# The Galois group of two irreducible polynomials

What information is necessary to determine the Galois group of two irreducible polynomials? If I know the Galois group of $p(x)$ is $S_3$ and the Galois group of $f(x)$ is $S_2$ can I say the Galois group of $p(x)f(x)$ is $S_3 \times S_2$ ?

• Warning: You need the two splitting fields to be linearly disjoint. Consider $p(x)=x^3-2$ and $f(x)=x^2+x+1$ over $\Bbb{Q}$. The splitting field of $f$ is contained in the splitting field of $p$, so the Galois group of $pf$ is equal to that of $p$. Similarly, with $p(x)=x^3-2$ and $f(x)=x^3-3$, the splitting fields of both $p$ and $f$ contain the splitting field of $x^2+x+1$, and the Galois group of $pf$ will be smaller. – Jyrki Lahtonen May 12 '17 at 6:38

If I know the Galois group of $$p(x)$$ is $$S_3$$ and the galois group of $$f(x)$$ is $$S_2$$ can I say the Galois group of $$p(x)f(x)$$ is $$S_3 \times S_2$$ ?

This is true as long as the intersection of the splitting fields of $$p$$ and $$f$$ gives the base field$$^\dagger$$. If this is the case, and if the Galois group for $$f$$ is $$G_1$$ and the Galois group for $$p$$ is $$G_2$$, then the Galois group of $$pf$$ is $$G_1 \times G_2$$.

You can see this by considering the fact that the Galois group of a polynomial $$f$$ can be thought of as a subgroup of the permutation group $$S_{\deg(f)}$$ that acts on the set of that polynomial's roots. In particular, one can show that roots of $$f$$ can be sent only to other roots of $$f$$ (this action is transitive $$\iff$$ the polynomial is irreducible). Put another way, given any algebraic element $$\alpha \in \overline{F}$$ over a field $$F$$, automorphisms of $$\overline{F}$$ can send $$\alpha$$ only to other roots of its minimal polynomial (its Galois conjugates).

This has the implication that the action of $$\text{Gal}(pf)$$, given $$p$$ and $$f$$ are irreducible, sends roots of $$p$$ only to other roots of $$p$$, and likewise for $$f$$. Therefore, any element of $$\text{Gal}(pf)$$ is going to permute some roots of $$p$$, or it's going to permute some roots of $$f$$, or it's going to be some composition of those two options (roots of $$p$$ can't be sent to roots of $$f$$ or vice-versa). This is characteristic of a direct product.

$$^\dagger$$ If the intersection of the two splitting fields is larger than the base field, then we will have automorphisms that cannot be expressed as a mere composition of two automorphisms, one of which moves around only the roots of $$f$$, and the other only roots of $$p$$. For example, if $$f(x) = x^3 - 2$$ and $$p(x) = x^3 -3$$, then complex conjugation is such an automorphism. More generally, if the polynomials are irreducible over $$F$$ and the intersection of the two splitting fields is some $$E/F$$, then the elements of $$\text{Aut}(E/F)$$ cannot be decomposed in this way.

You can say this if $E_p\cap E_f=k$ where $k$ is the base field. $E_p$ and $E_f$ are the splitting fields of $p,f$ respectively.

• Good job! You can actually say that this holds if and only if $E_p\cap E_f=k$ with the intersection taken inside a fixed algebraic closure of $k$. – Jyrki Lahtonen May 12 '17 at 6:42