Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 136 / 3

For sufficiently large $r$, the convergent p-adic logarithm and p-adic exponential series are inverse homomorphismsTheir identities $\log (xy)=\log x+\log y$ and $\exp (x+y)=\exp x\exp y$ follow first formally and then by convergence. Multiplication by ${\pi }^r$ identifies $({\mathcal{O}}_K,+)$ with $({\pi }^r{\mathcal{O}}_K,+)$. Since $(1+{\pi }^r{\mathcal{O}}_K)$ has finite index in ${\mathcal{O}}_K^{\ast }$, the conclusion follows.

Solved by gpt-5.6-sol high.

Since ${\bar{a}}^{q-1}=1$ in the residue field, $u=a^{q-1}\in 1+\pi {\mathcal{O}}_K$. Powers $u^{q^n}$ tend to one, by the binomial theorem initially and the $p$-adic logarithm once they enter its convergence domain. Therefore $a^{q^n}$ is Cauchy; let its limit be $\omega$. Reduction modulo $\pi$ gives $\bar{\omega }=\bar{a}$, whileThis is the Teichmuller representative of $\bar{a}$.

Solved by gpt-5.6-sol high.

For odd $p$, ${\mathbb{Q}}_p$ contains exactly the $p-1$ Teichmuller roots of unity; for $p=2$ it contains $\{\pm 1\}$, so there are two. For odd $p$, adjoining ${\zeta }_p$ adds the $p$ roots of unity of $p$-power order and no primitive $p^2$th root, because the latter would enlarge the degree by $p$. Combining the coprime-order groups givesFor $p=2$, ${\zeta }_2=-1$ already lies in ${\mathbb{Q}}_2$, so the answer remains two.

Solved by gpt-5.6-sol high.