I've been asked to prove the following statement in a Galois Theory Seminar after being introduced to Dedekind's Theorem. (I assume this could potentially help getting the answer.)

Let $f(x)$ be a a monic polynomial of degree $N$ in $\mathbb{Q}[x]$ and let $E_f$ be its splitting field over $\mathbb{Q}$. Then, there exists a monic polynomial $p(x) \in \mathbb{Z}[x]$ of degree $N$ that has the same splitting field $E_f$.

I can't find any way to prove it.

Any help is welcome!

EDIT: There used to be what I though was a counter example that has been answered already.

  • $\begingroup$ This is not a counter-example. The problem is that you're assuming that $f$ and $p$ have the same roots. This is not necessary in this case since any polynomial with integral coefficients and rational roots has the splitting field of $\mathbb{Q}$. So, you could take $p=1$ and get the same splitting field as $f$. $\endgroup$ – Michael Burr Nov 19 '18 at 12:16
  • $\begingroup$ BBC3, all integers are also elements of $\Bbb{Q}$. $\endgroup$ – Jyrki Lahtonen Nov 19 '18 at 12:23
  • $\begingroup$ You're both right! I was taking $\mathbb{Z}$ as its splitting field instead which makes no sense at all. I'll edit the question to the proof only. $\endgroup$ – BBC3 Nov 19 '18 at 12:24
  • 2
    $\begingroup$ Are you possibly taking the same course as the asker of this question. My comments there settle your question also. $\endgroup$ – Jyrki Lahtonen Nov 19 '18 at 12:27

Let $x^n + a_1 x^{n_1}+ \cdots + a_n=0$, and write $y=Nx$; then $y^n + a_1 N y^{n_1}+ \cdots + a_n N^n =0$. Write $a_j=p_j/q_j$ and $N=\prod_j q_j$ and you are done. This is basically Gauss's lemma isn't it?


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