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Let $f \in \mathbb{Q}[X]$ be irreducible and let $L$ be its splitting field. Can something be said about the Galois group of $L$ over $\mathbb{Q}$ without computing the roots of $f$ in $\mathbb{C}$?

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Sure. If I recall correctly, you can reduce $f$ modulo various primes and see what happens there, and then you can apply the Chebotarev density theorem to deduce something about the cycle structure of the elements of the Galois group of $f$ considered as a permutation group. – Zhen Lin Sep 13 '12 at 16:25
In fact, you can actually compute the group completely without knowing the roots. What more could one want? :-) – Mariano Suárez-Alvarez Sep 13 '12 at 16:27
Seems to me that it’s never necessary, indeed never helpful, to know the roots as complex numbers. – Lubin Sep 13 '12 at 16:41
Would any of you care to elaborate in an answer? It's strange if $G$ can be determined completely, since whenever I look fr examples I see automorphisms written explicitly using the roots. Also, isn't Chebotarev density theorem a thing of probability? How does that help? – Karolis Juodelė Sep 14 '12 at 13:50

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