2026-01-25-a-galois-theory-problem
Introduction
As the exam season approaches there is an intense test season preceding it1. One of those tests that I had to take was the one from the Algebra course, which had the following as the problem number three:
Prove that the splitting field L of $X^3+tX-1$ over $\mathbb Q(t)$2 is a Galois extension and that $\mathrm{Gal}(L/Q(t)) \simeq S_3$.
The setting
This is a fairly standard problem: proving that this indeed is an Galois extension is just noting that $0=\mathrm{char}\ \mathbb Q=\mathrm{char}\ \mathbb Q(t)$. Showing that the Galois group is isomorphic to $S_3$ is a bit more complicated. As the polynomial is of degree $3$, we can notice that it suffices to show that
\[|\mathrm{Gal}(L/\mathbb Q(t))|=3!.\]Most of the solutions I saw comprised of two steps:
- Proving that $3\mid [L:\mathbb Q(t)]$ (useful because $[L\colon \mathbb Q(t)]=|\mathrm{Gal}(L/\mathbb Q(t))|$). For example by noticing that $X^3+tX-1$ is irreducible.
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Proving that $2\mid |\mathrm{Gal}(L/\mathbb Q(t))|$. For example by stating the group explicitly or noticing that the discriminant of the polynomial is not a square3.
In this blog post I will show an interesting way to prove the latter.
The fun part
Let us remind ourselves of the construction of $\mathbb Q(t)$:
\[\mathbb Q(t)=\left\{\frac{P(t)}{Q(t)} \mid P, Q \in \mathbb Q[t]\right\}.\]We can notice from the construction that this object would be algebraically the same if in the place of the formal variable $t$ we put any transcendental number $\alpha$.
With that isomorphism we can assume $t$ to be some transcendental number $\alpha$ and thus $\mathbb Q(t) \subseteq \mathbb R$. For some $\alpha$s our original polynomial has only one real solution — this happens for example when the derivative is never zero. Let’s try to find such $\alpha$.
\[\begin{align*} (X^3+\alpha X-1)' &= 0 \iff\\ 3X^2+\alpha &= 0 \iff\\ X &= \pm i\sqrt{\alpha/3}. \end{align*}\]So for any transcendental $\alpha > 0$ (like $e$) our polynomial’s derivative is never $0$ over the reals, so it cannot have more than one real solution.
This in turn means that $\mathbb R \subset L \subset \mathbb C$. And we know that $\overline{(\cdot)}= (z \mapsto \overline z)$ is an $\mathbb R$-automorphism of $\mathbb C$ with the order $2$, and in particular it is an $\mathbb R$-automorphism of $L$ of the order two.
This means that for $\alpha > 0$, we get $2\mid |\mathrm{Gal}(L/\mathbb Q(\alpha))|$ as desired.
Footnotes:
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math courses at polish universities are usually divided into two parts — the lecture and the practicals. The lecture is theory and the practicals are problem-solving. Before being allowed to attempt the oral theory exam one must usually pass the practicals first, and to pass them, you have to write a test. ↩
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by $\mathbb Q(t)$ we mean the field of rational functions over $\mathbb Q$. ↩
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a well known lemma states that if $L$ is the splitting field of the polynomial $P$ of degree $n$ over $F$, then $\mathrm{Gal}(L/F) \subseteq A_n \iff $ the discriminant of $P$ is a square in $F$. ↩