Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In differential geometry, we know that given a smooth map between smooth manifolds $\phi:X\to Y$, such that $X$ is connected and $d\phi|_x\equiv0$ for all $x\in X$, then $\phi$ is constant (this follows directly from the fact that a real function on a connected domain whose derivative is identically zero is constant).

Now, we can ask whether the same is true about algebraic varieties. i.e where $X,Y$ algebraic varieties over some algebraically closed field $k$ and $\phi$ a morphism of algebraic varieties (with the differential interpreted suitably). Well, the general answer is NO with the canonical counterexample being the Frobenius morphism $x\mapsto x^p$ for $\mathbb{A}^1\to \mathbb{A}^1$ when $\mbox{char}(k)=p$ (it's not constant, but it has zero differential).

It seems though that in zero characteristic this problem can't happen, but I would like to see a proof of that (elementary as possible, even computational) or a counterexample.

My best try: Using standard arguments this can be reduced to the case where $X$ is irreducible affine and $Y=\mathbb{A}^1$. Embedding $X$ in $\mathbb{A}^n$ and translating to algebra we get the following formulation:

Let $R=k[x_1,...,x_n]$ , $f\in R$ and $P=(g_1,...,g_r)$ a prime ideal in $R$. Suppose that for all $a\in k^n$ such that $g_1(a)=...=g_r(a)=0$ we have $df|_a\in \mbox{Span}(dg_1|_a,...,dg_r|_a)$, then there is $c\in k$ such that $f-c\in P$.

It is known (though still not trivial) that if we had $df|_a\equiv0$ for all $a$ in the mutual zero set of $g_1,...,g_r$, which is equivalent to saying that all the partial derivatives of $f$ are in $P$, then we would have the desired conclusion. Perhaps there is a way to reduce the problem to this case.

Remark: This is a fairly natural question in my opinion and I was surprised that I couldn't find any reference to it in my algebraic geometry books or just by googling it.

share|cite|improve this question
up vote 4 down vote accepted

I don't know what you call by "algebraic varieties". Let's suppose they are integral for simplicity. The vanishing remains true if we replace $Y$ with the Zariski closure of $\phi(X)$, so we can suppose $\phi : X\to Y$ is dominant.

Suppose the generic fiber of $\phi$ is geometrically reduced (this is automatic when the base field has characteristic $0$). Equivalently, the generic fiber contains a dense open subset smooth over the function field of $Y$. Then you hypothesis implies that $\phi$ is constant, equivalently, that $\dim Y=0$.

Proof: Consider the canonical exact sequence $$ \Omega_{Y/k}^1\otimes_{O_Y} O_X\to \Omega_{X/k}^1\to \Omega_{X/Y}^1 \to 0.$$ The map at left is just $d\phi$. So the vanishing condition means that for all $x\in X$, $$ \Omega_{X/k}^1 \otimes k(x)\to \Omega_{X/Y}^1 \otimes k(x)$$ is an isomorphism. In particular, both sides have the same vector dimension on $k(x)$. The geometric reducedness implies that some open dense subset $U$ of $X$ is smooth over $Y$ (the smooth locus of $X\to Y$ is open, and contains a non-empty dense open subset of the generic fiber). Now for all $x\in U$, $$\dim\Omega_{X/k}^1 \otimes k(x)=\dim U=\dim X,$$ and $$\dim\Omega_{X/Y}^1 \otimes k(x)= \dim X -\dim Y.$$ So $\dim Y=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.