Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 20G (Number Fields)

The ring of integers of a quadratic field has integral basis $1,\sqrt{d}$ when the square-free integer $d$ is not congruent to one modulo four. Since $d=2$,This is the ring of integers of Q of square root two.

The Dirichlet unit theorem states that for a number field with $r$ real embeddings and $s$ conjugate pairs of complex embeddings,where $\mu (K)$ is its finite group of roots of unity. Here $K$ is real quadratic, so $(r,s)=(2,0)$ and $\mu (K)=\{\pm 1\}$. Thus its unit group has rank one.

The elementis a unit because its field norm isIt is the smallest unit greater than one in the chosen real embedding. Indeed, write such a unit as $u=a+b\sqrt{2}$ and let $u^{\prime }=a-b\sqrt{2}=N(u)/u$ be its algebraic conjugate. Since $|u^{\prime }|<1<u$, both $a=(u+u^{\prime })/2$ and $b=(u-u^{\prime })/(2\sqrt{2})$ are positive integers. Hence $a,b\ge 1$ and $u\ge 1+\sqrt{2}$. The rank-one logarithmic lattice in Dirichlet's theorem is therefore generated by $\log \epsilon$. Consequently the units of Q of square root two are

Let $\epsilon =1+\sqrt{2}$ and $P=(3+\sqrt{2})$. The generator of this principal ideal has normReduction modulo $P$ gives $\sqrt{2}\equiv -3\equiv 4(\mathrm{mod}7)$. The map below is surjective and has $P$ in its kernel; both the kernel and $P$ have index seven, the latter by the ideal norm, so they are equal. Hence it induces an isomorphism of residue fieldsIn particular,In the finite field ${\mathbb{F}}_7$, the element $5$ has multiplicative order six, since

By part (b), every unit is $(-1{)}^{\delta }{\epsilon }^n$ for $\delta \in \{0,1\}$ and $n\in \mathbb{Z}$. Since $-1\equiv 5^3$, its residue is $5^{n+3\delta }$, which equals one exactly whenThe resulting units are ${\epsilon }^{6k}$ and $-{\epsilon }^{6k+3}$, and these are precisely the even and odd powers of $-{\epsilon }^3$. Thereforewhich is the group of units congruent to one modulo 3 plus square root two.