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I'm reading a bit about rational maps, and I'm still trying to get get my head around birational maps.

Consider the map $f\colon\mathbb{A}^2_k\to\mathbb{A}^2_k$ on the affine $2$-space over $k$ algebraically closed, defined by $f(x,y)=(x,xy)$. I believe this map is regular, hence a rational map. Apparently this is also a birational isomorphism.

I computed the inverse map to be defined by $f^{-1}(u,v)=(u,v/u)$. However, I think the domain of definition of $f^{-1}$ is $\mathbb{A}^2\setminus\{(0,y):y\in k\}$ since $u$ can't be zero. Is this the correct domain of definition, or can it be extended to all of $\mathbb{A}^2_k$ somehow? I thought the domain would have to be all of $\mathbb{A}^2_k$ for it to be a rational inverse. Also, it's probably clear, but why is $f^{-1}$ a rational map?

Thanks for your explanations.

Ankita Desari
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1 Answers1

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You're right when you claim that the map $f$ is a regular map: in fact (the things you have to check may vary a bit, pending on where you're studying), if we consider each component of the function, this is just a regular function (because it is a polynomial).

It is also true what you say, that every regular map is rational. The inverse is indeed the map you computed. The domain of definition is indeed the whole plain minus the axis $x=0$. This is the definition of a rational map: a map which is defined only on an open set and not on the whole space, so $f^{-1}$ is a rational map. Hence $f$ is a regular map, but it only admits a rational inverse. Some books refer at these maps as to "rational morphisms". More generally, since $f$ is also a rational map, we can say that it is a birational map.

A reason to convince yourself that the function $f^{-1}$ (let's call it $g$) is not regular may come from algebra: you know that a map between affine varieties corresponds to a ring homomorphism between the rings of regular functions, with reversed arrows. In this case the map $g:(\mathbb{A}^2-Z(x))\to \mathbb{A}^2$ corresponds to the morphism \begin{equation} g_*:\mathcal{O}(\mathbb{A}^2)=k[x,y]\to k[x,y]_x=\mathcal{O}(\mathbb{A}^2-Z(x)) \end{equation} that sends $x$ to $x$ and $y$ to $y/x$. To check this, recall the properties of the correspondence: we have $g_*h=h\circ g$ for every regular function $h$ on the plane. You see then that there is no hope to extend the map to the whole plane, as it would mean that the ring homomorphism lands in the copy of $k[x,y]$ that lives inside $k[x,y]_x$, which is not true.

Franco
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