Can non-isomorphic smooth affine varieties dominate each other?
In the projective case one can take isogenous abelian varieties.
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Sign up to join this communityCan non-isomorphic smooth affine varieties dominate each other?
In the projective case one can take isogenous abelian varieties.
Let $X = \mathbb A^3$ and let $Y = SL_2$. The map $X \to Y$ sending $(a,b,c)$ to $$\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ b & 1 \end{pmatrix} \begin{pmatrix} 1 & c \\ 0 & 1 \end{pmatrix}= \begin{pmatrix} 1+ba & a+c + bac \\ b & 1 +bc\end{pmatrix} $$ is dominant because its image contains all matrices with lower-left corner nonzero.
The map $Y \to X$ sending $\begin{pmatrix} x & y \\ z & w \end{pmatrix}$ to $(x,y,z)$ is dominant because its image contains all triples with $x \neq 0$.
But $X$ and $Y$ are not isomorphic, for example because their complex points are nonisomorphic manifolds.
To complement the nice answer of Will Sawin, I give you an example in dimension $2$. Let $X=\mathbb{A}^2$ and let $Y=\mathbb{P}^1\times \mathbb{P}^1\setminus \Delta$, where $\Delta$ is the diagonal.
If you take $\ell\subset \mathbb{P}^1\times \mathbb{P}^1$ to be a fibre of one projection, then $Y\setminus \ell$ is isomorphic to $\mathbb{A}^2$, so you have an open embedding $X\hookrightarrow Y$, which is then a dominant morphism.
Conversely, you choose a point $p\in \Delta$ and consider the pencil of curves $C$ of $\mathbb{P}^1\times \mathbb{P}^1$ of bidegree $(1,1)$ such that $C\cap \Delta=\{p\}$, i.e, being tangent to $\Delta$. This pencil gives a rational map $\mathbb{P}^1\times \mathbb{P}^1\dashrightarrow \mathbb{P}^1$ and restricts to a morphism $Y\to \mathbb{A}^1$. Choosing two different points of $\Delta$, you obtain two morphisms $Y\to \mathbb{A}^1$ and thus a morphism $Y\to \mathbb{A}^2$. It is dominant because the general fibres of the two intersect into finitely many points. You can also see this by embedding $Y$ into $\mathbb{A}^3$, using the Segre embedding of $\mathbb{P}^1\times \mathbb{P}^1\hookrightarrow \mathbb{P}^3$, and project onto two factors.
The two surfaces $X$ and $Y$ are not isomorphic, because the Picard group of $X$ is trivial but the one of $Y$ is not.