Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 136 / 1

One form of the Hensel lemma is the following. Let $R$ be a complete discrete valuation ring with maximal ideal $\mathfrak{m}$, and let $f\in R[X]$. If $a_0\in R$ satisfiesthen there is a unique $a\in R$ with $f(a)=0$ and $a\equiv a_0(\mathrm{mod}\mathfrak{m})$.

Define $a_{n+1}=a_n-f(a_n)/f^{\prime }(a_n)$. Since $f^{\prime }(a_n)$ remains a unit, Taylor expansion givesso the valuations of the errors at least double. The corrections tend to zero, making $(a_n)$ a Cauchy sequence; completeness gives a limit $a$, and continuity gives $f(a)=0$. If $a,b$ are two such roots, thenwith $u\in \mathfrak{m}$, so the second factor is a unit and $a=b$.

Put $f(X)=X^3+5X-12$. Any root in ${\mathbb{Q}}_p$ is a p-adic integer: if its valuation were negative, $X^3$ would be the unique term of least valuation.

For $p=2$, reduction modulo two has roots zero and one. The root zero is simple because $f^{\prime }(0)=5$ is odd, so it lifts uniquely. An odd integer $x$ satisfies $x^2\equiv 1(\mathrm{mod}8)$, and henceThus there is no odd $2$-adic root and the number of roots is one.

For $p=3$,All three roots are simple because $f^{\prime }(X)=3X^2+5\equiv 2(\mathrm{mod}3)$. Each lifts uniquely, giving three roots in ${\mathbb{Q}}_3$.

For $p=5$, reduction gives $X^3-2$, whose unique root is $X=3$; it is simple because $3X^2\not\equiv 0(\mathrm{mod}5)$. It lifts uniquely, so there is one root in ${\mathbb{Q}}_5$.

For odd $p$ there is a decompositioninto the valuation factor, the Teichmuller representative factor, and the group of principal units. If $x$ is a $((p-1)m)$th power for every $m$ coprime to $p$, its valuation is divisible by every such $(p-1)m$, and is therefore zero. Its residue in ${\mathbb{F}}_p^{\times }$ is a $(p-1)$st power, hence is one. Thus $x\in 1+p{\mathbb{Z}}_p$.

Conversely, exponentiation by any integer $n$ coprime to $p$ is an automorphism of $1+p{\mathbb{Z}}_p$. This follows either from the principal-unit logarithm, under which it becomes multiplication by $n$, or by applying the Hensel lemma to $Y^n-x$. Since $(p-1)m$ is coprime to $p$, every $x\in 1+p{\mathbb{Z}}_p$ has a $((p-1)m)$th root for every allowed $m$.