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How can I compute the Galois group of the polynomial $fg\in K[x]$ assuming that I know the Galois groups of $f\in K[x]$ and $g\in K[x]$? Let's suppose for simplicity that the field $K$ is perfect.

flag	asked Feb 8 at 20:11 roger123 355●6

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Apparently this is not trivial: see <a href = "citeseerx.ist.psu.edu/viewdoc/…; (if the link doesn't work, google for "Galois Theory And Reducible Polynomials" by Annick Valibouze). – Franz Lemmermeyer Feb 8 at 20:28

Johannes Hahn · Answer 1 · 2010-02-08 21:21:38Z UTC

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Let $L,M$ be the splitting field of $f,g$ over $K$. Then the compositum $LM$ is the splitting field of $fg$. Now your question translates with the help of the main theorem of galois theory into the following group question: How can one compute $G$ if one knows normal subgroups $N_1, N_2$, the quotients $G/N_i$ and the fact that $G=N_1 N_2$. This seems pretty hard to me in general.

edited Feb 8 at 21:21

answered Feb 8 at 20:24

Johannes Hahn
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The Galois group of $LM$ is simply the subgroup of $N_1\times N_2$ made up of items of the form $(\phi_1,\phi_2)$, where $\phi_1$ agrees with $\phi_2$ on $L\cap M$. – Steve D Feb 8 at 20:29

The Problem is, that you don't know $L\cap M$ in general. – Johannes Hahn Feb 8 at 21:20

You mean $G/N_1 \times G/N_2$, right? – David Speyer Feb 8 at 21:31

@David: Yes, sorry. @Johannes: Right, but I don't think your characterization of $G$ works very well either, since it doesn't uniquely determine $G$. – Steve D Feb 8 at 22:33

The problem is: What does "I know the Galois groups" mean? If we just know their isomorphism classes as groups, then we can't say much. It would be natural to require that we know their actions on the roots, but then again, where do the roots lie? If the roots of both polynomials lie in one given field extension of $k$, then Steve's answer applies. If not, we are left with trouble: How do we find out whether the two field extensions are "the same" (e. g., one injects into the other) or completely disjoint (i. e., the compositum has maximal dimension). – darij grinberg Feb 9 at 0:00

Galois group of a product of polynomials

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