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Suppose that $f:X\rightarrow Y$ is a morphism between two affine varieties over an algebraically closed field $K$.

I believe that if the corresponding morphism of $K-$algebras, $f^\ast:K[Y]\rightarrow K[X]$ is injective, it is not necessarily true that $f:X\rightarrow Y$ must be surjective but I have yet to come up with a counterexample.

Is there such a counterexample?

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    My definition of an affine variety is a variety that is isomorphic to an irreducible algebraic set.2012-10-24

2 Answers 2

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Consider the inclusion $k[x] \subset k[x, x^{-1}]$. The corresponding map on $k$-points ($k$ algebraically closed) is also an inclusion, namely $\mathbb{A}^1 \setminus \{ 0 \} \subset \mathbb{A}^1$.

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    You're absolutely right, I was momentarily confused by the fact that $\mathbb{A}^2\setminus\{(0,0)\}$ is not irreducible.2012-10-24
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Given a morphism of rings $\phi:A\to B$ and the corresponding morphism of affine schemes $\sideset {^a}{} \phi=f:Spec (B)\to Spec(A)$, we have the equivalence: $ f (Spec(B))\: \text {is dense in}\: Spec(A)\iff \text {Ker } (\phi) \subset \text {Nil} (A)$ From this it is very easy to find injective morphisms of rings $\phi:A\to B$ with associated non surjective morphisms $ f:Spec (B)\to Spec(A)$.

For example if $A$ is a domain and $0\neq a\in A$, then the inclusion morphism $\phi:A\to A_a=A[\frac {1}{a}]$ yields the inclusion $\sideset {^a}{} \phi=f:Spec (A_a)\to Spec(A)$, which is not surjective as soon as $a$ is not an invertible element of $A$. (Qiaochu's example is of that type)