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.
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2 Answers
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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.
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$\begingroup$ 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. $\endgroup$– CortizolSep 12, 2013 at 9:03
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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. $\endgroup$– user8268Sep 12, 2013 at 9:14
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$\begingroup$ Ahhh, yes. It's known as Dedekind theorem. Silly of me. Thank you. (+1) $\endgroup$– CortizolSep 12, 2013 at 9:20
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$\begingroup$ Note the importance of signs. Had we said $x^4+3x+3=0$ then the roots would be cobstructible. $\endgroup$ Feb 10, 2021 at 14:02
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$\begingroup$ For that equation, its resolvent cubic is $-8z^3-(24\sqrt{3})z^2-24z+9=0$. does this have rational solutions? $\endgroup$ Jan 29 at 20:36
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As has been proven, for a quartic's solutions to be constructible, the roots of the resolvent cubic must be rational. Given this equation, the cubic coefficient of its resolvent is $-8$ and its resolvent constant coefficient is $9$. All rational solutions must be of the form $p/q$, where $p$ and $q$ are coprime integers, $p$ is a factor of 9 and $q$ is a factor of 8 (either $p$ and/or $q$ may be negative.)
This equation's resolvent is $-8z^3-(24i\sqrt{3})z^2+48z+9=0$. We can see that there are no rational solutions, because the imaginary term in the cubic's quadratic coefficient can not vanish with any real (let alone rational) nonzero $z$.
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$\begingroup$ More generally, of quartics of the form $ax^4+dx+e$, with all coefficients rational, $e$ must be a perfect square for there to be constructible roots (though this is not guaranteed). $\endgroup$ Jan 30 at 0:17