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.
Best Answer
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 $187/54 \pm \sqrt{-1957/108}$ . . .]