Galois group of a palindromic polynomial is not $S_n$?

Let $f(x) = a_nx^n+\cdots+a_0 \in \mathbb{Q}[x]$ be a palindromic polynomial; that is, the coefficients of $f$ satisfy $a_n = a_0$, $a_{n-1} = a_1$, and more generally $a_{n-i} = a_i$ for all $0\leq i\leq n$. Is it true that if $n > 2$, then the Galois group of $f$ (i.e. the group $Gal(K/Q)$ where $K$ is a splitting field of $f$) is not $S_n$ ?

Solution 1:

This is actually very fun. So the reason for this is that a palindromic polynomial is symmetric under exchange $r \to r^{-1}$ (i.e. $p(r) = p(r^{-1})$) if $p(r) = 0$. This implies that for all $s \in G \subseteq S_n$ we have that $s(a)s(a^{-1}) = 1$ and thus we know that, for instance, one cannot send $a$ to its inverse while sending $a^{-1}$ to another root, and thus for polynomials of size at least $3$ we have that $(a, a^{-1}, b)$, $b \neq a$ is not a possible Galois group element.

Solution 2:

Yes. This is the case.

If $n$ is even, say $n=2m$, then we can do the following. Induction on $m$ shows that there exists a polynomial $g(x)\in\Bbb{Q}[x]$ of degree $m$ such that $$ x^{-m}f(x)=g(x+\frac1x). $$ We then get the splitting field $L$ of $f$ (inside $\Bbb{C}$) as follows. Let $\beta_1,\beta_2,\ldots,\beta_m$ be the zeros of $g(x)$. Then $K=\Bbb{Q}(\beta_1,\ldots,\beta_m)$ is a splitting field of $g$, and the usual argument shows that $[K:\Bbb{Q}]\le m!$. If we denote by $\alpha_i$ and $\alpha_{i+m}$ the solutions of $$ x+\frac1x=\beta_i,\qquad(*) $$ $i=1,2,\ldots,m$, then the numbers $\alpha_i,i=1,\ldots,n$, are the zeros of $f$. As $\beta_i=\alpha_i+1/\alpha_i$ for all $i=1,\ldots,n$, we see that

• $K\subset L=\Bbb{Q}(\alpha_1,\alpha_2,\ldots,\alpha_n)$, and
• $L=K(\alpha_1,\alpha_2,\ldots,\alpha_m)$.

So we get $L$ by adjoining a carefully selected half of the zeros of $f$ to $K$. But the equations $(*)$ are all quadratic. Therefore $[L:K]\le 2^m$.

Consequently $$ [L:\Bbb{Q}]\le 2^m\cdot m!. $$ As $2^m\cdot m!<(2m)!$ when $m>1$, we can conclude that the Galois group of $f$ must be a proper subgroup of $S_{2m}=S_n$.

If $n$ is odd, then the palindromic condition implies that $f(-1)=0$. Therefore $f$ cannot be irreducible, and hence we can immediately conclude that its Galois group is a proper subgroup of $S_n$.