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Here it is claimed that most irreducible quartic polynomials with four real roots over $\mathbb Q$ have galois group $A_4$. What do they mean by this and (why) is what they mean true?

Carla_
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    Compare with this post. If the discriminant is a square then the Galois group is $A_4$ in this theorem from the answer (and $S_4$, if it is not a square). I would say, that being a square is not "generic", so usually I would expect $S_4$ as Galois group (but I don't know what they mean exactly). – Dietrich Burde Jul 25 '23 at 18:49
  • @DietrichBurde Why does the discriminant not being a square imply the Galois group to be $S_4$? – Carla_ Jul 27 '23 at 01:01
  • This is the cited Theorem in the answer here, as I said, with the reference to a detailed proof in Conrad's notes, namely Corollary $4.3$ on page $12$ with table $8$. – Dietrich Burde Jul 27 '23 at 08:47
  • @DietrichBurde In that table there are also a rows where the discriminant is not a square and the Galois groups is $\mathbb Z/(4)$ or $D_4$ rather than $S_4$. – Carla_ Jul 27 '23 at 12:55
  • I mean, with the same conditions of the answer in the duplicate, for $A_4$ (in comparison to $S_4$), so with $R_3(x)$ irreducible. In other words, we are discussing only the first two rows in table $8$. We are not interested in $D_4$, $C_4$ or $V$. So my first comment only says, that $S_4$ should be more likely than $A_4$. – Dietrich Burde Jul 27 '23 at 12:59

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