# A dominate morphism is not an isomorphism

Here is a Theorem: Let X is a smooth variety of dimension 1, and let f:P$^1$$\to X be a dominate morphism. Then X is an isomorphism to P^1(although f need not be an isomorphism). I want to find an example that the above-mentioned f is a dominate morphism, but not an isomorphism. - ## 2 Answers In characteristic zero, this is a consequence of Hurwitz's theorem$$2g_ {P^1}-2=n(2g_ X-2)+deg( Ram(f))$$which in your case yields$$-2=n (2g_ X-2)+ \text {nonnegative number }$$This equality forces$g_X$=0 (else the right-hand side is nonnegative) and since the only smooth projective curve of genus zero is (up to isomorphism)$\mathbb P^1$, your first question is answered. As to your second question, just take$X=\mathbb P^1$and$f([z:w])=[z^2:w^2]$- Dear Georges: one has to treat separately the case of purely inseparable morphisms :). This is done for instance in Hartshorne when the base field is algebraically closed. – user18119 Oct 1 '12 at 22:41 Dear @QiL, ah yes , I was thinking in characteristic zero, and I have edited my answer to make that explicit. Thanks a lot for drawing my attention to that . I would be extremely happy (like everybody else, I'm sure) if you'd let us share your expertise by writing an answer surveying the characteristic$p$case, maybe even evoking the non algebraically closed situation. – Georges Elencwajg Oct 1 '12 at 23:52 As requested by Georges, here are a few words about the positive characteristic situation. Let$k$be a field of characteristic$p>0$. Any finite morphism of smooth projective curves$f: X\to Y$over$k$decomposes into$X\to Z\to Y$where$Z\to Y$is separable (meaning that the extension of functions fields$k(Z)/k(Y)$is separable, and where$X\to Z$is purely inseparable. This corresponds to the decomposition of any finite algebraic extensions into separable then purely inseparable extensions, and results from the correspondance between functions fields and projective normal curves. One can show that$Z$is smooth because$X$is smooth (as$Z$is regular of dimension$1$,$X\to Z$is flat, so$X_{\bar{k}}\to Z_{\bar{k}}$is flat, then the regularity of$X_{\bar{k}}$implies that of$Z_{\bar{k}}$). Proposition. We have$g(X)=g(Z)$as curves over$k$. Proof. The purely inseparable extension$k(X)/k(Z)$can be decomposed into successive inseparable extensions of degree$p$. So we are reduced to the case$[k(X) : k(Z)]=p$. As$X$is smooth,$k(X)$is a finite separable extension of some$k(t)$:$k(X)=k(t)[\theta]$. Consider$k(t^p)[\theta^p]$. It is contained in$k(Z)$because any$p$-th power of elements of$k(X)$belongs to$k(Z)$. It is not hard to see that$k(t^p)[\theta^p]$has index$p$in$k(X)$. Therefore$k(Z)=k(t^p)[\theta^p]$. This implies that$X\to Z$is in fact the relative Frobenius map$X\to X^{(p)}$. In other words,$Z=X\otimes_k k'$where$k'=k$is viewed as an extension of$k$by$\lambda\mapsto \lambda^p$. Since the genus in invariant by field extensions, we get$g(Z)=g(X)$. Now suppose we have a finite morphism$X=\mathbb P^1_k\to Y$. Then$g(Z)=0$, hence$Z$is a smooth conic with a rational point (because$X$has a rational point), so$Z\simeq \mathbb P^1_k$. Therefore we get a finite separable morphism$\mathbb P^1_k\to Y$. This implies$Y$is isomorphic to$\mathbb P^1_k\$ using Hurwitz as did Georges.

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Thank you so much, QiL. – Georges Elencwajg Oct 2 '12 at 19:29