When two k-varieties with the same underlying topological spaces isomorphic? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T15:08:41Z http://mathoverflow.net/feeds/question/12767 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12767/when-two-k-varieties-with-the-same-underlying-topological-spaces-isomorphic When two k-varieties with the same underlying topological spaces isomorphic? Jinhyun Park 2010-01-23T20:18:44Z 2010-01-24T15:08:16Z <p>I have a little problem. I'm probably being just so careless..... Here k-varieties are all integral separated k-schemes of finite type over k, where k is a field.</p> <p>Suppose $X, Y$ are $k$-varieties, and let $f :X \to Y$ be a morphism of $k$-varieties that is one to one and onto. Then, when can we say this $f$ is an isomorphism of $k$-varieties?</p> <p>If this is too vague, let me add that the case I would like to see is when each fibre of $f$ (which is a singleton) is reduced. Under this assumption, would this give an isomorphism?</p> http://mathoverflow.net/questions/12767/when-two-k-varieties-with-the-same-underlying-topological-spaces-isomorphic/12773#12773 Answer by Ben Webster for When two k-varieties with the same underlying topological spaces isomorphic? Ben Webster 2010-01-23T21:15:08Z 2010-01-23T21:15:08Z <p>No, algebraic structures are more subtle than that. For example, consider the map of rings $k[x^2,x^3]\to k[x]$. This is obviously not an isomorphism of rings, but it is one-to-one and onto on prime ideals (since there are no elements of any field that have the same square <em>and</em> same cube but are different). This is also a homeomorphism in the Zariski topology, since open subsets of the Zariski toplogy are those with finite complement, and this is obviously preserved by any bijection.</p> http://mathoverflow.net/questions/12767/when-two-k-varieties-with-the-same-underlying-topological-spaces-isomorphic/12776#12776 Answer by Pete L. Clark for When two k-varieties with the same underlying topological spaces isomorphic? Pete L. Clark 2010-01-23T21:34:02Z 2010-01-23T22:39:19Z <p>[This is a situation where things have been rewritten several times in response to an ongoing discussion. Let me try to reconstruct some of the temporal sequence here.]</p> <p><b>ROUND ONE</b>:</p> <p>Ben's answer gives one way that your statement can fail: $Y$ can be singular and $f$ can be a birational morphism which does not induce an isomorphism of local rings at at least one of the singular points. </p> <p>Here is something else that can go wrong: in positive characteristic, $f$ can be purely inseparable, e.g. the $p$-power Frobenius map. </p> <p><b>ROUND TWO</b></p> <p>I edited my response to point out that Ben's example has a nonreduced fiber over the singular point. I also said "I think" that mine does not, but this was pointed out by Kevin Buzzard to be false. [Or rather, the statement is false. I truly did think it was true for a little while.] </p> <p>I also suggested that the following modification might be true: </p> <blockquote> <p>Suppose $X$ and $Y$ are geometrically irreducible and $Y$ is <b>nonsingular</b> (together with all of the questioner's hypotheses, especially reducedness of the fibers!). Then if $f:X \rightarrow Y$ is a bijective morphism with reduced fibers, it is an isomorphism. </p> </blockquote> <p><b>ROUND THREE</b></p> <p>I typed up a counterexample over an imperfect ground field when the varieties are not geometrically integral (g.i. = the base change to the algebraic closure is reduced and irreducible: the reduced business has to be taken more seriously when the ground field is imperfect, since taking an inseparable field extension can introduce nilpotent elements). But Kevin Buzzard posted a simpler counterexample, so I deleted my answer.</p> <p><b>ROUND FOUR</b></p> <p>Kevin's answer also includes a beautifully simple example to show that the question is false even over $\mathbb{C}$ without some nonsingularity hypotheses: use nodes instead of cusps and remove one of the preimages of the nodal point.</p> <p>I still wonder if my attempted reparation of the statement above is correct.</p> http://mathoverflow.net/questions/12767/when-two-k-varieties-with-the-same-underlying-topological-spaces-isomorphic/12779#12779 Answer by H. Hasson for When two k-varieties with the same underlying topological spaces isomorphic? H. Hasson 2010-01-23T21:48:33Z 2010-01-23T21:48:33Z <p>Here is a possible statement: <p> If f:X->Y and g:X->Y are two finite morphisms of schemes that agree topologically, and they give the same homomorphisms on all the residue fields (if x goes to y, f and g should give the same homomorphisms from the residue field of y to the residue field of x), and X is reduced (which it is in your case); then f=g.