Connection between polynomial map permuting roots and automorphisms in Galois group

Question

I'm hoping someone can help patch a bit of a gap in my knowledge of Galois theory. As an example, take $x^4 - 5x^2 + 5$, which has a root $\alpha = \sqrt{\frac{5 + \sqrt{5}}{2}}$, and Galois group isomorphic to $\mathbb{Z}_4$. I notice that there is a polynomial map $f(x) = x^3 - 3x$, which has the nice property that $f(\alpha)$, $f(f(\alpha))$, and $f(f(f(\alpha)))$ are exactly the conjugates of $\alpha$. Clearly in $f$ there is a connection to the Galois group here, but what exactly is that connection? Can I explicitly construct an automorphism of the splitting field from $f(x)$?

Answer 1

Yes, $f$ defines an automorphism of the splitting field and hence a generator of the Galois group.

In general, suppose $p(x)$ is an irreducible polynomial over a field $F$ of degree $n$ which has the property that, in its splitting field $K$, there exists a root $\alpha$ of $p$ such that every other root is a polynomial in $\alpha$. This says exactly that

$$K \cong F[\alpha] \cong F[x]/p(x)$$

is already the splitting field of $p$, which is equivalent to the statement that the Galois group of $p(x)$ has order $n$. Note that this is a very special condition; in general the Galois group can have order between $n$ and $n!$.

If furthermore the Galois group is cyclic, let $g$ be a generator of it. Then $g \alpha$ is some polynomial $f(\alpha)$ in $\alpha$. But $\alpha$ can be any root of $p(x)$; said another way, in the isomorphism $F[\alpha] \cong F[x]/p(x)$ we can choose to send $x$ to any root of $p(x)$, not just the chosen root $\alpha$. So what this actually means is that if $\alpha$ is any root of $p(x)$ then so is $f(\alpha)$, and so we can iterate $f$ to obtain all the conjugates $\alpha, f(\alpha), f(f(\alpha)), \dots $ as desired (since $g$ generates the Galois group by hypothesis).

Example 1. The simplest examples of this behavior occur over finite fields $\mathbb{F}_q$; here every irreducible polynomial has cyclic Galois group and it is always generated by the Frobenius map $x \mapsto x^q$.

Example 2. The next simplest examples of this behavior occur for the cyclotomic polynomials $\Phi_p(x) = \frac{x^p - 1}{x - 1}$ over $\mathbb{Q}$ ($p$ a prime), which have cyclic Galois group $\mathbb{Z}/p\mathbb{Z}^{\times}$ acting on the roots via $\zeta_p \mapsto \zeta_p^k$, where $k \in \mathbb{Z}/p\mathbb{Z}^{\times}$ and $\zeta_p$ is any primitive $p^{th}$ root of unity.

Answer 2

I want to bring up a few extras that are too long to fit into a comment. The starting point is that $$\alpha=2\cos(\pi/10).$$ This allows us to unravel the mystery.

The roots of the polynomial $g(x)=x^4-5x^2+5$ are $x_j=2\cos(j\pi/10)$ for $j=1,3,7,9$. In other words, the splitting field of $g$ over $\Bbb{Q}$ is the real subfield of the twentieth cyclotomic field. See this related answer of mine for the derivation of $g(x)$ (actually I needed the reciprocal polynomial there, but the calculation is the same).

This leads us to the following explanation of the automorphisms. The key is the trigonometric formula $$ \cos 3t=4\cos^3t-3\cos t. $$ Multiplying that identity by two allows a rewrite $$ (2\cos3t)=f(2\cos t) $$ with $f(x)=x^3-3x$. Just what the doctor ordered!

We thus see that $$ f(x_1)=x_3,\quad f(x_3)=x_9,\quad f(x_9)=x_7\quad\text{and}\quad f(x_7)=x_1. $$ It follows that the same recipe gives us the Galois group of the real subfield of the cyclotomic field of conductor $m$ whenever $\Bbb{Z}_m^*=\langle 3\rangle\times \langle -1\rangle$. Here $m=20$, but $m=16$ also works.

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This can also be seen as a modification of the recurring theme/exercise that whenever $r$ is a root of a carefully selected cubic, $r^2-2$ is another. The two cubics that appear are $x^3+x^2-2x-1$ and $x^3-3x+1$, with respective splitting fields the real subfields of $\Bbb{Q}(\zeta_7)$ and $\Bbb{Q}(\zeta_9)$. This time the automorphism is based on the trig identity $$(2\cos 2t)=(2\cos t)^2-2.$$ So if $r=2\cos t$ for some angle $t$, then $r^2-2=2\cos2t$.