Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 26 / 3 / ii / Solution

Logarithms on sufficiently deep units. Suppose $\mathrm{char}K=0$, let $\ell$ be the residue characteristic, and put $e=v(\ell )$. For every integer $r>e/(\ell -1)$ the p-adic logarithm and p-adic exponential function, evaluated in $K$, converge on the required domains:The bounds $v(m!)=e{v}_{\ell }(m!)\le e(m-1)/(\ell -1)$ and $v(m)\le e(m-1)/(\ell -1)$ show that, for $x,y\in {\pi }^r{\mathcal{O}}_K$, every term beyond the linear one has valuation strictly greater than $v(x)$ or $v(y)$. The series therefore converge and preserve these lattices; in particular $\exp (y)\in 1+{\pi }^r{\mathcal{O}}_K$. Their formal composition and addition identities are valid by convergence, giving inverse continuous group homomorphismsThus the logarithm isomorphism on deep principal units isThe right-hand group is torsion-free.

Roots of unity in ${\mathbb{Q}}_p$. For odd $p$, take $\ell =p,e=1,r=1$. The principal unit group $1+p{\mathbb{Z}}_p$ is torsion-free by the logarithm isomorphism on deep principal units. The Teichmuller representative splitting leaves precisely the $p-1$ roots in ${\mu }_{p-1}$.

For $p=2$, take $r=2$. Each odd unit is uniquely $\pm u$ with $u\in 1+4{\mathbb{Z}}_2$, since its residue modulo four is either one or minus one. This latter group is torsion-free. ConsequentlyThis computes the roots of unity in the p-adic numbers.

The units of ${\mathbb{Q}}_2(i)$. Set $\pi =1-i$. Its polynomial $T^2-2T+2$ is Eisenstein, soThe Eisenstein polynomial facts justifying this are stated in the next solution. Write $U_s=1+{\pi }^s{\mathcal{O}}_K$. Reduction gives ${\mathcal{O}}_K^{\times }=U_1$, and the successive quotients of principal-unit groupshave order two for $s\ge 1$. Hence $[U_1:U_3]=4$. The four roots in ${\mu }_4=\{1,i,-1,-i\}$ have distinct images modulo $U_3$, sinceThey exhaust the quotient and intersect $U_3$ trivially. Since $r=3>2/(2-1)$, the logarithm isomorphism on deep principal units gives $U_3\cong ({\mathcal{O}}_K,+)$. Therefore the unit decomposition of the 2-adic Gaussian field isIn particular these four elements are all its roots of unity.

Quadratic extensions. A nonzero element of $K$ is uniquely a power of $\pi$ times a unit, so $K^{\times }\cong {\pi }^{\mathbb{Z}}\times {\mathcal{O}}_K^{\times }$. Squaring on the decomposition above acts as squaring on ${\mu }_4$ and multiplication by two on the additive ring ${\mathcal{O}}_K={\mathbb{Z}}_2\oplus {\mathbb{Z}}_{2}i$. Thus the square-class group of the 2-adic Gaussian field isBy quadratic extensions from square classes, in characteristic different from two the nontrivial square classes classify quadratic extensions $K(\sqrt{a})$: two such extensions are $K$-isomorphic exactly when $a/b$ is a square. There are therefore $\boxed{2^4-1=15}$ quadratic extensions up to $K$-isomorphism.