Timeline for Galois group of a product of polynomials
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
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| Feb 9, 2010 at 0:00 | comment | added | darij grinberg | 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). | |
| Feb 8, 2010 at 22:33 | comment | added | Steve D | @David: Yes, sorry. @Johannes: Right, but I don't think your characterization of $G$ works very well either, since it doesn't uniquely determine $G$. | |
| Feb 8, 2010 at 21:31 | comment | added | David E Speyer | You mean $G/N_1 \times G/N_2$, right? | |
| Feb 8, 2010 at 21:21 | history | edited | Johannes Hahn | CC BY-SA 2.5 |
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| Feb 8, 2010 at 21:20 | comment | added | Johannes Hahn | The Problem is, that you don't know $L\cap M$ in general. | |
| Feb 8, 2010 at 20:29 | comment | added | Steve D | 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 $L\cap M$. | |
| Feb 8, 2010 at 20:24 | history | answered | Johannes Hahn | CC BY-SA 2.5 |