# 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 $S_n$ ?

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.

• Nice! LOL! Why did I bother to bound the degree of the splitting field :-) – Jyrki Lahtonen Jun 24 '16 at 6:27
• Thanks! Though I should say for posterity that a student probably learns a lot more about Galois theory from reading your answer than mine. – Sempliner Jun 24 '16 at 6:38
• Not necessarily. Thinking about the elements of the Galois group as permutations of the zeros is at the core here. The group is $S_n$, iff there are no hidden relations among the roots, and they can be permuted in an unconstrained manner. Here you use the constraint that the roots are reciprocals of each other ... – Jyrki Lahtonen Jun 24 '16 at 6:43
• Very true. Though it's interesting to see how a (very closely related) cousin of this idea about algebraic constraints cutting down the size of the possible group appears below in your answer. – Sempliner Jun 24 '16 at 7:00

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 $S_n$.