Let $f$ be an irreducible polynomial over a field K, and $\deg f = 4$, with roots $a,b,c,d$.

Let $g$ be a cubic resolvent with roots $\alpha,\beta,\gamma$. And $\alpha=ab+cd, \beta=ac+bd, \gamma=ad+bc$. Let $V$ be a Klein group. $G$ - Galois group of f. How to prove that field $K(\alpha,\beta,\gamma)$ is invariant to $G\cap V$, and there is no other field $L$ larger then $K$, being invariant too. How do I find coefficients of resolvent? And how to get known about $G$, for each $f$?.

I know that $K(\alpha,\beta,\gamma)$ is invariant but dont know why there is no other larger, I think that coefficients can be found through Viete formulas, I think that $Gal(g) = G/G\cap V$, but how to get information of G, if $Gal(g)$ can be empty, $A_{3}, S_{3}$ only?


The resolvent's roots are the combinations of $\alpha,\beta$ and $\gamma,$ so that's right you can get $g$ by the relation of roots and coefficients.

The clues to find $G$ is from the polynomial $(x-\alpha)(x-\beta)(x-\gamma),$ by its irreducibility, its discriminant and if it splits over the base field, which would be a direct check!

P.S. For your $f$ is irreducibe, G is transitive hence the only choices are just $S_4,A_4,V,D_4$ and $C_4.$


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