Prove that $f(x)$ is irreducible over $\mathbb{Q}$

Question

Let $f(x)\in \mathbb{Q}[x]$ with $\gcd(f(x),f'(x))=1$. If for all $a,b\in \mathbb{C}$ with $a,b$ roots of $f(x)$ there exists $\sigma\in \operatorname{Gal}(K,\mathbb{Q})$ which $\sigma(a)=b$ with $K\subset \mathbb{C}$ splitting field of $f(x)$ over $\mathbb{Q}$ prove that $f(x)$ is irreducible over $\mathbb{Q}$. It's obvious that every two roots of $f(x)$ are different but I couldn't find way to use it to find the answer.

Answer 1

As Rob Arthan noticed, we need the hypotesis for all $a,b\in \mathbb C$ roots of $f(x)$. With this assuption the following works.

Suppose $f(x)$ has degree $n$ and it is monic. Since $(f,f')=1$ it has $n$ different roots. Let's call them $a_1,...,a_n$.

Take now $\sigma_i$ such that $\sigma_i(a_1) = a_i$ that exist by hypotesis. The minimal polynomial $\mu_{a_1}(x)$ of $a_1$ has degree at least $n$ because $a_1$ has $n$ conjugates.

Since $a_1$ is also a root of $f(x)$ we have $\mu_{a_1}(x)| f(x)$, but $f(x)$ has degree $n$, so $\mu(x)=f(x)$ and $f(x)$ is irreducible.

Answer 2

EDIT: As Rob Arthan stated in the comments, the problem as stated is false. If you meant to say $a, b \in \mathbb C$ rather than $a, b \in \mathbb R$ then the following discussion holds.

First of all, as you stated, the condition that $gcd(f, f') = 1$ is equivalent to $f$ having no repeated roots. We can therefore factor $f = \prod f_i$ where the $f_i$ are pairwise coprime irreducible polynomials in $\mathbb Q[x]$ sharing no roots. To show that $f$ is irreducible we need to show that only one such polynomial appears in this factorization. The key to this problem is, of course, to understand how to Galois group acts on the roots of a polynomial.

Let $a$ be a root of $f$. Then it must be a root of some $f_i$ dividing $f$. Let $\sigma \in G = Gal(K/F)$. Then $\sigma(a)$ is not only a root of $f$, it's a root of $f_i$. To see this, let $f_i = \sum a_{ij} x^j$. Then $f_i(\sigma(a)) = \sum a_{ij} \sigma(a)^j = \sigma(\sum a_{ij} a^j) = 0$ as $\sigma$ is a field automorphism fixing $\mathbb Q$ and the $a_{ij} \in \mathbb Q$. This is a very important fact. The Galois group not only acts on the roots of $f$, but on its factors. With this, we gain a finer understanding of the action of the Galois group.

Now, let's let $Z(f)$ be the roots of $f$ in $K$ and $Z(f_i)$ be the roots of $f_i$ in $K$. By the condition on the derivative, $Z(f) = \coprod Z(f_i)$ a disjoint union. The above discussion tells us that the action of $G$ on $Z(f)$ is partitioned into the $Z(f_i)$. That is, if $a \in Z(f_i)$ and $\sigma \in G$ then $\sigma(a) \in Z(f_i)$. Suppose now that the factorization of $f$ contained two distinct polynomials, say $f_1, f_2$. Then let $a_i \in Z(f_i)$. By the condition you stated, there exists a $\sigma \in G$ such that $\sigma(a_1) = a_2$. But then $a_2 \in Z(f_1) \cap Z(f_2) = \emptyset$, a contradiction. Hence, only one factor appears in $f = \prod f_i$ so $f$ is irreducible.

To discuss this a bit more, the converse is true as well. The condition on the group action of $G$ on $Z(f)$ you stated is called transitivity and is equivalent to there being only one orbit. The converse I have in mind is that the Galois group of an irreducible polynomial with distinct roots acts transitively on the roots. With this in mind, the above discussion actually shows that the partition $Z(f) = \coprod Z(f_i)$ is precisely the partition into orbits from the action of $G$.