# Non-isomorphic smooth affine varieties dominating each other

Can non-isomorphic smooth affine varieties dominate each other?

In the projective case one can take isogenous abelian varieties.

• Projective varieties? – Ben McKay Apr 29 at 13:43
• For quasi-affine varieties this would be easy: Take $\mathbb A^2$ minus a point and $\mathbb A^2$. The dominant map from $\mathbb A^2$ minus a point to $\mathbb A^2$ is the inclusion. To go the other way, map $(x,y)$ to $(x,xy)$, and take the missing point to be $(0,1)$. – Will Sawin Apr 29 at 14:29

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.