Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 28 / 1

Use the simple-root form of Hensel lemma. Let $R$ be a complete discrete valuation ring, let $\pi$ be a uniformizer, and let $F\in R[T]$. If $\bar{a}\in R/(\pi )$ is a root of $\bar{F}$ with ${\bar{F}}^{\prime }(\bar{a})\ne 0$, then there is exactly one root $a\in R$ reducing to $\bar{a}$.

Choose a lift $a_1$ of $\bar{a}$. Suppose $F(a_r)\in {\pi }^{r}R$. To construct a lift satisfying the next congruence, write $a_{r+1}=a_r+{\pi }^r{t}_r$. The polynomial expansion givesSince $F^{\prime }(a_r)$ is a unit, its residue is invertible in the residue field. There is a unique residue class of $t_r$ making this zero modulo ${\pi }^{r+1}$. Thus $a_{r+1}\equiv a_r(\mathrm{mod}{\pi }^r)$ and $F(a_{r+1})\equiv 0(\mathrm{mod}{\pi }^{r+1})$. This produces a Cauchy sequence in the complete discrete valuation ring. Its limit $a$ satisfies $F(a)=0$ by continuity of polynomial evaluation, and $a\equiv a_1(\mathrm{mod}\pi )$.

For uniqueness, suppose $a,b$ are two such roots. Polynomial division of their difference givesBecause $b-a\in \pi R$, the second factor is a unit. Hence $b-a=0$. The simple-root hypothesis is used both for constructing successive lifts and for this uniqueness argument.

First reduce the polynomial modulo seven. At residues $0,1,\dots ,6$, its values are respectively $4,2,6,1,0,2,6$. Its only residue root is therefore $4$, and $f^{\prime }(4)=45\equiv 3(\mathrm{mod}7)$ is a unit. By Hensel lemma, this residue root lifts uniquely to the 7-adic integers; any other root would have to reduce to the same residue root. Thus there is exactly one root in ${\mathbb{Z}}_7$, congruent to $4$ modulo seven.

Modulo five the values at $0,1,\dots ,4$ are $4,2,1,2,1$, so there is no residue root and hence no root in ${\mathbb{Z}}_5$. Modulo three the only possible residue root is $-1$. But writing $x=-1+3t$ givesThus that residue root cannot lift even modulo nine, and there is no root in ${\mathbb{Z}}_3$. A repeated residue root alone would not establish this obstruction; the second congruence is needed.

Modulo two both residues are roots. The even residue is simple because $f^{\prime }(0)=-3$ is a 2-adic unit, so Hensel lemma gives exactly one even root. For any odd 2-adic integer, $x^3\equiv x(\mathrm{mod}4)$, whence $f(x)\equiv -2x\equiv 2(\mathrm{mod}4)$. No odd root can exist. Consequently the full answer is

The two cosets of $2{\mathbb{Z}}_2$ are clopen sets in the 2-adic integers. The function is constant on each, so it is a locally constant function and therefore continuous. Its values on the nonnegative integers are $f(j)=(-1{)}^j$.

The Mahler coefficient at index $n$ is the nth forward difference operator applied at zero. HereThus the Mahler expansion of the parity function isTo verify the expansion directly, each binomial polynomial $\left(\genfrac{}{}{0ex}{}{x}{n}\right)$ takes values in ${\mathbb{Z}}_2$: it does so at nonnegative integers, which form a dense subset, and polynomial evaluation is continuous into ${\mathbb{Q}}_2$. Hence the nth summand has 2-adic absolute value at most $2^{-n}$ throughout the domain. The ultrametric inequality proves uniform convergence, and the sum is continuous. At each nonnegative integer $m$, the series becomes the finite binomial theorem identity $(1-2{)}^m=(-1{)}^m$. Equality on a dense subset and continuity prove equality everywhere. Finally, the coefficients in any such expansion are determined successively by its values at $0,1,2,\dots$, since $\left(\genfrac{}{}{0ex}{}{m}{n}\right)=0$ for $n>m$ and $\left(\genfrac{}{}{0ex}{}{m}{m}\right)=1$. This also verifies uniqueness without needing the general Mahler theorem.