Aesthetic proofs that involve Field Theory / Galois Theory

Question

I am preparing for an oral exam on Abstract Algebra, especially Field Theory and Galois Theory.

Now, I'm looking for some aesthetic proofs that involve Galois Theory/Field Theory for two reasons.

1. I may be asked to point out the basic concepts of Galois Theory and their fields of applications and consequences. Therefore it might be useful to know an non-standard example.

2. I am just interested in the field of Abstract Algebra and I'm looking forward to find some topics and fields, in which I can intensify my knowledge.

I know already two very basic examples:

1. The application of Field Theory to clarify the classical antique problems on Straightedge and Compass Construction (Squaring the Circle, Doubling the Cube, Angle Trisection, Construction of a regular Polygon).

2. The application of Galois Theory to determine whether a polynomial is solvable in radicals.

Which further aesthetic proofs are there, that involve the basic concepts and theorems of Field Theory / Galois Theory? Of course, the number of such proofs is huge, so I'm looking for good examples in the sense pointed out above.

Many thanks in advance.

Answer 1

I'll separate out examples in different answers as this is a big-list question. I gave what I think is the FTA proof Dietrich alludes to here. Quote:

Suppose $K$ is a Galois extension of $\mathbb{R}$. We'll aim to show that either $K = \mathbb{R}$ or $K = \mathbb{C}$. (In particular, $\mathbb{C}$ itself must therefore be algebraically closed.) Let $G$ be its Galois group and let $H$ be the Sylow $2$-subgroup of $G$.

By Galois theory, $K^H$ is an odd extension of $\mathbb{R}$. But $\mathbb{R}$ has no nontrivial odd extensions: any such extension has primitive element something with an odd degree minimal polynomial over $\mathbb{R}$, but any such polynomial has a root by the intermediate value theorem. Hence $K^H = \mathbb{R}$, or equivalently $H = G$, so $G$ has order a power of $2$.

But now $K$ is an iterated quadratic extension of $\mathbb{R}$, and it's easy to explicitly show using the quadratic formula that the only nontrivial quadratic extension of $\mathbb{R}$ is $\mathbb{C}$, which itself has no nontrivial quadratic extensions.

One of the many lovely features of this proof is that it reveals that the only analytical / topological fact you need about $\mathbb{R}$ to prove the FTA is that every polynomial of odd degree has a root. Generally speaking you can classify FTA proofs by what fundamental analytical / topological fact they use; see the old MO question listing proofs of the FTA for more. This proof also appears there (I probably learned it there!) and is attributed to Emil Artin.

Answer 2

The so-called irreducible case of the cubic. The so-called Cardan formula for the cubic gives complex expressions for the roots, when there are three real roots and the coefficients are all real. The form is: $$\sqrt[3]{\alpha}+\sqrt[3]{\alpha^*}$$ so we have a sum of two complex conjugates. Rafael Bombelli pointed this out in 1572.

Perhaps the first instance of Painlevé's adage (often attributed to Hadamard), "between two truths of the real domain, the easiest and shortest path quite often passes through the complex domain".

But in this case, the only route passes through the complex domain: Galois theory proves that if $f(x)$ is an irreducible cubic with rational coefficients and three real roots, then it is impossible to find a root of $f(x)$ via real radicals. The basic idea of the proof: the splitting field of $f(x)$ has a three-fold symmetry, which cannot happen inside $\mathbb{R}$ since $\mathbb{R}$ contains only one cube root of unity. Cox's Galois Theory (for example) contains a formal proof.

Answer 3

You mention the application to the classical geometrical constructions. There is a modern addition to this: origami constructions.

Cox's Galois Theory (Chapter 1o) gives the details. Here is the main result:

Theorem 10.3.6. Let $\alpha\in\mathbb{C}$ be algebraic over $\mathbb{Q}$ and let $\mathbb{Q}\subset L$ be the splitting field of the minimal polynomial of $\alpha$ over $\mathbb{Q}$. Then $\alpha$ is an origami number if and only if $[L:\mathbb{Q}]=2^a 3^b$ for some integers $a,b\geq0$

Two equivalent ways of describing the field of origami numbers: they are the numbers that can be constructed with a marked ruler; they are the numbers that can be constructed using the intersections of conics.

Answer 4

You can use Galois theory over finite fields to prove the following congruence for the Fibonacci numbers:

$$F_{p - \left( \frac{p}{5} \right)} \equiv 0 \bmod p$$

where $p$ is prime and $\left( \frac{p}{5} \right)$ is the Legendre symbol. I give the proof here. Quote:

Recall that $$F_n = \frac{\phi^n - \varphi^n}{\phi - \varphi}$$

where $\phi, \varphi$ are the two roots of $x^2 = x + 1$. Crucially, this formula remains valid over $\mathbb{F}_{p^2}$ where $p$ is any prime such that $x^2 = x + 1$ has distinct roots, thus any prime not equal to $5$. We distinguish two cases:

• $x^2 = x + 1$ is irreducible. This is true for $p = 2$ and for $p > 2, p \neq 5$ it's true if and only if the discriminant $\sqrt{5}$ isn't a square $\bmod p$, hence if and only if $\left( \frac{5}{p} \right) = -1$, hence by QR if and only if $\left( \frac{p}{5} \right) = -1$. In this case $x^2 = x + 1$ splits over $\mathbb{F}_{p^2}$ and the Frobenius map $x \mapsto x^p$ generates its Galois group, hence $\phi^p \equiv \varphi \bmod p$. It follows that $\phi^{p+1} \equiv \phi \varphi \equiv -1 \bmod p$ and the same is true for $\varphi$, hence that $F_{p+1} \equiv 0 \bmod p$.

• $x^2 = x + 1$ is reducible. This is false for $p = 2$ and for $p > 2, p \neq 5$ it's true if and only if $\left( \frac{p}{5} \right) = 1$. In this case $x^2 = x + 1$ splits over $\mathbb{F}_p$, hence $\phi^{p-1} \equiv 1 \bmod p$ and the same is true for $\varphi$, hence $F_{p-1} \equiv 0 \bmod p$.

The case $p = 5$ can be handled separately. Maybe this is slightly ugly, though.