# Existence of polynomials in $\mathbb{Q}[x]$ and $\mathbb{Z}[x]$ with same splitting fields.

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.

• 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 $f$. – Michael Burr Nov 19 '18 at 12:16
• BBC3, all integers are also elements of $\Bbb{Q}$. – Jyrki Lahtonen Nov 19 '18 at 12:23
• 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. – BBC3 Nov 19 '18 at 12:24
• Are you possibly taking the same course as the asker of this question. My comments there settle your question also. – 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?