Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 20H / b / i / Solution

For $K=\mathbb{Q}(\sqrt{10})$, the ring of integers of a quadratic field isThe elementhas field norm $9-10=-1$, so it is a unit. We claim that

Let $u$ be any unit. Its norm is $\pm 1$. After changing its sign and replacing it by its inverse if necessary, its first real embedding satisfies $u\ge 1$. Choose $n\in \mathbb{Z}$ so thatWrite $v=a+b\sqrt{10}$ with $a,b\in \mathbb{Z}$. Its conjugate is $v^{\prime }=\pm v^{-1}$, so $|v^{\prime }|\le 1$. ThereforeThus $|a|\le 3$ and $|b|\le 1$. The equation $a^2-10b^2=\pm 1$ now shows directly that the only value in $[1,\epsilon )$ is $v=1$: if $b=0$ then $a=\pm 1$, while if $|b|=1$ then $a=\pm 3$ and the positive possibilities are either below $1$ or equal to $\epsilon$. Hence $u={\epsilon }^n$ after normalization, proving the claimed description of the units of Q of square root ten.

Solved by gpt-5.6-sol high.