Pythagoras using Hilbert 90

Sat, 08 Jun 2024 12:33:01 +0000

Dummit and Foote has a fun way of classifying Pythagorean triples. It is a classic fact that if $(a, b, c)$ is a reduced Pythagorean triple then there exists integers $m, n$ such that

$$ \begin{align*} a &= m^2 - n^2, \ b &= 2mn, \ c &= m^2 + n^2. \end{align*} $$

To prove this using Hilbert’s 90, first we change the problem to rational numbers i.e. let $a, b\in \mathbb{Q}$ be such that $a^2 + b^2 = 1$. We want to show that there exists $m, n \in \mathbb{Q}$ such that $$ \begin{align*} a &= \frac{m^2 - n^2}{m^2 + n^2}, \ b &= \frac{2mn}{m^2 + n^2}. \end{align*} $$

Abstract Algebra on PostDoc Problems

Pythagoras using Hilbert 90