,1)$.
Edit: The below is wrong. Sorry.
The minimal generators of the module $\mathrm {coker}\ M$ are the column vectors $(v,x)^T$ and $(y,z)^T$. They generate a two-dimensional vector space of all the minimal generators of the module. This is just $X/mX$, where $X = \mathrm{coker}\ M$ and $m=(x,y,z,v)$. If there is going to be a direct summand isomorphic to $k$, there must be a minimal generator which is annihilated by the maximal ideal. But one can write down a generic minimal generator $(av+by, ax+bz)^T$ and the 8 $k$-linear equations saying that it is annihilated by $x,y,z$ and $v$. Two of them are $v(av+by)=0$ and $z(ax+bz)=0$. The relations in the ring imply $bvy=0=avx$. Since $vy$ and $vx$ are nonzero in $R$, this means $a=0=b$, and so there is no such direct summand.
Would you tell us why you thought there should be such a direct summand?
answered Jun 26, 2013 at 18:19
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$\begingroup$ Having already embarrassed myself with silly mistakes both above and in private email, I'm posting this comment with considerable trepidation, but --- aren't $(v,x)^T$ and $(y,z)^T$ both zero in $coker(M)$? $\endgroup$ Commented Jun 27, 2013 at 0:49
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$\begingroup$ Doh! Turnabout is fair play indeed. Quite right. $\endgroup$ Commented Jun 27, 2013 at 1:23
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$\begingroup$ So let me go farther out on a limb here. I claim $Coker(M)=(R/J_{xv}\oplus R/J_{yv})/((v,x),(y,z))$, with the ideals $J_t$ as in my answer above. In particular, this implies $(xv,yv)=0$. And I claim further that the map $k\rightarrow coker(M)$ given by $1\mapsto (x,y)$ is an $R$-map. (Check that $x,y,z,v$ all kill $(x,y)$ in $coker(M)$.) This map appears to split the projection from $coker(M)$ onto $k$. Have I managed to make another mistake yet? $\endgroup$ Commented Jun 27, 2013 at 3:12
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$\begingroup$ Is the projection onto $k$ an $R$-map? I don't see why. $\endgroup$ Commented Jun 27, 2013 at 10:20
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$\begingroup$ @Graham Leuschke: Your explanation implies that my thought is false. I understand, thank you and Steven. $\endgroup$– TmobiusXCommented Jun 27, 2013 at 12:14
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I did this by hand and got $$ coker(f)=(R/J_{xv}\oplus R/J_{yv})/(xv,yv)$$, where $J_t$ is the ideal generated by all quadratic monomials except for $t$.
(Sorry for the garbled version of this I briefly posted earlier.)
Edit: Graham Leuschke has helped me realize that I failed to mod out by the images of $(1,0)$ and $(0,1)$, so one should also mod out $(v,x)$ and $(y,z)$.
answered Jun 25, 2013 at 15:20
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$\begingroup$ Thank you, Steven. In your decomposition, where is the direct summand $k$? $\endgroup$– TmobiusXCommented Jun 26, 2013 at 7:04
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$\begingroup$ TmobiusX: I'm reluctant to answer this because it's beginning to seem like maybe this is a homework problem. Can you explain why you expect such a direct summand? $\endgroup$ Commented Jun 26, 2013 at 13:09