# Singular points of an irreducible polynomial

Let $k$ be a field, and consider an irreducible polynomial $f \in k[x,y]$. 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, then I can prove $S(f)$ is finite. Also, I can prove that if the field has characteristic $0$, then $S(f)$ is finite.

But what if the field has characteristic $p$ and is not algebraically closed? Is it true that $S(f)$ is finite?

I asked this question to my algebraic geometry professor last semester and stumped him! Hopefully one of you can think of a counterexample or proof.

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What do you mean by irreducible and what do you mean by $S(f)$?

Does irreducible mean absolutely irreducible (ie irreducible over the algebraic closure of $k$)? Is $S(f)$ considered as a scheme or as a set of rational points? If the latter, then is $S(f) := \{ (a,b) \in k^2: f(a,b) = 0 = \frac{\partial f}{\partial x}(a,b) = \frac{\partial f}{\partial y}(a,b) \}$? Or is it the set of singular points over the algebraic closure of $k$?

If by irreducible, you mean absolutely irreducible, then as Douglas Zare suggests, you can pass to the algebraic closure and prove that $S(f)$ is finite.

If irreducible is to be read over $k$, but you are considering $S(f)$ scheme theoretically or are evaluating the points in the algebraic closure of $k$, then the assertion is false. Consider for instance $k$ of characteristic $p$ with $a \in k$ a non-$p^\mathrm{th}$ power and $f(x,y) = x^p + y^p + a$.

Finally, if by $S(f)$ you mean the $k$-rational points, then if $S(f)(k)$ were infinite, then the set of $k$-rational points on the curve defined by $f$ would be infinite and $f$ would then be absolutely irreducible so that by the first case considered, $S(f)$ would be finite.

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Do you know what happened to Douglas Zare's comment? –  Charles Chen Apr 6 '10 at 3:11

Dear Charles, here is a more general statement:

Let $R$ be a reduced affine algebra over $k$ of dimension 1. Then the set of singular points of $Spec(R)$ is finite.

Proof: Let $V$ be the set of singular point, i.e $V=\{p \| R_p \text{is singular} \}$. Then V does not contain any minimal prime of $R$, since $R$ localize at them are fields (as $R$ is reduced). So $\dim V=0$, and $V$ is closed, therefor $V=Spec(R/J)$ with $dim R/J=0$. But it is well known that an Artinian ring has only finitely many primes.

In your situation, let $R=k[x,y]/(f)$. Since $f$ is irreducible, $R$ is a domain and therefore reduced (reducedness = $f$ has no repeated factors which is weaker then irreducible). Any point that is $0$ on $f,f_x,f_y$ will give a maximal singular point in $Spec(R)$. So $S(f)$ is finite.

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I think my second paragraph needs more work. Specifically, I believe the italic claim is right, but to show that a k-point which is 0 on $f,f_x,f_y$ gives a maximal ideal which becomes singular when localizing might be more complicated than I thought. I am trying to fix it. –  Hailong Dao Apr 6 '10 at 3:02

Even though there are two complete answers, I would like to point out a proof which in my opinion is the most elementary. Let $R$ be an integral domain and $R[x]$ be the ring of polynomials in one variable over $R$. Then for every pair of polynomials $f, g \in R[x]$, one can define the resultant $R(f,g)$ of $f$ and $g$ (if necessary, see e.g. Griffiths' "Introduction to algebraic curves" for a very elementary treatment of resultants). The only properties you need is that if $f$ and $g$ are nonzero polynomials (and if at least one of them is a "real polynomial", i.e. does not belong to $R$), then $R(f,g)$ is a nonzero element of $R$ and there exist $p, q \in R[x]$ such that $R(f,g) = pf + qg$.

Now let us get back to your scenario. Let $R = k[x]$. Then $f \in R[y]$. If $f \in R$, then it is trivial to check that $S(f)$ is finite. So assume $f \not\in R$. Then $g := \partial f/\partial y \neq 0 \in R[y]$, which implies that $h := R(f, g)$ is a nonzero element of $k[x]$. Since $h$ is a linear combination of $f$ and $\partial f/\partial y$, it follows that for all $(x,y) \in S(f)$, $h(x) = 0$. But $h$ has only finitely many roots - voila!

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