4
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.