# kernel and cokernel of a morphism

Let $\phi: A=\mathbb{C}[x,y]/(y^2-x^3) \to \mathbb{C}[t]=B$ be the ring morphism defined by $x\mapsto t^2, y \mapsto t^3.$ Let $f:Y=\text{Spec} B \to X=\text{Spec} A$ be the associated morphism of affine schemes.

Since $\phi$ is injective, then $f^{\sharp} : \mathcal{O}_X \to f_{\star}\mathcal{O}_Y$ is injective too.

1.What is the cokernel of $f^{\sharp}?$

I know that it will be isomorphic to the quotient sheaf $f_{\star}\mathcal{O}_Y/\mathcal{O}_X$ but I'm unable to find it explicitly.

Here I computed the cotangent sheaf $\Omega_X$ of $X.$

2.How can I find the kernel and the cokernel of $f^{\star} \Omega_X \to \Omega_Y?$

What I've tried out:

The sheaf morphism corresponds to a $B$-module morphism $\Omega_A \otimes_A B \to \Omega_B$ or $Adx \oplus Ady/(2ydy-3x^2dx) \otimes B \to Bdt$ which is in fact, the map $(t)dt \oplus (t^2)dt \to Bdt$ where $(tg(t)dt,t^2h(t)dt)\mapsto (tg(t)+t^2h(t))dt$ for $g,h \in B.$ The kernel is isomorphic to $B$ since $g(t)=-th(t)$ and the cokernel is $B/(t).$

Is this correct?

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Hopefully you know that coherent sheaves over $\mathrm{Spec} A$ are equivalent to finitely generated $A$-modules. The structure sheaf $\mathcal O_X$ corresponds to $A$ as a module over itself and $f_\ast\mathcal O_Y$ corresponds to $\mathbb C[t]$ with the $A$-module structure given by restricting through $\phi$.

So $f^\#$ corresponds to $\phi$ as a map of $A$-modules, the image of $\phi$ consists of all polynomials with no degree $1$ term so the cokernel is $(t)/(t^2)$. The $A$-module action on this is that $x$ and $y$ act as $t^2$ and $t^3$ respectively, hence $x$ and $y$ act as $0$. So the cokernel is a trivial $1$-dimensional $A$-module.

Edit: I forgot about $1$, the cokernel should be $1$-dimensional, not $2$.

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By $(t)$ you mean the $A$-module generated by $t$ right? –  Ehsan M. Kermani Feb 4 '13 at 3:54
By $(t)$ I meant the principal ideal generated by $t$ in $\mathbb C[t]$. –  Jim Feb 4 '13 at 4:38
So why the cokernel is not $\mathbb{C}[t]/(t^2)?$ –  Ehsan M. Kermani Feb 4 '13 at 5:42
Because the unit $1 \in A$ maps to the unit $1 \in \mathbb C[t]$. So when taking the cokernel you have to factor that out. –  Jim Feb 4 '13 at 5:45
Yes, but that $A$-module homomorphism is exactly the map $\phi$ and because $\phi$ also happens to be a ring homomorphism we know that it sends $1 \mapsto 1$. –  Jim Feb 4 '13 at 6:03