Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 136 / 3

One useful form of Hensel lemma is this: if $R$ is a complete discrete valuation ring, $f\in R[X]$, andthen there is a unique $a\in R$ such that $f(a)=0$ and $a\equiv a_1(\mathrm{mod}\pi )$. Indeed, after constructing $a_r$ with $f(a_r)\equiv 0(\mathrm{mod}{\pi }^r)$, choose the unique $t$ modulo $\pi$ for whichand put $a_{r+1}=a_r+{\pi }^{r}t$. The resulting Cauchy sequence converges by completeness, and the same first-order congruence proves uniqueness.

Apply this to $X^{p-1}-1$. Every nonzero class in ${\mathbb{F}}_p$ is a simple root, so it has a unique Teichmuller representative in ${\mathbb{Z}}_p$. These give all roots of unity of order prime to $p$. For odd $p$, the group $1+p{\mathbb{Z}}_p$ has no nontrivial torsion: if $u\ne 1$, then the binomial theorem gives $v_p(u^p-1)=v_p(u-1)+1$, which is incompatible with finite $p$-power order. HenceFor $p=2$, the subgroup $1+4{\mathbb{Z}}_2$ is torsion-free by the same argument, while $-1$ supplies the extra torsion element. Thus $\mu ({\mathbb{Q}}_2)=\{\pm 1\}\cong C_2$. This describes the roots of unity in a p-adic field for $K={\mathbb{Q}}_p$.

Solved by gpt-5.6-sol high.

Suppose first that $x^p+y^p=z^p$ with $x,y,z\in {\mathbb{Z}}_p^{\times }$. Divide by $y^p$ and put $u=x/y$, $w=z/y$. Reduction modulo $p$ gives $w\equiv u+1(\mathrm{mod}p)$. Choose an integer $a\equiv u(\mathrm{mod}p)$; then $p\nmid a(a+1)$. A unit's $p$th power modulo $p^2$ depends only on its residue modulo $p$, because $(a+pt{)}^p\equiv a^p(\mathrm{mod}{p}^2)$. Therefore

Conversely, suppose such an integer $a$ exists. The congruence saysFor odd $p$, the $p$th-power map sends $1+p{\mathbb{Z}}_p$ onto $1+p^2{\mathbb{Z}}_p$: under the p-adic logarithm it becomes multiplication by $p$. Hence the displayed ratio is $t^p$ for some $t\in 1+p{\mathbb{Z}}_p$. Takingproduces the required solution in ${\mathbb{Z}}_p^{\times }$.

Solved by gpt-5.6-sol high.