Isomorphism between varieties of char 0 - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T00:07:55Z http://mathoverflow.net/feeds/question/73321 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73321/isomorphism-between-varieties-of-char-0 Isomorphism between varieties of char 0 ernest 2011-08-21T11:33:37Z 2011-08-22T09:14:52Z <p>Hi,</p> <p>the following statement appeared implicitly in a text I read and maybe you could just give me a hint how to see this resp. give a reference:</p> <p>If you have two k-varieties $X$ and $Y$ (sufficiently nice) and you have a morphism</p> <p>$f:\ X \rightarrow Y$</p> <p>between them, which is surjective and injective, then it is an isomorphism if $k$ is of characteristic zero.</p> <p>Thank you!</p> http://mathoverflow.net/questions/73321/isomorphism-between-varieties-of-char-0/73324#73324 Answer by Wilberd van der Kallen for Isomorphism between varieties of char 0 Wilberd van der Kallen 2011-08-21T12:23:58Z 2011-08-21T12:23:58Z <p>They must be extremely nice, as the projection of the cusp $y^2=x^3$ onto the $x$-axis is already a counterexample. You want the varieties normal?</p> http://mathoverflow.net/questions/73321/isomorphism-between-varieties-of-char-0/73325#73325 Answer by Georges Elencwajg for Isomorphism between varieties of char 0 Georges Elencwajg 2011-08-21T12:26:21Z 2011-08-22T09:14:52Z <p>This is false. </p> <p>Consider a characteristic zero field $k$ and the cusp $C\subset \mathbb A^2_k$ with equation $y^2=x^3$ .<br> Its normalization $n: \mathbb A^1_k \to C: t\mapsto (t^2, t^3)$ is bijective but not an isomorphism.</p> <p>"Ah, but Georges", you will say, "be attentive! The OP said <em>nice</em> varieties. Yours is ugly!"</p> <p>In that case Zariski's main theorem will come to your rescue. One version says that a birational morphism $f:Y\to X$ of $k$-varieties (in any characteristic) with finite fibers and $X$ normal is an isomorphism of $Y$ onto an open subset of $X$, hence an isomorphism if $f$ is bijective.</p> <p>"Aw, come on Georges, admit that you just dug up this birational stuff to make yourself look important!"</p> <p>Well, the theorem no longer holds without some such hypothesis, even in dimension zero.<br> Just consider the bijective $\mathbb Q$-morphism $Spec(\mathbb Q(\sqrt 2)) \to Spec(\mathbb Q)$ of (singleton!) smooth schemes, which is not an isomorphism because it is not birational.(Of course you can inflate this to counterexamples in all dimensions)</p> <p>Some other counter-examples of bijective morphisms which are not isomorphisms, even over $\mathbb C$, are $Spec \mathbb C[\epsilon] \to Spec(\mathbb C)$ and $\mathbb G_m \bigsqcup Spec(\mathbb C)\to \mathbb A^1_\mathbb C$ (the evident morphism from the disjoint sum of a punctured affine line and a point onto the affine line).However the sources of those morphisms are respectively non reduced and reducible. </p> <p><strong>Edit</strong>: Our friend Akhil gives a great argument (see his answer) showing that in ernest's case birationality is automatic. So, to sum up, we have the precise statement answering ernest's question:<br> <strong>Proposition</strong> <em>Let $k$ be an an algebraically closed field of characteristic zero and $f:X\to Y$ a morphism between integral $k$-schemes of finite type over $k$. Then if $f$ is bijective and $Y$ normal the morphism $f$ is an isomorphism.</em> </p> <p><strong>The case of characteristic $p$</strong> As Akhil remarks, the Proposition is false in characteristic $p$. Consider an algebraically closed field $k$ of characteristic $p$ and the Frobenius morphism $f:\mathbb A^1_k \to \mathbb A^1_k:x\mapsto x^p$ with associated ring morphism $\phi:k[T] \to k[T]:P(T)\mapsto P(T^p)$ The morphism $f$ is bijective, but all fibers at closed points are non reduced of degree $p$ over $k$ .Indeed, let me denote for clarity by $A$ the $k$-algebra $\phi:k[T] \to A=k[T]$ above. Then the fibre of $f$ at the closed point $(T-a)$ in $\mathbb A^1_k$ is the affine $k$-scheme with algebra $A\otimes _{k[T]} \frac{k[T]}{(T-a)}=\frac{k[T]}{(T^p-a)}= \frac{k[T]}{(T-\sqrt[p] a)^p}$, so that the fiber is a single point but with non-reduced structure.<br> This is an example where Grothendieck's introduction of non-reduced schemes helps dissipate a mystery: how can a <em>bijective</em> morphism have degree $p\gt 1$ ? </p> http://mathoverflow.net/questions/73321/isomorphism-between-varieties-of-char-0/73348#73348 Answer by Akhil Mathew for Isomorphism between varieties of char 0 Akhil Mathew 2011-08-21T18:34:32Z 2011-08-21T18:34:32Z <p>According to Georges Elencwajg's request, here is an expanded version of my comment:</p> <p>Given a dominant morphism $f: X \to Y$ of varieties over the algebraically closed field $k$ whose <em>degree</em> is $d$ (that is, $[k(X): k(Y)] = d$) and which is <em>separable</em> (i.e., the same field extension is separable), then the degree of the generic fiber $f^{-1}(y)$ consists of $d$ points. </p> <p>(I think of this is the analog of the fact in algebraic number theory that, given a finite extension of Dedekind domains with one the integral closure of the other in a finite sep. extension, almost all primes are unramified. Of course, not all primes split into the full number of factors possible -- the degree of the extension of quotient fields --- but that's because the extension on <em>residue</em> fields might be nontrivial. That doesn't happen in the case of varieties over an alg. closed field: the residue fields on closed points are all the same.)</p> <p>Proof: $f$ is generically finite, so shrinking $X, Y$, we can assume $f$ finite. Shrinking $Y$, we can also assume $Y$ normal (e.g. smooth). Shrinking further, we can assume $k[X]$ is generated by one element over $k[Y]$, so $k[X] = (k[Y])/(P)$ for some polynomial $P$ of degree $d$. Then at all maximal ideals $\mathfrak{M}$ of $k[Y]$ (i.e. closed points of the scheme $Y$ -- or points of the variety $Y$), such that the discriminant of $P$ does not lie in said ideal, the fiber consists of $d$ points: this follows because the fiber is $\mathrm{Spec} k[Y]/(\mathfrak{M}, P)$. </p> <p>So this implies that a morphism of varieties in characteristic zero which is injective is birational. In particular, the argument given by Georges Elencwajg shows that a bijective morphism of varieties over an alg. closed field of characteristic zero, with the target normal, is an isomorphism.</p> <p>Slightly fancier argument: an injective morphism of varieties over an algebraically closed field is <em>radicial</em>: consequently, if it is dominant, the map from the generic point of $X$ to the generic point of $Y$ is radicial. But radicial extensions are trivial in characteristic zero.</p>