Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

3 Answers 3

up vote 7 down vote accepted

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.

share|improve this answer
    
Do you know what happened to Douglas Zare's comment? –  Charles Chen Apr 6 '10 at 3:11
add comment

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.

share|improve this answer
    
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
add comment

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!

share|improve this answer
add comment

Your Answer

 
discard

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.