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.



Is Lehmer's polynomial solvable?
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 $187/54 \pm \sqrt{-1957/108}$ . . .]


