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Consider the regular map $f: \mathbb{A}^2 \to \mathbb{A}^2$ defined by $f(x,y)= (x,xy)$. Find the image $f(\mathbb{A}^2)$. Is it open in $\mathbb{A}^2$? Is it dense? Is it closed?

I really don’t know how to tackle this problem and I think I’m not even understanding what it’s asking: isn’t the image already described by $f(\mathbb{A}^2) = \left\{ (x,xy) : x,y \in \mathbb{A}^1 \right\}$? Any help would be very appreciated, thank you

cip
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Notice that if $x=0$, then $(x,xy) = (0,0)$. So, the only point of the image lying on the $y$-axis is the origin. However, if $x \neq 0$, we can choose $y$ to make $xy$ take any value in $\mathbb{A}^1$. Therefore, the image can be described as $$ f(\mathbb{A}^2) = \mathbb{A}^2 - (V(x) - \{(0,0)\}). $$ Since $V(x) - \{(0,0)\}$ is neither open nor closed, $f(\mathbb{A}^2)$ is neither. However, it is dense.

Notice that this image can be expressed as a union of $\mathbb{A}^2 - V(x)$ and $\{(0,0)\}$. So, it is the union of an open set and a closed set, and is therefore constructible (i.e. it is a finite union of locally closed sets, those which can be written as the intersection of a closed set and an open set). That images of maps of algebraic varieties have this property is a theorem of Chevellay.

MarkM
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  • Very clear, thank you. One last question: why is it dense? – cip Aug 04 '20 at 16:12
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    @cip It contains the non-empty open set $\mathbb{A}^2 - V(x)$, and all non-empty open subsets of $\mathbb{A}^2$ are dense ($\mathbb{A}^2$ is irreducible). – MarkM Aug 04 '20 at 16:36
  • Ok, I see. It suffices to have just one dense subset of $f(\mathbb{A}^2)$ to conclude that $f(\mathbb{A}^2)$ is dense, right? I think that is true by definition of density? – cip Aug 05 '20 at 18:57
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    @cip Yes. In general, if $S \subset T$ is an inclusion of subsets of any space, then there is an inclusion $\overline{S} \subset \overline{T}$ of closures. – MarkM Aug 06 '20 at 15:48