I'm studying methods of computation of Galois group of irreducible polynomials over $\mathbb{Q}$. In case of fifth degree there are 5 variants of Galois group:$S_5,A_5,AGL_1(\mathbb{F}_5).D_5,\mathbb{Z}_5$. I have troubles with determining wether Galois group is $\mathbb{Z}_5$ or $D_5$ if given that it is a subgroup o $D_5$ because my algorithm requires too big computational power and I wonder if there are an easy way to solve this particular problem.
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4$\begingroup$ The practical answer is to reduce $f(x)$ modulo many primes and see if you ever get a factorization of the form $(\mbox{linear})(\mbox{quadratic})(\mbox{quadratic})$. If so, you know that your Galois group contains an element of that cycle structure and so must be $D_5$; by Cebatarov density, if the group is $D_5$, you should see such a factorization for $1/2$ of all primes, so if you don't see one soon you have excellent probabilistic evidence that the group is $C_5$. Finding a provably correct criterion will be messier. $\endgroup$– David E SpeyerCommented May 19, 2014 at 16:19
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$\begingroup$ Lenstra's notes websites.math.leidenuniv.nl/algebra/Lenstra-Chebotarev.pdf are a good introduction to this method. $\endgroup$– David E SpeyerCommented May 19, 2014 at 16:22
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$\begingroup$ @DavidSpeyer Yes, I know about this method and there are known some error estimates in Chebotarev theorem(so this algorithm is deterministic), but I thought that there are some simple way to distinguish these particular variatns if Galois group. $\endgroup$– SashaPCommented May 19, 2014 at 16:25
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1 Answer
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Section 6.3 of Henri Cohen's "A Course in Computational Algebraic Number Theory" is about computing Galois groups using resolvents. Subsection 6.3.4 is specifically about the quintic case. Cohen gives an algorithm which addresses the case of $\mathbb{Z}_{5}$ versus $D_{5}$ in step 6 (once a 5-cycle contained in the Galois group is known). This algorithm is implemented in PARI/GP (see this page).
In light of David Speyer's remarks, the difficulty is in proving the Galois group is $\mathbb{Z}_{5}$ when it appears that the only possibilities for the factorization of $f(x)$ modulo $p$ are irreducible, and a product of $5$ linear factors.
answered May 19, 2014 at 17:15