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Steven Landsburg
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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 $(y,z)$.

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.)

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 $(y,z)$.

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Steven Landsburg
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I did this by hand and got $$ coker(f)=(R/J_{xv}\oplus R/J_{yv})/(xv,yv)$$, where $coker(f)=[Rxv\oplus Ryv]/(xv,yv)$$J_t$ is the ideal generated by all quadratic monomials except for $t$.   

(Sorry for the garbled version of this I briefly posted as a commentearlier.)

I did this by hand and got $coker(f)=[Rxv\oplus Ryv]/(xv,yv)$.  (Sorry for the garbled version of this I briefly posted as a comment.)

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.)

Source Link
Steven Landsburg
  • 23k
  • 5
  • 95
  • 153

I did this by hand and got $coker(f)=[Rxv\oplus Ryv]/(xv,yv)$. (Sorry for the garbled version of this I briefly posted as a comment.)