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 $\mathbb{C}$?
Determining a Galois group without factoring
1
$\begingroup$
galois-theory
-
0Would 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? – 2012-09-14