10
$\begingroup$

Starting from a problem in combinatorics, I ended up with a very simple problem about polynomials, which, unfortunately, I am not able to solve.

Say we work over $\mathbb C$. Fix $d>1$. Is it possible to find 3 homogenous polynomials $F_0,F_1$ and $F_2$ of degree $d$ in $\mathbb C[x,y,z]$ without any non-trivial zeroes in common and such that $$F_1\cdot F_2=F_0^2- x^{d+1}\cdot G$$ for some homogeneous polynomial $G$ of degree $d-1$?

My guess is that the answer is no, but I couldn't rephrase it to a more classical algebraic geometry problem. It is easy to show that the answer is negative if we take $G=0$ or $G=x^{d-1}$ or if $d=2$.

$\endgroup$
  • 2
    $\begingroup$ The analogous problem for polynomials in one variable, say $t$, has a trivial affirmative answer, because one can choose freely $f_0(t)$, and for a generic $g(t)$ of degree $d-1$, the polynomial $f_0(t)-g(t)$ has simple zeros and distinct from those of $f_0$, so that there are $f_1(t)$ and $f_2(t)$ as wanted. Then, can't you substitute $t=P(y,z)/x^r$, for a polynomial $P(y,z)$ of degree $r$, and get your $F_0, F_1, F_2, G$? $\endgroup$ – Pietro Majer Jan 4 '18 at 17:20
  • $\begingroup$ Thanks for your reply, but how do you make sure that there are no common zeroes at infinity (i.e. of the form (0,y,z) for some y,z not both equal to zero)? $\endgroup$ – mrw Jan 4 '18 at 17:55
  • 2
    $\begingroup$ actually, I think that what I mentioned always happens: indeed if p(y,z)=0 then (0,y,z) is a common solution for F_0, F_1, and F_2. $\endgroup$ – mrw Jan 4 '18 at 18:17
  • $\begingroup$ @Somos: The question is about polynomials in 3 variables. $\endgroup$ – user6976 Jan 5 '18 at 0:04
  • $\begingroup$ I am curious, which kind of combinatorial problem could lead to such question? Even a vague domain specification could be interesting. Citation of an exemplary paper is above all the dreaming :) $\endgroup$ – Sergey Dovgal Jan 6 '18 at 21:12
9
$\begingroup$

This equation has no solutions when $d$ is odd. (EDIT: See below for the general case.)

Actually, for $d$ odd, there are no triples $(F_0,F_1,F_2)$ with $F_1\cdot F_2-F_0^2$ a multiple of $x$, let alone $x^{d+1}$. One can see this in a completely hands-on way by setting $x=0$ and looking at the resulting triple of polynomials in $y$ and $z$, which can be seen to have a common root. Here is a more geometric argument.

The problem is equivalent to asking for a map (necessarily finite) $\mathbb{P}^2\rightarrow\mathbb{P}^2$ such that the pullback of $\mathcal{O}(1)$ is $\mathcal{O}(d)$ and the pullback of the conic defined by $xy=z^2$ contains the line $x=0$ with multiplicity $d+1$; the strengthening I gave above says that for $d$ odd, this divisor cannot contain the line at all. Indeed, if it does, the pushforward of the line must be some multiple of the conic. However, the pushforward of the line is of degree $d$, and so this can't happen.

EDIT: Here is an attempt to settle the problem for even $d$ as well. This argument uses a surprising amount of machinery , considering how elementary the problem is to state. This makes me a little suspicious that this argument is either wrong or more complicated than necessary, but here goes.

As above, we have a map $f:\mathbb{P}^2\rightarrow\mathbb{P}^2$ defined by $(F_0,F_1,F_2)$ satisfying $f^*\mathcal{O}(1)\cong\mathcal{O}(d).$ Let $X$ be the $d$th formal neighborhood of the line $L$ defined by $x=0$, or in other words, $X$ is the subscheme defined by $x^{d+1}$. Our hypothesis implies that the restriction of $f$ to $X$ lands (scheme-theoretically) in the conic $yz=x^2.$ Treating this conic as an abstract $\mathbb{P}^1$, we get a map $g:X\rightarrow\mathbb{P}^1.$

Now as $\mathcal{O}_{\mathbb{P}^2}(1)$ restricts to $\mathcal{O}_{\mathbb{P}^1}(2)$ under the embedding of this conic, we know that $g^*\mathcal{O}_{\mathbb{P}^1}(2)\cong\mathcal{O}_X(d)$. We would like to know the stronger statement that $g^*\mathcal{O}_{\mathbb{P}^1}(1)\cong\mathcal{O}_X(d/2),$ which would follow if we knew that the Picard group of $X$ is torsion free.

