When two k-varieties with the same underlying topological spaces isomorphic?

Question

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.

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?

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?

Answer 1

The condition you are looking for is seminormality. A variety (or a reduced scheme) $Y$ is seminormal if any proper bijective morphism $f:X\to Y$, with $X$ reduced, inducing isomorphisms on residue fields $k(y)=k(x)$ for points $x\in X$, $y=f(x)\in Y$, is an isomorphism. A basic fact is that any variety has a unique seminormalization.

A related notion which differs only in positive characteristic is weak normality for which $k(y)\to k(x)$ is required to be purely inseparable and an isomorphism for each generic point $x\in X$.

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.

For clarity, let me add the standard fact: $f$ is proper and bijective $\iff$ it is finite and bijective (as opposed to quasifinite = finite fibers, which of course follows from bijective).

Answer 2

With the examples of Kevin, we should suppose $k$ algebraically closed and $Y$ normal. Then the answer is yes:

1. 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 $d\ge 1$.

2. 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).

3. 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.

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.

Liu

Answer 3

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.

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.

Answer 4

[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.]

ROUND ONE:

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.

Here is something else that can go wrong: in positive characteristic, $f$ can be purely inseparable, e.g. the $p$-power Frobenius map.

ROUND TWO

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.]

I also suggested that the following modification might be true:

Suppose $X$ and $Y$ are geometrically irreducible and $Y$ is nonsingular (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.

ROUND THREE

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.

ROUND FOUR

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.

I still wonder if my attempted reparation of the statement above is correct.

Answer 5

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 and 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.

Answer 6

Here is a possible statement:

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.

Proof: It suffices to prove this affinely. Say we have a two maps f,g:R->S of rings. If f^-1(P)=Q then for all a∈R: a/1+QR_Q goes to g(a)/1+PS_P. So: (f(a)-g(a))/1=p/1 for some p∈P. So ∃d∉P such that (f(a)-g(a))d=pd, which is in P. So f(a)-g(a)∈P. Do this for every prime P of S. Then f(a)-g(a) is in ∩P, for P prime in S, which is 0 (by the reduced assumption). So f(a)=g(a). So f=g.

Hope this helps.