# Morphism of algebraic varieties with vanishing differential

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.

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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 $\dim Y=0$.