To see that $\operatorname{Pic}(X)$ is torsion-free, note that we have a short exact sequence of sheaves $0\rightarrow I\rightarrow \mathcal{O}_X^*\rightarrow\mathcal{O}_{\mathbb{P}^1}^*\rightarrow 0.$ This in turn gives us the exact sequence $0\rightarrow H^1(I)\rightarrow H^1(\mathcal{O}_X^*)\rightarrow H^1(\mathcal{O}_{\mathbb{P}^1}^*)\rightarrow H^2(I).$ But $I$ is in fact a sheaf of $\mathbb{C}$-vector spaces, so its cohomology groups are also $\mathbb{C}$-vector spaces. (One can actually compute all of the cohomology groups of $I$, but we will not need this here.) As $H^1(\mathcal{O}_{\mathbb{P}^1}^*)\cong\operatorname{Pic}(\mathbb{P}^1)\cong\mathbb{Z},$ we see immediately that $\operatorname{Pic}(X)\cong H^1(\mathcal{O}_X^*)$ is torsion-free.

Now that we know $g^*\mathcal{O}_{\mathbb{P}^1}(1)\cong\mathcal{O}_X(d/2),$ this tells us that there are sections $s_1,s_2$ of $\mathcal{O}_X(d/2)$ and a unit $u$ of $\mathcal{O}_X$ such that we have $(F_0,F_1,F_2)=(us_1s_2,us_1^2,us_2^2).$ A quick cohomology calculation shows that all sections of $\mathcal{O}_X(d/2)$ are restrictions of sections of $\mathcal{O}_{\mathbb{P}^2}(d/2)$ and that all units of $\mathcal{O}_X$ are constants. So we can take $u=1$ and identify $s_1,s_2$ with two homogeneous degree $d/2$ polynomials (which we will again call $s_1,s_2.$)

All in all, we have $(F_0,F_1,F_2)\equiv (s_1s_2,s_1^2,s_2^2)\pmod{x^{d+1}}.$ But this implies in fact that $(F_0,F_1,F_2)=(s_1s_2,s_1^2,s_2^2),$ and so that $F_1F_2-F_0^2=0,$ aka that $f$ in fact comes from a morphism $\mathbb{P}^2\rightarrow\mathbb{P}^1.$ But no non-constant such morphisms exist, a contradiction.

Just for fun: The above argument breaks down in characteristic $2$. In fact, the original problem has an affirmative answer in characteristic $2$, even for $d$ as small as $2$. Take $F_0=xy+xz+yz, F_1=x^2+y^2, F_2=x^2+z^2.$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for your reply. And sorry, I forgot to mention that I arrived to the same conclusion before. I am interested to know the answer for $d$ even. $\endgroup$ – mrw Jan 4 '18 at 16:09
  • $\begingroup$ @mrw: I have edited this answer to also deal with the case $d$ even. $\endgroup$ – dhy Jan 5 '18 at 3:27
2
$\begingroup$

It is a comment which is too long for a comment. You probably tried this but here it is anyway. You can eliminate $x$ and the condition that $F_i$ are homogeneous by substitution $y'=y/x, z'=z/x$. So you want to prove that there are no polynomials in two variables $F_i(y',z'), i=1,2,3$ of degree $d$ without zeroes in common such that $F_1F_2-F_3^2$ is a polynomial $G$ of degree $d-1$. If you take the highest degree homogeneous components $H_1, H_2, H_3$ of $F_1,F_2, F_3$, then you have $H_1H_2=H_3^2$. So the question is:

Given three homogeneous polynomials in 2 variables of the same degree $d$, satisfying $H_1H_2=H_3^2$ can you deform $H_1, H_2, H_3$ by adding polynomials in the same two variables of degree $d-1$ so that the resulting polynomials do not have common roots?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, I agree with your point of view. Thank you. $\endgroup$ – mrw Jan 4 '18 at 16:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.