# Domain of definition of a rational automorphism

I am a bit confused about the following problem from Liu's book on Algebraic geometry. Let $X= \mathbb{P}^1_\mathbb{Z}$ and let $f: X_\mathbb{Q} \rightarrow X_\mathbb{Q}$ be an automorphism corresponding to a matrix of $PGL_2 (\mathbb{Q})$. Determine the domain of definition of the rational map $X \rightarrow X$ induced by f.

My question is: How does f induce a rational map? I thought that, by definition , a rational map is defined on an open set. Should I just "clear the denominators" ? Second - how do I determine the domain of definition? My intuition says defined everywhere, but I am probably wrong.

## 1 Answer

If $X$ and $Y$ are irreducible, then rational maps from $X$ to $Y$ correspond to injections of fields $K(Y) \hookrightarrow K(X)$. Now $K(X) = K(X_{\mathbb Q}) = \mathbb Q(t)$, and $f$ induces an automorphism of $\mathbb Q(t)$ by the usual linear fractional transformation. This will extend to a (regular, not just rational) automorphism of $\mathbb P^1_{\mathbb Q}$ (i.e. $X_{\mathbb Q}$), but it may not extend regularly over all of $X$, and the problem is to determine whether or not it does.

Also, the process of converting a map $K(Y) \hookrightarrow K(X)$ to a rational map $X \to Y$ is essentially one of clearing denominators, as you say. But computing the maximal domain of definition of the rational map can be more subtle, since clearing denominators in different ways may lead to extensions over different open pieces of $X$. (E.g. for rational maps between smooth projective varieties, the complement of the domain of definition is always codimension $2$; and although you are not quite in this context, your scheme $X$ is regular of dimension two, and it is fairly analogous; so one thing you'll need to work out is whether in your context the complement of the maximal domain of definition of your rational map has codimension one or two.)

• Thank you for your answer. If I am thinking correctly - if the map is nor defined somewhere, it must be at a closed point. So the complement is of dimension 2 - I fail to see any further though. – Heidar Svan Jun 9 '13 at 14:26
• @HeidarSvan: Dear Heidar, Why don't you try some examples? Regards, – Matt E Jun 9 '13 at 15:06
• Dear Matt,I tried to play around with some matrices, but I don't see how I can "check" whether a point is in the domain of definition . Sorry for being dense! – Heidar Svan Jun 10 '13 at 12:25
• @HeidarSvan: Dear Heidar, Let's take the matrix $(2 0, 0 1),$ so in terms of the usual coordinate on $\mathbb A^1$, this gives the transformation $x \mapsto 2 x$. Can you see that this formula might become slightly problematic in characteristic $2$? Regards, – Matt E Jun 10 '13 at 18:18
• Right - since it won't be an automorphism there. Am I correct in that the domain is defined everywhere except where 2 isn't invertible? – Heidar Svan Jun 16 '13 at 12:44