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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 $S_3 \times S_2$ ?

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    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen May 12 '17 at 6:38
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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.

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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.

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  • $\begingroup$ 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 $k$. $\endgroup$ – Jyrki Lahtonen May 12 '17 at 6:42

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