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If $f(x,y)\in \mathbb{Q}[x,y]$ factors non-trivially (as a product of two non-constant polynomials) in some field extension of $\mathbb{Q}$, then does it also factor non-trivially in some finite field extension of $\mathbb{Q}$?

user26857
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Cezar
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  • What is $F=\overline{G} \overline{H}$ supposed to mean? A field $F$ is the product of two polynomials? – Dietrich Burde Apr 10 '24 at 19:17
  • @DietrichBurde, yes professor, its a very bad writing, I have rewritten the post, hopefully correct, and hope it is still of interest – Cezar Apr 10 '24 at 21:53
  • @DietrichBurde , there is the following fast argument, let $\mathbb{Q}\subseteq F$ a field extension where $f$ factors, then it also factors in $F^a$ the algebraic closure of $F$, and since the theory of algebraic closed fields of char=0 is complete, then this $f$ would also factor in $\mathbb{Q}^a$(algebraic closure of $\mathbb{Q}$), hence $f$ factors in $\mathbb{Q}^a$, and then obviously in a finite extension of $\mathbb{Q}$. But this doesn't make me understand why this fact has to happen, and what exactly stands in the way of an elementary proof of this. – Cezar Apr 11 '24 at 14:42

1 Answers1

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I think it is true.

For every field extension $E$ of $\mathbb{Q}$, we still have that $E[x,y]$ is an UFD. Thus there are only finitely many monic factors of $f\in\mathbb{Q}[x,y]$ in $E[x,y]$.

Thus by considering the coeffecients of $g,h$, we can rewrite $f=gh$ as a set of polynomial equations with coefficients in $\mathbb{Q}$. If we fix the leading terms of $g,h$ and require them to be monic, this set of equations only has finitely many solutions. Thus the solutions (=coeffecients of $g,h$) are algebraic.

This is because every automorphism $\sigma$ of $E$ fixes $\mathbb{Q}$, thus if $(g,h)$ is a solution, $(\sigma(g),\sigma(h))$ is also a solution. But a transcendental element can be mapped to infinitely many elements, contradicting the finiteness of solutions. For example, $\pi\mapsto\pi+1$ gives an automorphism of $\mathbb{Q}(\pi)\to\mathbb{Q}(\pi)$. We still need $E$ to be algebraically closed to extend this to $\operatorname{Aut}(E)$, though.

More precisely, for $x$ transcendental on $\mathbb{Q}$, we can find some transcendental basis $B\ni x$ of $\overline E$. That is, $\overline E$ is the algebraic closure of $\mathbb{Q}(B)$. Apply the following lemma:

Lemma. For any field $F$, any automorphism $f\in\operatorname{Aut(F)}$ can be extended to an element in $\operatorname{Aut}(\overline F)$.

To see this, Zorn's lemma on $\left\{(E,\phi:E\to\overline F):E\subset\overline F\text{ and }\phi|F=f\right\}$ shows the existence of a maximal element $(E,\phi)$. We must have $E=\overline F$. Otherwise, there is some $a\in\overline F-E$ with mimimal polynomial $P$, and we can extend $\phi$ to $E(a)$ by mapping $a$ to some element with minimal polynomial $f(P)$. On the other hand, we must have $\phi(E)=\overline F$, because it's algebraically closed (being the isomorphic image of $\overline F$) and contains $F$. In conclusion, $\phi\in\operatorname{Aut}(\overline F)$.

This proof doesn't look like what I expected, but there's nothing I can do nothing about it.

William Turner
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  • Yes it is true, but this proof I am not sure I follow. First, what you mean monic in the context of multivariate polynomial, but this aside, why if a set of equations has finitely many solutions then the solutions are algebraic? – Cezar Apr 11 '24 at 14:53
  • @Cezar By "monic" I'm just trying to get rid of the scaling by an element in $E$. You can just first compare $x$-degree and then compare $y$-degree, and make the first term have coefficient 1. – William Turner Apr 11 '24 at 15:01
  • @Cezar I've editted to add a paragraph in the answer. – William Turner Apr 11 '24 at 15:10
  • yes, hahaha, didn't expect to be that obvious : D, thank you – Cezar Apr 11 '24 at 15:22
  • I undid the correct answer, because I am not so sure this is true.

    So what if it has finite solutions, the field extension is a general one, so it can also be a non-Galois extension, meaning the invariant field given by $Aut(E / \mathbb{Q})$ can be bigger than $\mathbb{Q}$, and we don't know if the invariant field is algebraic over $\mathbb{Q}$. And I don't rly buy the transcendental can be mapped to infinitely many elements, the example you give is ok, but what we need is an automorphism from $E $ to $E$.

     – Cezar Apr 12 '24 at 15:01
  • @Cezar Yes, there is a problem there. I just realized that $\mathbb{R}$ has no automorphisms over $\mathbb{Q}$, so anything like $\pi\mapsto\pi+1$ does not work. But replacing $E$ with its algebraic closure solves that problem. We are guaranteed to have many automorphisms if $E$ is algebraically closed. See the answer here. With this we can extend the elements of $Aut(\mathbb{Q}(\pi))$ to elements of $Aut(\overline{E})$. – William Turner Apr 13 '24 at 00:12
  • I have to admit don't know to well transcendental basis. That lemma I know, just didn't know one can always extend a partial isomorphism(with domain a field from $\overline{E}$ and image in $\overline{E}$) in an algebraically closed field to an automorphism of the whole algebraically closed field. I was a bit hasty, need to convince myself of this transcendence basis, and will accept this answer, once post is open again. Thank you. For an unknown reason to me it got closed with no explanation. – Cezar Apr 13 '24 at 09:43
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    Instead of using automorphisms, you can use the Nullstellensatz as in this answer of mine to a similar question. – Eric Wofsey Apr 13 '24 at 17:57