12
$\begingroup$

The degree 10 polynomial $$\displaystyle x^{10} + x^9 - x^7 - x^6 - x^5 - x^4 - x^3 + x + 1$$

given by D.H. Lehmer in 1933 has the property that its largest real root, $\beta = 1.176280 \cdots$ is the smallest known Salem number. Moreover, it is a folklore conjecture that $\beta$ is in fact the smallest Salem number.

However, it is curious that one cannot find a reference for the explicit value of $\beta$. I suspect that this is because Lehmer's polynomial is not solvable. Is this the case? If so, is there a reference/relatively simple argument? Furthermore, if Lehmer's polynomial is in fact not solvable, then what is its Galois group?

Thanks for your time.

Improve this question
$\endgroup$
12
  • 10
    $\begingroup$ Magma quickly computes that it's a nonsolvable group of order 1920. $\endgroup$
    – rlo
    Commented Dec 1, 2015 at 21:44
  • 2
    $\begingroup$ @PerAlexandersson I was hoping that such a nice polynomial would be solvable, and therefore one can find an explicit algebraic expression for $\beta$ $\endgroup$
    – Stanley Yao Xiao
    Commented Dec 1, 2015 at 21:57
  • 2
    $\begingroup$ Quintic equations are solvable by radicals and modular functions. As for Galois groups $G$ of irreducible polynomials $P \in \mathbb{Z}[x]$ with small Mahler measure, there are results to the effect that either $G$ is "large" or $P = Q(x^m)$ for a large $m$. So no surprise that the polynomial with smallest known Mahler measure has a large Galois group (as large as possible for a reciprocal polynomial). $\endgroup$
    – Vesselin Dimitrov
    Commented Dec 1, 2015 at 22:37
  • 3
    $\begingroup$ @rlo I am surprised that the Galois group is not the whole hyperoctahedral group! $\endgroup$
    – Igor Rivin
    Commented Dec 1, 2015 at 23:22
  • 4
    $\begingroup$ @igor If you run computations on Galois groups of reciprocal polynomials of degree 2n, you'll find that having a "half full" Galois group is quite common. The "full" group is $C_2 \wr S_n$, corresponding to the "full permutation module" for $S_n$ over $\mathbb{F}_2$, whereas the "half full" group corresponds to the "standard module" (of dimension $n-1$). It was shown by McKee and Christopoulos that these two situations are the only ones that can happen for Salem numbers (although the $S_n$ can be replaced by a smaller transitive group of degree $n$. $\endgroup$
    – Bobby Grizzard
    Commented Dec 7, 2015 at 14:04

1 Answer 1

Reset to default
26
$\begingroup$

Lehmer's polynomial is symmetrical, so $x + x^{-1} =: y$ satisfies a polynomial of half the degree. It turns out that this is the quintic $y^5 + y^4 - 5y^3 - 5y^2 + 4y + 3 = 0$, whose Galois group is the unsolvable $S_5$ (for instance, it's irreducible $\bmod 2$ and decomposes as $(y^2-2y-1)(y^3-2y^2+2y+2)$ $\bmod 5$, so the Galois group is a subgroup of $S_5$ of order divisible by $30$ that contains an odd permutation, and the only such subgroup is $S_5$ itself). Hence Lehmer's polynomial is not solvable either.

[It turns out that $y$ generates the totally real quintic field of third-smaller discriminant $36497$. By the way, even if a polynomial is solvable, exhibiting a solution in radicals may not be of much use; for instance, the Salem root of $x^8 - x^5 - x^4 - x^3 + 1 = 0$ satisfies $x + x^{-1} = y$ where $y$ is a solution of the quartic $y^4 - 4y^2 - y + 1 = 0$ with Galois group $S_4$, but even though this group (and thus also the octic $x^8 - x^5 - x^4 - x^3 + 1$) is solvable I doubt that you really want to ponder the explicit formula for $x$ involving things like the cube roots of $187/54 \pm \sqrt{-1957/108}$ . . .]

Improve this answer
$\endgroup$
6
  • 3
    $\begingroup$ "a subgroup of $S_5$ of order divisible by $30$, and the only such subgroup is $S_5$ itself".. I guess such things happen even to the best of us :) $\endgroup$
    – Jarek Kuben
    Commented Dec 1, 2015 at 22:53
  • 1
    $\begingroup$ Oops, corrected now. I noticed this soon after posting (but by then I was e-incommunicado for an hour or so). I had already checked that the Galois group is not contained in $A_5$ (either because the discriminant is not square, or because the cycle structure $\{2,3\}$ is odd), and forgot to mention this argument that the Galois group cannot be $A_5$. Not that it matters for the present application because $A_5$ is also unsolvable . . . $\endgroup$
    – Noam D. Elkies
    Commented Dec 1, 2015 at 23:58
  • $\begingroup$ Why is the homomorphism from the Galois group of the degree 10 Lehmer polynomial P(x) given by the OP to the Galois group of your degree 5 polynomial Q(y) is surjective? $\endgroup$
    – YCor
    Commented Dec 2, 2015 at 1:17
  • 8
    $\begingroup$ If you insist on fully expanding them in all their glory, expressions such as the Salem root $x$ in the second paragraph are indeed rather fearsome, but they are not so bad if obtained in a sequence of steps, such as (I hope I got this right): $$ \eqalign{a_1 &= (748 + 12 i \sqrt{5871})^{1/3}\cr a_2 &= \sqrt{a_1 + 16 + 112/a_1}\cr a_3 &= \sqrt{12 \sqrt{6}/a_2 + 32 - a_1 - 112/a_1}\cr y &= (a_2 + a_3) \sqrt{6}/12\cr x &= \left(y + \sqrt{y^2-4}\right)/2\cr}$$ $\endgroup$
    – Robert Israel
    Commented Dec 2, 2015 at 1:35
  • 3
    $\begingroup$ @YCor already answered by David Speyer, but not necessary for the original question once you know that solvable Galois $\Leftrightarrow$ solvable by radicals: clearly if $x$ satisfied this condition then so would $y = x + x^{-1}$. $\endgroup$
    – Noam D. Elkies
    Commented Dec 2, 2015 at 1:40

You must log in to answer this question.

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