8
$\begingroup$

The number $b:=\frac{\sqrt{2a}+\sqrt{4\sqrt{a^2-3}-2a}}{2}$ with $a:=\frac{\sqrt[3]{18+2\cdot\sqrt{65}}}{2}+\frac{2}{\sqrt[3]{18+2\cdot\sqrt{65}}}$ is a root of the irreducible polynomial $x^4-6x+3$.The algebraic number $a$ is a root of the irreducible polynomial $2x^3-6x-9$ and therefore not constructible with ruler and compass.

I claim that $b$ isn't constructible either. If this is the case, how can this be shown?

Improve this question
$\endgroup$
5
  • 4
    $\begingroup$ A constructible number necessarily arises from a tower of quadratic extensions (which are automatically Galois), so the Galois group of its minimal polynomial has a composition series consisting of copies of $C_2$. The rest is a straightforward exercise in Galois theory and not appropriate for MO; perhaps you should ask at math.stackexchange.com. $\endgroup$
    – Qiaochu Yuan
    Commented Mar 13, 2012 at 1:08
  • $\begingroup$ @GH: that is false. For example the root of an irreducible polynomial of degree $4$ with Galois group $S_4$ is not constructible. $\endgroup$
    – Qiaochu Yuan
    Commented Mar 13, 2012 at 1:09
  • $\begingroup$ @Qiaochu: You are right. I should have said "if and only if the splitting field of its minimal polynomial has degree equal to a power of 2". So one needs to find the minimal polynomial of $b$ and the degree of its splitting field. $\endgroup$
    – GH from MO
    Commented Mar 13, 2012 at 1:24
  • 2
    $\begingroup$ @Joel: Where in the world is this number/question coming from? It sounds odd to claim a number isn't constructible but then not be sure this is the case. It'd be better to say "I think" or "I suspect" the number is not constructible. By the way, since $A_4$ has no subgroup of index 2, if $f(x)$ is an irreducible quartic in ${\mathbf Q}[x]$ whose Galois group over $\mathbf Q$ is isomorphic to $A_4$, the roots of $f(x)$ are not constructible. An example is $x^4 + 8x + 12$. $\endgroup$
    – KConrad
    Commented Mar 13, 2012 at 23:42
  • 1
    $\begingroup$ The point of my previous comment was that the roots of $f(x)$ are numbers of degree 4 over ${\mathbf Q}$ that are not constructible. The field generated over ${\mathbf Q}$ by a root has no quadratic subfield since that would lead to a subgroup of the Galois group with index 2, and in $A_4$ there are no such subgroups. $\endgroup$
    – KConrad
    Commented Mar 14, 2012 at 1:33

1 Answer 1

Reset to default
7
$\begingroup$

As explained in the comments, one needs to check if the splitting field of $x^4-6x+3$ has degree equal to a power of $2$ or not. The resolvent cubic of this polynomial equals $x^3-12x+36$ which is irreducible (over $\mathbb{Q}$), because it has no root modulo $7$. Hence the splitting field in question contains a subfield of degree $3$, so it cannot have degree equal to a power of $2$. This shows that $b$ is not constructible. You can read about the resolvent cubic here.

Improve this answer
$\endgroup$
3
  • 3
    $\begingroup$ One can also observe that the polynomial has an irreducible cubic factor modulo $7$, so by Dedekind's theorem the Galois group contains a $3$-cycle. $\endgroup$
    – Qiaochu Yuan
    Commented Mar 13, 2012 at 2:04
  • $\begingroup$ @Qiaochu: Thank you! I observed that, too, but I did not know of Dedekind's theorem! I only conjectured this statement quickly, but then I was too sleepy to verify it formally. Do you have a good reference? $\endgroup$
    – GH from MO
    Commented Mar 13, 2012 at 2:07
  • $\begingroup$ Found Dedekind's theorem in jmilne.org/math/CourseNotes/FT400.pdf $\endgroup$
    – GH from MO
    Commented Mar 13, 2012 at 2:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .