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Let $\mathbb{F}_q$ be a finite field where $q$ is odd. Let $f \in \mathbb{F}_q[x,y]$ be the following polynomial

$$f:=(x^2y^2 - 2x^2y - 2xy^2 - x^2 + xy - y^2 - 2x - 2y + 1).$$

How to prove that $f$ is irreducible or absolutely irreducible over $\mathbb{F}_q$? Is there any general criterion to deal with such polynomials? (Absolutely irreducible means irreducible over every algebraic extension of the base field.)

Viktor Vaughn
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Rohit Gupta
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  • What do you mean by absolutely irreducible? As a start; one way is to simply attempt to write $f$ as a product of two polynomials, and find constraints on the coefficients. – Servaes May 14 '19 at 19:03
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    @Servaes absolutely irreducible means irreducible in every algebraic extensions of our field. – user10354138 May 14 '19 at 19:07
  • Trying the above shows that $f$ does not have a linear factor for any $q$, including even $q$. – Servaes May 14 '19 at 19:27
  • It is easy to show that $$g(u,v)=u^2-2uv-v^2+3u-2v+1$$ is absolutely irreducible. Wonder whether that implies the claim given that A) $f(x,y)=g(xy,x+y)$ (exploiting the symmetry of $f$) and B) $\phi:(x,y)\mapsto(xy,x+y)$ is an onto mapping from $k^2$ to itself ($k$ an algebraic closure of $\Bbb{F}_q$). If the zero loci of $f$ were a union of two or more irreducible curves, wouldn't pulling it back by $\phi$ imply that the same holds for the zero locus of $g$? God, I'm so rusty in basic algebraic geometry :-( – Jyrki Lahtonen May 15 '19 at 06:12
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    A tool algebraic geometry gives is that if the projective closure of the zero locus curve is smooth, then the polynomial is absolutely irreducible. This is because, by Bezout, the distinct components (that would exist if the polynomial were not absolutely irreducible) intersect somewhere. And those points of intersection will then be singularities. This gives one-way logic only: a curve with a singularity can still be irreducible. If $q=3^n$ then this curve has a singularity at $(x,y)=(-1,-1)$ for $f$ and both partial derivatives vanish there. Yet, we have absolute irreducibility. – Jyrki Lahtonen May 15 '19 at 06:42
  • Unless I made a mistake the curve $f(x,y)=0$ is non-singular in (odd) characteristic $\neq 3,7$ or $31$. I'm not sure I want to reproduce all the steps though :-) – Jyrki Lahtonen May 15 '19 at 07:10
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    @JyrkiLahtonen The factors of a symmetric polynomial are not necessarily symmetric; a simple example is given by $P(x,y)=xy$, and more generally $$P(x,y)=Q(x,y)Q(y,x),$$ for any polynomial $Q\in k[x,y]$. – Servaes May 16 '19 at 11:23
  • @Servaes Yes, I was aware of that possibility. It is not unlike the $x\leftrightarrow -x$ symmetry of order two that I utilized here, when it was possible to deduce that the eventual factors must obey the same symmetry (rather than be symmetric images of each other). But at the time of writing the comment I couldn't figure out whether that necessarily ruins the approach. Of course, the polynomial $f(x,y)=xy$ is a simple counterexample as you pointed out :-) – Jyrki Lahtonen May 16 '19 at 12:21

1 Answers1

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Since this $f$ has a unique top (total) degree term $x^2y^2$, we can just examine all possible factorization and show they don't work.

Any factorization $f=gh$ also give $f_{top}=g_{top}h_{top}$, so the top-degree piece of possible least-degree factors are $x,y,x^2,xy,y^2$. So we have to examine the four cases

  1. $(x-c)(xy^2+\dots)$
  2. $(y-c)(x^2y+\dots)$
  3. $(x^2+\dots)(y^2+\dots)$
  4. $(xy+\dots)(xy+\dots)$

For example, if $x-c$ is a factor of $f$, then $$ \begin{cases} 0=c^2-2c-1 & (\text{coeffs of }y^2)\\ 0=-2c^2+c-2 & (\text{coeffs of }y)\\ 0=-c^2-2c+1 & (\text{coeffs of }1)\\ \end{cases} $$ which is impossible (if $2\mid q$ then the second equation implies $c=0$ but that doesn't satisfy the other two. Otherwise, adding the first and third equation gives $c=0$, which again doesn't work). Same reason works for $y-c$.

For the quadratic factors, note that case (3) must completely separate $x$ and $y$ (because there are no $x^3$ or $y^3$), so it reduces to case 1 if we work over a quadratic extension. For case (4), use the $x\leftrightarrow y$ symmetry of $f$ to further break into (4a) $(xy+a(x+y)+b)(xy-(2+a)(x+y)+b')$, or (4b) $(xy+ax+by+c)(xy+bx+ay+c)$, $a\neq b$, etc.

user10354138
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