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.
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2$\begingroup$ 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). $\endgroup$– Franz LemmermeyerCommented Feb 8, 2010 at 20:28
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$\begingroup$ This link should work better: citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.27.7409 $\endgroup$– Zach TeitlerCommented Mar 14, 2020 at 8:37
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2 Answers
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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.
answered Feb 8, 2010 at 20:24
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2$\begingroup$ 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 $L\cap M$. $\endgroup$– Steve DCommented Feb 8, 2010 at 20:29
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$\begingroup$ The Problem is, that you don't know $L\cap M$ in general. $\endgroup$ Commented Feb 8, 2010 at 21:20
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$\begingroup$ You mean $G/N_1 \times G/N_2$, right? $\endgroup$ Commented Feb 8, 2010 at 21:31
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$\begingroup$ @David: Yes, sorry. @Johannes: Right, but I don't think your characterization of $G$ works very well either, since it doesn't uniquely determine $G$. $\endgroup$– Steve DCommented Feb 8, 2010 at 22:33
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3$\begingroup$ 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). $\endgroup$ Commented Feb 9, 2010 at 0:00
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Here you can find the proof of $Gal(E_1 E_2) = Gal(E_1) \times_{Gal(E_1 \cap E_2)} Gal(E_2)$ and an example how this can be used to compute the Galois group of a product of two polynomials.
answered Mar 27, 2010 at 15:46
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