Is $f$ reduced if and only if the derivations $\gcd(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})=1$ under some conditions?

I have encountered the following problem. I have no ideas to prove it or disprove it.

Suppose that $f\in \mathbb{C}[x,y]$, $f(0,0)=0$, $\frac{\partial f}{\partial x}(0,0)=0, \frac{\partial f}{\partial y}(0,0)=0$. Suppose $f$ is square-free—for simplicity we may first assume that $f$ is irreducible. Then $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ have no non-trivial common factor, in other words, the greatest common factor of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ is 1.

Thanks.

5
2022-07-25 20:43:48
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Answers: 1

This is not true. Consider for instance $$f(x,y)=(x^2+1)^2(y^2+1)^3-1.$$ But you can correct the statement by assuming $f(x,y)+c$ is square-free for all $c\in \mathbb C$ (no condition at $(0,0)$).

Proof: Suppose $\gcd(f_x, f_y)\ne 1$. Let $p(x,y)$ be an irreducible factor and let $C=Z(p(x,y))$ be the irreducible algebraic curve in $\mathbb C^2$ defined by $p(x,y)$. It is contained in the set of critical points of $f(x,y)$. By Sard's theorem, $f(C)$ has measure $0$. By Chevalley's theorem (or the openess of non-constant homolorphic functions defined on the smooth part of $C$), $f(C)$ either contains an open subset or is finite. So it is finite (and connected because $C$ is connected). Therefore $f(C)=c$ for some $c\in\mathbb C$.

This implies that $p(x,y) \mid f(x,y)-c$. Replacing $f$ by $f-c$ (which doesn't change the partial derivatives), we get $p(x,y)\mid f(x,y)$. So $p$ is an irreducible factor of $f$. Write $f=pg$ with $g$ prime to $p$ ($f$ is square-free after any translation). As $f_x=p_xg+pg_x$ and $f_y=p_yg+pg_y$, we get $p \mid \gcd\{ p_x, p_y\}$. But $p(x,y)$ is irreducible, so $p_x=p_y=0$ and $p(x,y)$ is constant, impossible.

Remark: this remains true over any field of characteristic $0$. But the example $f(x,y)=x^{p+1}+y^p$ shows that it fails in characteristic $p>0$.

Edit With some algebraic geometry background, the statement is clear: consider $f$ as a morphism from the affine plane to the affine line. Then the set of common zeros of the its partial derivatives is the set $S$ of non-smooth points of $f$. As we are in characteristic $0$, the generic fiber of $f$ is smooth, so $S$ is contained in the union of finitely many fibers of $f$. So $S$ is finite if and only if every fiber has only finitely many singular points. Equivalently, every fiber is reduced. But a fiber $f^{-1}(c)$ is just the scheme defined by $f(x,y)-c=0$. This fiber is reduced if and only if $f(x,y)-c$ is square-free. Finally, $S$ is finite iff $\gcd\{ f_x, f_y\}=1$.

6
2022-07-25 21:36:59
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