Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question
    
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 $1,4,13,40,121,364,...$. –  Graham Leuschke Jun 25 '13 at 14:39
    
Thank you, Graham. Your idea is good. But I know little of Macaulay2. –  TmobiusX Jun 26 '13 at 7:06
    
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. –  Graham Leuschke Jun 26 '13 at 11:18
    
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? –  Steven Landsburg Jun 26 '13 at 13:16
    
As sure as one can ever be with computer algebra -- I checked it on Singular and got the same answer. –  Graham Leuschke Jun 26 '13 at 15:26

2 Answers 2

up vote 1 down vote accepted

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?

share|improve this answer
    
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 $coker(M)$? –  Steven Landsburg Jun 27 '13 at 0:49
    
Doh! Turnabout is fair play indeed. Quite right. –  Graham Leuschke Jun 27 '13 at 1:23
    
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? –  Steven Landsburg Jun 27 '13 at 3:12
    
Is the projection onto $k$ an $R$-map? I don't see why. –  Graham Leuschke Jun 27 '13 at 10:20
    
@Graham Leuschke: Your explanation implies that my thought is false. I understand, thank you and Steven. –  TmobiusX Jun 27 '13 at 12:14

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 $(y,z)$.

share|improve this answer
    
Thank you, Steven. In your decomposition, where is the direct summand $k$? –  TmobiusX Jun 26 '13 at 7:04
    
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? –  Steven Landsburg Jun 26 '13 at 13:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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