as you may know it is known in Galois theory that if $f\in\mathbb{Q}[T]$ and $\partial f=p$, with $p$ prime, $f$ irreducible and $f$ has $p-2$ real roots and $2$ complex roots, then $\operatorname{Gal}(f)=S_p$ and therefore, $f$ is not resoluble if $p\ge5$

So, my question is, is there an $f\in\mathbb{Q}[T]$ of degree $n$ with $n-2$ real roots, $2$ complex roots and Galois group not being $S_n$ for some composite $n\in\mathbb{N}$?

Moreover, is there an example of such polynomial when $n=4$?

  • 1
    $\begingroup$ This is a really good question to ask. In general, when some theorem has a bunch of conditions, they're all necessary, and finding out exactly what MAKES each one necessary is a really good thing to do to help understanding. Sometimes, of course, they conditions aren't all necessary -- it's just that no one knows how to prove the less-constrained version of the theorem. But those theorems don't tend to come up much in beginning courses. $\endgroup$ – John Hughes Jul 7 at 23:28

Sure: for instance, $f(T)=T^4-2$ works. Its splitting field is $\mathbb{Q}(\sqrt[4]{2},i)$ which has degree $8$ and thus the Galois group is not $S_4$.


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