A cubic polynomial which contains a linear factor with irreducible residual quadratic form - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T13:17:46Z http://mathoverflow.net/feeds/question/106309 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106309/a-cubic-polynomial-which-contains-a-linear-factor-with-irreducible-residual-quadr A cubic polynomial which contains a linear factor with irreducible residual quadratic form Bill B 2012-09-04T08:04:54Z 2012-09-04T09:26:47Z <p>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 $\lambda g \in \mathbb{Z}[x_{1},\dots,x_{n}]$? </p> <p>I have seen this several times (for example the comment before Lemma 4.3 in <a href="http://www.springerlink.com/content/g0583k343q720095/" rel="nofollow">this paper</a>) but cannot really understand why. I feel that this can be proven by some Galois theory. </p> http://mathoverflow.net/questions/106309/a-cubic-polynomial-which-contains-a-linear-factor-with-irreducible-residual-quadr/106322#106322 Answer by Peter Mueller for A cubic polynomial which contains a linear factor with irreducible residual quadratic form Peter Mueller 2012-09-04T09:26:47Z 2012-09-04T09:26:47Z <p>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.</p>