Example of non-surjective dominant morphism of affine varieties

Question

I would like to see an example of a morphism $f:X\rightarrow Y$ of affine varieties for which the corresponding morphism of $k$-algebras $f^{*}:\mathcal{O}_{Y}(Y)\rightarrow \mathcal{O}_{X}(X)$ is injective, but for which $f(X)\neq Y$. I know that $f(X)$ is dense in $Y$, so such an example would need that $f(X)$ is not closed in $Y$.

Answer 1

E.g. $Y = \mathbb{A}^1$ and $X = Y \backslash \{0\}$.

$f:X \rightarrow Y$ is the inclusion.

$f^*:\mathcal{O}_Y(Y)\rightarrow\mathcal{O}_X(X)$ is the injective map from $k[t]$ to $k[t, t^{-1}]$.

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Depending on your definition of affine variety, you might want to say $X = \{(x_1, x_2) \in \mathbb{A}^2:x_1x_2 = 1\}$.

The map $f$ then becomes $f:X\rightarrow Y, (x_1, x_2)\mapsto x_1$.

Answer 2

Consider $X = Y = \mathbb{A}_{\mathbb{R}}^1$ the affine line and the morphism

$$ f: \mathbb{A}_{\mathbb{R}}^1 \rightarrow \mathbb{A}_{\mathbb{R}}^1, ~~ \xi \mapsto \xi^2$$

Then the image $f(\mathbb{A}_{\mathbb{R}}^1) = \{x \in \mathbb{R} \vert x \geq 0\}$ of this map is not Zariski closed, since the only Zariski closed subsets of $\mathbb{A}_{\mathbb{R}}^1$ are the finite sets. However, its closure is all of $\mathbb{A}_{\mathbb{R}}^1$ i.e. $f$ is dominant.

The induced morphism on the coordinate rings is

$$ \varphi: \mathbb{R}[x] \rightarrow \mathbb{R}[x], ~~ x \mapsto x^2$$ This map is injective.

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Also notice that $f$ is not injective, hence $\varphi$ is not surjective.