</p> <p>Proof: It suffices to prove this affinely. Say we have a two maps f,g:R->S of rings. If f<sup>-1</sup>(P)=Q then for all a&#8712;R: a/1+QR<sub>Q</sub> goes to g(a)/1+PS<sub>P</sub>. So: (f(a)-g(a))/1=p/1 for some p&#8712;P. So &#8707;d&#8713;P such that (f(a)-g(a))d=pd, which is in P. So f(a)-g(a)&#8712;P. Do this for every prime P of S. Then f(a)-g(a) is in &#8745;P, for P prime in S, which is 0 (by the reduced assumption). So f(a)=g(a). So f=g.</p> <p>Hope this helps.</p> http://mathoverflow.net/questions/12767/when-two-k-varieties-with-the-same-underlying-topological-spaces-isomorphic/12781#12781 Answer by Kevin Buzzard for When two k-varieties with the same underlying topological spaces isomorphic? Kevin Buzzard 2010-01-23T22:02:06Z 2010-01-23T22:10:50Z <p>Here's another type of counterexample not (as I write) ruled out by the hypotheses of the question: consider an inclusion of fields $K\to L$, with $K$ and $L$ finite extensions of $k$. Now take the spec. Integral schemes, reduced fibres, bijective on points.</p> <p>EDIT: aah, but even $k$ alg closed of char zero doesn't save you! Consider a nodal curve $Y$ and its normalisation $X'$. Now $X'\to Y$ isn't bijective, it's 2-1 at the singularity. So let $X$ denote $X'$ minus one of the points mapping to the singularity. I think $X\to Y$ is a counterexample.</p> http://mathoverflow.net/questions/12767/when-two-k-varieties-with-the-same-underlying-topological-spaces-isomorphic/12788#12788 Answer by Qing Liu for When two k-varieties with the same underlying topological spaces isomorphic? Qing Liu 2010-01-23T22:36:43Z 2010-01-23T22:36:43Z <p>With the examples of Kevin, we should suppose $k$ algebraically closed and $Y$ normal. Then the answer is yes:</p> <ol> <li><p>The condition that the fibers of $f$ are reduced (over closed points) means $f$ is unramified (here we need $k$ be perfect) at closed points. The condition is equivalent to $\Omega_{X/Y}$ vanishes at closed points. So it vanishes on $X$, in particular at the generic point of $X$. Therefore the extension of fields of rational funcitons $k(X)/k(Y)$ is finite separable, of degree $d\ge 1$. </p></li> <li><p>It is easy to see that then over a general closed point of $Y$ there are $d$ distinct points ($k$ is algebraically closed), so $d=1$ and $f$ is birational and quasi-finite (one-one). </p></li> <li><p>As $Y$ is normal and $f$ is separated (because $X$ is separated by defintion) birational and quasi-finite, it is an open immersion (this is a form of Zariski Main Theorem). Thus $f$ is an isomorphism as it is onto. </p></li> </ol> <p>While I am typing, I see Pete's new answer (Hi Pete!). Probably $k$ algebraically close could be replaced by his condition, but I don't see why the one-one condition could stay after field extension. </p> <p>Liu </p> http://mathoverflow.net/questions/12767/when-two-k-varieties-with-the-same-underlying-topological-spaces-isomorphic/12795#12795 Answer by VA for When two k-varieties with the same underlying topological spaces isomorphic? VA 2010-01-23T23:37:34Z 2010-01-24T15:08:16Z <p>The condition you are looking for is <strong>seminormality</strong>. A variety (or a reduced scheme) $Y$ is <em>seminormal</em> if any proper bijective morphism $f:X\to Y$, with $X$ reduced, <em>inducing isomorphisms on residue fields</em> $k(y)=k(x)$ for points $x\in X$, $y=f(x)\in Y$, is an isomorphism. A basic fact is that any variety has a unique seminormalization. </p> <p>A related notion which differs only in positive characteristic is <strong>weak normality</strong> for which $k(y)\to k(x)$ is required to be purely inseparable and an isomorphism for each <em>generic point</em> $x\in X$.</p> <p>One basic reference for this notion is the appendix to Chapter 1 of Koll'ar's "Rational curves on algebraic varieties", where you will find many standard facts and examples such as: normal implies seminormal; in dim 1 seminormal means analytically isomorphic to the $n$ axes in $A^n$; irreducible components of seminormal schemes need not be normal, etc. You will also find references to many papers where this notion was comprehensively investigated.</p> <p>For clarity, let me add the standard fact: $f$ is proper and bijective $\iff$ it is <em>finite</em> and bijective (as opposed to quasifinite = finite fibers, which of course follows from bijective).</p>