Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 20H (Number Fields)

Minkowski's lemma states that if $\Lambda$ is a full lattice in ${\mathbb{R}}^n$ and $C\subset {\mathbb{R}}^n$ is measurable, convex, and centrally symmetric, thenimplies that $C$ contains a nonzero point of $\Lambda$.

Let ${\sigma }_1,{\sigma }_2:K\to \mathbb{R}$ be the two real embeddings. The Minkowski embedding of a real quadratic field identifies ${\mathcal{O}}_K$ with a latticeof covolume $\sqrt{|D_K|}$. Choose a constant $c$ satisfyingFor $R>0$, consider the closed diamondIt is convex and centrally symmetric, and its area isMinkowski's lemma supplies a nonzero ${\alpha }_R\in {\mathcal{O}}_K$ whose two embeddings lie in $C_R$.

For $(x,y)\in C_R$, the arithmetic-geometric mean inequality givesSince the coordinate product is the field norm,Finally, $|{\sigma }_2({\alpha }_R)|\le c/R\to 0$ as $R\to \infty$. No fixed nonzero algebraic integer can occur for arbitrarily large $R$, because its second embedding is nonzero. Consequently the elements ${\alpha }_R$ obtained along an unbounded sequence of $R$ contain infinitely many distinct values. Here $N(\alpha )$ denotes the absolute norm, as usual in this inequality.

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.

For $K=\mathbb{Q}(\sqrt{-3})$, putThen the ring of integers of a quadratic field is ${\mathcal{O}}_K=\mathbb{Z}[\omega ]$, andA unit must have norm $1$. Multiplying by four givesso $b\in \{0,\pm 1\}$, and checking these cases gives exactlyThese are precisely the sixth roots of unity, proving the description of the units of Q of square root minus three.

Assume that $\zeta$ is a primitive fifth root of unity. The cyclotomic field $K=\mathbb{Q}(\zeta )$ has degree four and signature $(r_1,r_2)=(0,2)$. The Dirichlet unit theorem therefore gives free rankThe roots of unity in a rational cyclotomic field are the ten elements $\pm {\zeta }^j$, so

For explicit units, $-\zeta$ has order ten, whileHence $1+\zeta$ is a unit. It has infinite order because in the standard complex embeddingwhereas every root of unity has modulus one. Its image in the free $\mathbb{Z}$ factor is therefore nonzero and generates a finite-index subgroup of that factor. Consequentlyhas finite index in ${\mathcal{O}}_K^{\times }$, as described by the unit group of the fifth cyclotomic field.