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Can anyone give me an example of a finite Galois extension whose Galois group is $A_4$?

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More advanced sources: 1) Topics in Galois theory of Serre, page 43 and 2) Inverse Galois Theory of Malle and Matzat, see the tables toward the end. –  Lior B-S Nov 7 '12 at 12:13

1 Answer 1

lhf's links in the comments give good examples over the base field of $\mathbb{Q}$. If you're willing to accept a slightly larger base field (your question doesn't specify one), you can get a whole slew of examples with very little work.

Namely, take a random quartic polynomial $f$ with, say, discriminant $\Delta$. With very high probability (actually, essentially 100%, in a sense that can be made precise), its splitting field $K$ has Galois group $S_4$ over $\mathbb{Q}$. Now since $\sqrt{\Delta}$ is the product of the differences of the roots of $f$, it lives in $K$, and so whenever $\Delta$ is not a perfect square, $\mathbb{Q}(\sqrt{\Delta})$ is a quadratic extension of $\mathbb{Q}$ contained in $K$. But now $\operatorname{Gal}(K/\mathbb{Q}(\sqrt{\Delta}))$ is an index-2 subgroup of $\operatorname{Gal}(K/\mathbb{Q})\cong S_4$, the unique example of which is $A_4$.

So we conclude that $\operatorname{Gal}(K/\mathbb{Q}(\sqrt{\Delta}))\cong A_4$ for almost every quartic polynomial with non-square discriminant.

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