Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 136 / 4

One strong form of the Hensel lemma is this: if a complete discretely valued field $K$, a polynomial $f\in {\mathcal{O}}_K[X]$, and $a\in {\mathcal{O}}_K$ satisfythen there is a unique root $\alpha$ in the ball $v(\alpha -a)>v(f^{\prime }(a))$.

Set $a_{n+1}=a_n-f(a_n)/f^{\prime }(a_n)$. Taylor expansion shows that the valuation of the error at least doubles at each step, while $v(f^{\prime }(a_n))$ remains constant. Thus the corrections tend to zero geometrically, so completeness gives a limit $\alpha$. Continuity gives $f(\alpha )=0$. Applying the same Taylor estimate to two roots in the stated ball proves uniqueness. This is Newton iteration over a valued field.

Write $x={\pi }^{r}u$ with $r\in \mathbb{Z}$ and $u\in {\mathcal{O}}_K^{\times }$. If $x$ is an $m$th power, then $m$ divides $r$. Divisibility by infinitely many $m$ forces $r=0$, so $x$ is a unit.

Conversely, if $m$ is coprime to both the residue characteristic $p$ and $q-1$, exponentiation by $m$ is an automorphism on the finite residue-unit group and on every principal unit quotient; equivalently, use Hensel's lemma on $Y^m-u$. Hence every unit is an $m$th power for infinitely many such $m$. Thereforeas recorded by elements that are powers of infinitely many degrees in a local field.

Put $Y=X^2$. Modulo two, $Y^2+9Y-2$ has the two simple roots zero and one, so Hensel lemma lifts them to roots $y_0,y_1\in {\mathbb{Z}}_2$, with $v_2(y_0)=1$ and $y_1\equiv 1(\mathrm{mod}8)$. The 2-adic unit criterion makes $y_1$ a square, so $X^2-y_1$ splits into two linear factors. The polynomial $X^2-y_0$ is Eisenstein and remains irreducible. Thus $X^4+9X^2-2$ has three irreducible factors, of degrees $1,1,2$, as in factorization of X4 plus 9X2 minus 2 over the 2-adic numbers.