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let $k$ be a field, and consider an irreducible polynomial $f∈k[x,y]$. Let $S(f)$ denote the singular points of $f$ (points that are simultaneously zero on $f$, the $x$-derivative of $f$, and the $y$-derivative of $f$.)

If $k$ is algebraically closed, How can I prove that $S(f)$ is finite?

Thanks, Alex

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    The locus of singular points is a closed subset of the curve and is not the whole curve, hence it is finite. – Zhen Lin Jun 23 '14 at 11:01

1 Answers1

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The singularities of the curve $V(f)$ form the closed set $S(f)=V(f,\partial f/\partial x, \partial f/\partial y)\subset V(f)$ . Closedness of $S(f)$ is obvious but finiteness is not, contrary to what you often read on this site and elsewhere.

The basic (non-trivial!) result is that two affine plane curves $V(f),V(g)\subset \mathbb A^2(k)$ intersect in finitely many points if $f,g$ have no common irreducible factor.
So in our case, since $f$ is irreducible, it suffices to prove that for example $f$ does not divide one of $\partial f/\partial x, \partial f/\partial y$, which looks pretty obvious for reasons of degree.
End of story? No, because $\partial f/\partial x$ (for example) might be zero if $char. k=p$ !
For example $f=x^py+1$ is irreducible but nevertheless satisfies $\partial f/\partial x=0$ in characteristic $p$.
However we are saved because $\partial f/\partial y=x^p\neq 0$
This will always be the case: an irreducible polynomial $f(x,y)$ with $\partial f/\partial x=\partial f/\partial y=0$ is a polynomial $f(x,y)=\phi(x^p,y^p)$ in $x^p,y^p$ and thus over an algebraically closed field is a $p$-th power $f(x,y)=\phi(x^p,y^p)=\psi(x,y)^p$ and so is not irreducible.
Contrapositively, if $f$ is irreducible then it is guaranteed to not divide at least one of $\partial f/\partial x,\partial f/\partial y$ and so, indeed, $S(f)=V(f,\partial f/\partial x, \partial f/\partial y)\subset V(f)$ is a finite (maybe empty) set.

Conclusion
Yes, an irreducible plane curve has only finitely many singularities but you have to work to prove it!

Georges Elencwajg
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    +1 for saying the finiteness is not as obvious as people seem to enjoy claiming. Also note that we don't require the field to be algebraically closed. Here is the Stacks reference. Also, for the benefit of students I'll leave this link to a proof that two affine plane curves intersect in finitely many points if $f$ and $g$ have no common factor. – Ragib Zaman Jun 23 '14 at 12:05
  • Thanks for your interest and the reference, @Ragib. – Georges Elencwajg Jun 23 '14 at 13:31
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    Why is $f$ a p-power? – Zero Feb 16 '19 at 20:22
  • @Zero If there is some power of $x$ or $y$ which is not a multiple of $p$, then the derivative will be nonzero, for example $(d/dx) (x^{p-1}) = (p - 1)x^{p - 2} \neq 0$. – Forrest Flesher Apr 14 '19 at 00:59
  • I think @Zero is referring to $f(x,y)=\psi(x,y)^p$. I also did not follow this step. Could you enlighten us @Georges? – Kadmos May 09 '24 at 15:59
  • I think I figured how it is done! We write $f(x,y)=\sum_{k,j} a_{kj}x^{pk}y^{pj}$ and $a_{k,j}$ has $p-$roots in $f$ (because $F$ is algebraically closed). Define $\psi(x,y)=\sum_{k,j} a_{k j}^{1/p}x^k y^j$. In this case $\psi(x,y)^p=f(x,y)$ (we are in a characteristic $p$ field). – Kadmos May 09 '24 at 16:11
  • He adds a condition,algebraiclly closed field. In this case every coefficient has a p-power root. – kuanglei May 06 '25 at 02:52