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I cannot think of a non-constructible algebraic number of degree $4$ over $\Bbb Q$ so far. I wish if I can find such an example. Could some one tell me some such numbers with justification? Also I would like to know the track of working out such an example. Any help or reference would be appreciate. Thanks in advance!

Y.X.
  • 4,449
  • see https://math.stackexchange.com/questions/490879/is-the-real-root-of-x43-x-3-constructible – user8268 May 28 '17 at 18:01
  • Given a quartic equation with integer coefficients, all its roots are constructible if and only if the resolvent cubic has a rational root. Most quartic equations with integer coefficients do not have that feature, and there you are. – Oscar Lanzi Feb 20 '20 at 02:20

2 Answers2

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Take an $S_4$ extension which is the splitting field of a quartic polynmial, say $f(x)=x^4-4x+2$ with splitting field $K$. If the roots of $f(x)$ were constructible, then all the elements of $K$ would be constructible. For $G$ a Sylow $2$-subgroup of $S_4$, the fixed field of $G$, $K^{G}$ has odd degree over $\Bbb{Q}$, so the elements of $K\setminus\Bbb{Q}$ can't be constructible. From Milne, Remark 3.26, Fields and Galois Theory.

sharding4
  • 5,027
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One problem which leads to such a number is Alhazen's billiard problem.

José Carlos Santos
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  • When was it proven that the solution to Alhazen's problem is not soluble by construction even if the coefficients are constructible? – Michael Ejercito Jan 29 '24 at 17:51
  • @MichaelEjercito It was Jack M. Elkin; see A deceptively easy problem, The Mathematics Teacher, Vol. 58, No. 3, pp. 194–199. – José Carlos Santos Jan 29 '24 at 18:09
  • I am giuessing Elkin figured out that the Alhazen quartic's resolvent cubic does not always have rational solutions. In particular, if the linear coefficient for a quartic equation is $c$ and the quartic coefficient is $a$, for all rational solutions $p/q$ of this cubic, $p$ and $q$ being coprime, $p$ is a factor of $c$ and $q$ is a factor of $a$. – Michael Ejercito Jan 29 '24 at 19:31