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Let $K$ be a field with $char(K)=0$ and $f\in K[t]$ an irreducible polynomial which Galois group $G_{K}(f)$ is cyclic. Show the discriminant $\Delta(f)$ of $f$ is a square of an element of $K$ if and only if $o(G_{K}(f))$ is odd.

I have been able to prove $o(G_{K}(f))$ odd $\Rightarrow$ $\Delta(f)$ is a square of an element of $K$. I do not know how to prove the other implication.

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1 Answer 1

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If $\Delta(f)$ is a square of an element of $K$, then the Galois group, regarded as a group of permutations on the roots, is a subgroup of the alternating group. Hence a generator $g$ of the group must be an even permutation. Now $g$ is a single cycle (as $f$ is irreducible, so that the Galois group is transitive on the roots), so $g$ has to be a cycle of odd length.


The following are actually equivalent in characteristic different from $2$:

  • $\Delta(f)$ is a square of an element of $K$, and
  • the Galois group is a subgroup of the alternating group,

so you can use this for both implications.

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Do you know a website where that is prove? Thank you –  Gauloises Jun 23 at 11:49
    
[Thanks.] Not sure about a web site, but it's the corollary on page 250 of the first edition of Jacboson's Basic Algebra I, for instance. –  Andreas Caranti Jun 23 at 11:51
    
OK, it appears to be Theorem 4.7 here math.uconn.edu/~kconrad/blurbs/galoistheory/galoisaspermgp.pdf –  Andreas Caranti Jun 23 at 11:53
    
Thank you very much –  Gauloises Jun 23 at 11:55

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