I don't know how to solve the following:

Let $\alpha$ be a real root of $f(x)=x^4+3x-3\in Q[x]$. Is $\alpha$ a constructible number?

Any help is welcome.


I tried to get some info about the Galois group $G$ of $f$ (over $\mathbb Q$). It turns out that $f$ has a root $-1$ in $\mathbb F_5$ and $f(x)/(x+1)$ is irreducible in $\mathbb F_5[x]$. The group $G\subset S_4$ thus contains a $3$-cycle (Frobenius at $p=5$), in particular the order of $G$ is not a power of $2$, so the roots of $f$ are not constructible.

  • $\begingroup$ Can we say directly that $G \le S_4$ because $f$ is irreducible in $\mathbb{Q}$ (Eisenstein criterion for $p=3$)? And can you explain me why $G$ contains $3$-cycle (Frobenius at $p=5$)? I don't understand that part. Thank you. $\endgroup$ – Cortizol Sep 12 '13 at 9:03
  • 1
    $\begingroup$ @Cortizol: Yes, I should have said that $f$ is irreducible in $\mathbb Q[x]$ (I took it as somewhat implicit in the question); $G$ then permutes the roots of $f$. As for Frobenius: if $f\in\mathbb Z[x]$ is a monic irreducible polynomial, and if $p$ is a prime such that $f$ is multiplicity-free in $\mathbb F_p[x]$, with irreducible factors $f_i$, then the Galois group of $f$ over $\mathbb Q$ contains a permutation with the cycles of length $\deg f_i$. For more info look for Frobenius automorphism in Algebraic number theory. $\endgroup$ – user8268 Sep 12 '13 at 9:14
  • $\begingroup$ Ahhh, yes. It's known as Dedekind theorem. Silly of me. Thank you. (+1) $\endgroup$ – Cortizol Sep 12 '13 at 9:20

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