Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 20F (Number Fields)

Embed the ring of integers into ${\mathbb{R}}^2$ by its two real embeddings. Its image is a full Euclidean lattice of some covolume $d>0$. Fix $C>d$ and consider the symmetric rectangle $|{\sigma }_1(\alpha )|\le T$, $|{\sigma }_2(\alpha )|\le C/T$. Its area is $4C>4d$, so the Minkowski convex body theorem gives a nonzero algebraic integer ${\alpha }_T$ in it. Its algebraic norm is a nonzero integer with $|N({\alpha }_T)|\le C$.

As $T\to \infty$, infinitely many distinct ${\alpha }_T$ must occur: a finite collection of nonzero elements has a positive minimum of $|{\sigma }_2(\alpha )|$, whereas here that coordinate tends to zero. The principal ideals $({\alpha }_T)$ have norm $|N({\alpha }_T)|\le C$. There are only finitely many ideals of bounded norm: they are subgroups of bounded index in a rank-two free abelian group, and integer Hermite normal forms give finitely many possibilities at each such index. Thus infinitely many of these elements generate one ideal. Ratios of two generators of the same ideal are units, because inclusion in both directions supplies inverse algebraic integers. Among infinitely many distinct generators choose two whose ratio is not $\pm 1$. This gives a non-trivial unit without assuming Dirichlet's theorem.

The Dirichlet unit theorem states that for a number field with $r_1$ real embeddings and $r_2$ conjugate pairs of nonreal embeddings,where $\mu (K)$ is its finite cyclic group of roots of unity. Equivalently there exist $r_1+r_2-1$ multiplicatively independent fundamental units such that every unit is a unique product of a root of unity and integer powers of those units.

Since $7\equiv 3(\mathrm{mod}4)$, the ring of integers is $\mathbb{Z}[\sqrt{7}]$. A unit $a+b\sqrt{7}$ has norm $a^2-7b^2=\pm 1$. Norm minus one is impossible: it would give $a^2\equiv -1(\mathrm{mod}7)$, whereas the squares modulo seven are $0,1,2,4$. The element $\epsilon =8+3\sqrt{7}$ has norm one and exceeds one.

Suppose a positive unit $u$ in the chosen real embedding satisfies $1<u<\epsilon$. Its conjugate is $1/u>0$, soThis expression is strictly increasing in $u$, and at $u=\epsilon$ it equals three. Thus the integer $b$ is one or two. But then $a^2=7b^2+1$ is eight or twenty-nine, neither a square. Therefore there is no positive unit strictly between one and $\epsilon$.

For any positive unit $u$, choose an integer $k$ such that $1\le u{\epsilon }^{-k}<\epsilon$. The remaining unit must be one, so $u={\epsilon }^k$. Multiplication by minus one handles negative units. Hencewhich proves that $8+3\sqrt{7}$ is fundamental, using no continued fractions.