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Let $f(x)\in\mathbb{Z}[x_{1},\dots,x_{n}]$ be a cubic homogeneous polynomial, which factors as $f(x)=g(x)h(x)$ over $\mathbb{C}$ with $\mathrm{deg}(g)=1$ and $h$ irreducible over $\mathbb{C}$. Assume that $n>3$. How can one prove that $\exists \lambda \in \mathbb{C}^{\times}$ such that $\lambda g \in \mathbb{Z}[x_{1},\dots,x_{n}]$?

I have seen this several times (for example the comment before Lemma 4.3 in this paper) but cannot really understand why. I feel that this can be proven by some Galois theory.

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  • $\begingroup$ The link to the paper at springerlink.com appears to be broken. I am also not able to find a copy on the Wayback Machine… $\endgroup$
    – The Amplitwist
    Commented Apr 3, 2022 at 11:22

1 Answer 1

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By unique factorization in $\mathbb C[x_1,\ldots,x_n]$, you have $g^\sigma=\mu g$ for each $\sigma\in\text{Aut}(\mathbb C)$ and $\mu$ depending on $\sigma$. Now pick $g$ such that one of its coefficients is $1$. Then $\mu=1$ for each $\sigma$. So all coefficients of $g$ are fixed under $\text{Aut}(\mathbb C)$, hence they are rational.

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  • $\begingroup$ How do you conclude that $\mu=1$ for any $\sigma \in \mathrm{Aut}(\mathbb{C})$ $\endgroup$
    – user2013
    Commented Sep 4, 2012 at 19:16
  • $\begingroup$ @Muon: Not sure what the problem is: By $g^\sigma$ I mean the field automorphism $\sigma$ applied to the coefficients of $g$. So if one coefficient of $g$ is $1$, then applying $\sigma$ doesn't change this coefficient. So if $g^\sigma$ is a scalar multiple of $g$, we must have $g^\sigma=g$. $\endgroup$
    – Peter Mueller
    Commented Sep 5, 2012 at 7:29
  • $\begingroup$ @Peter, I misunderstood your answer. You are totally right. $\endgroup$
    – user2013
    Commented Sep 5, 2012 at 15:15

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