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Let $X={\rm Spec}~k[x,y,t]/<yt-x^2>$ and let $Y={\rm Spec}~ k[t]$. Let $f:X \rightarrow Y$ be the morphism determined by $k[t] \rightarrow k[x,y,t]/<yt-x^2>$.

Is f surjective> If f is surjective, why??

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Do you mean the inclusion of rings there? – Zhen Lin Feb 11 '13 at 11:01
Yes, $f$ is induced by $k[t] \hookrightarrow k[x,y,t] \rightarrow k[x,y,t]/<yt-x^2>$. – Sang Cheol Lee Feb 11 '13 at 11:55
up vote 1 down vote accepted

I'm assuming your map of rings comes from the natural inclusion: $i:k[t]\rightarrow[x,y,t]\rightarrow k[x,y,t]/<yt-x^2>=A$.

A prime of $k[T]$ is of the form $(F(t))$ where $F(t)$ is an irreducible polynomial over $k$. Show that the $I=F(t)A$ is not the whole ring $A$ (This amounts to showing that $yt-x^2$ and $F(t)$ don't generate $k[x,y,t]$). In fact it is even prime but we won't need that. We just need the fact that $I$ is contained in a prime ideal $P\in Spec(A)$. So $i^{-1}(P)$ is a prime of $k[t]$ that contains $F(t)$ and hence equal to $(F(t))$.

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$k[t] \to k[x,y,t]/(yt-x^2)$ is a split monomorphism with retraction $k[x,y,t]/(yt-x^2) \to k[t], t \mapsto t, y \mapsto 1, x \mapsto 0$. It follows that the induced morphism of schemes is a split epimorphism; in particular surjective.

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