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Given a cubic polynomial $f(x) \in \mathbb{Z}[x_{1},\dots,x_{n}]$. Suppose that $f(x)$ does not factor into three linear polynomials but contains a linear factor. Is the linear factor defined over $\mathbb{Z}$? I think this can be solved by Galois theory.

Thanks in advance.

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Your question is unclear. What ring does $f$ not factor into linear factors over, and what do you mean it "contains a linear factor" (which I would assume means factors into $(x-a)g(x)$ over $\mathbb Z$, but the next sentence suggests otherwise)? –  Alex Becker Sep 3 '12 at 12:17
I think the question could be stated: suppose $f\in\mathbf Z[x_1,\ldots,x_n]$ with $\deg f=3$ factors $f=gh$ in $\mathbf C[x_1,\ldots,x_n]$ with $\deg g=1$ and $h$ irreducible, then does there necessarily exist $\lambda\in\mathbf C^\times$ such that $\lambda g\in\mathbf Z[x_1,\ldots,x_n]$? –  Marc van Leeuwen Sep 3 '12 at 12:23
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Thanks for the comments. For some reason I cannot login my account and reply to the comments here. Marc is right, so please let me restate my question:


I have seen the claim several times before and am pretty sure that it is true, but I cannot really prove it. Thanks in advance.

Edit I forgot to say an important thing; I am thinking about cubic homogeneous polynomial $f(x)$.

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How about if you type in your own answers for $n=1$ and $n=2$ so we have some idea what you are talking about and what work you may have done. –  Will Jagy Sep 3 '12 at 18:42
I am basically interested in $n>3$ case, but I will write my idea up soon. –  Bill Sep 3 '12 at 18:48
This content should be added to the question, not put into an answer. –  robjohn May 10 '13 at 15:19
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Let me describe my idea. In case $n=1$, $f=nx_{1}^3$ for some $n\in \mathbb{Z}$ and the assumption (that the residual quadric is irreducible) is not satisfied.

Let's now assume $n=2$. By inhomogenizing $f$, we reduce our problem to the case where $f\in\mathbb{Z}[x]$ is (not necessarily homogeneous) cubic polynomial. But $f$ decomposes into three linear factors over $\mathbb{C}$, so the assumption is not satisfied.

I have no idea for $n>2$ case at this point.

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Bill, I have merged all the various accounts that I've seen you use, so that a cookie from any of them will allow you to stay logged in. It might be easier for you to register. You should now be able to edit your answers into your question and comment. –  mixedmath Sep 4 '12 at 3:54
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