Decomposition of a quotient module 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! 
 A: 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 $(y,z)$.   
A: 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?
