MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer
How do you conclude that $\mu=1$ for any $\sigma \in \mathrm{Aut}(\mathbb{C})$ – user2013 Sep 4 '12 at 19:16
@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 $g^\sigma=g$. – Peter Mueller Sep 5 '12 at 7:29
@Peter, I misunderstood your answer. You are totally right. – user2013 Sep 5 '12 at 15:15

Your Answer


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.