Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 28 / 1 / b

Write $\mathcal{O}=\{x:|x|\le 1\}$. First, $\mathcal{O}$ is compact. A compact neighborhood of zero contains a sufficiently small closed ball $t^m\mathcal{O}$, since $|t{|}^m\to 0$; that ball is closed in the compact neighborhood. Scaling back proves compactness of $\mathcal{O}$. Also $K$ is complete: a Cauchy sequence has a tail in a translate of a compact neighborhood, hence a convergent subsequence, and the Cauchy condition makes the whole sequence converge. This is the completeness of locally compact nontrivially valued fields argument.

The additive subgroup $t\mathcal{O}$ is open: a sufficiently small ball around any of its points stays in $t\mathcal{O}$ by the ultrametric inequality. Thus the compact discrete quotient $\mathcal{O}/t\mathcal{O}$ is finite. Choose a set $A\subset \mathcal{O}$ of representatives, with zero representing the zero coset. For $x\in \mathcal{O}$, recursively choose $a_n\in A$ and $x_{n+1}\in \mathcal{O}$ such that $x_n=a_n+t{x}_{n+1}$, beginning with $x_0=x$. ThenFor arbitrary $x\in K$, multiply first by a large power of $t$ to put it in $\mathcal{O}$, then divide the resulting expansion by that power. This gives the digit expansion with a nonuniformizer.

For uniqueness, take the first exponent $r$ at which two expansions differ. After dividing their difference by $t^r$, all the later terms belong to the closed subgroup $t\mathcal{O}$, whereas $a_r-b_r$ does not, because the representatives are distinct. The difference cannot be zero. Conversely every series with coefficients in $A$ and only finitely many nonzero negative-index terms converges by completeness and $|t|<1$. Every element therefore has a unique expansion $\boxed{x=\sum _{n\ge r}{a}_n{t}^n}$, with the harmless initial zero coefficients understood as part of the same indexed series. No assumption that $t$ is a uniformizer is needed.

The compactness and completeness proved above apply again. The finite ring $\mathcal{O}/t\mathcal{O}$ has a positive integer characteristic $m$, so $m\in t\mathcal{O}$ and $|m|<1$. Prime factorization supplies a prime $p$ with $|p|<1$. In characteristic zero the restriction to $\mathbb{Q}$ is therefore a positive power of the p-adic absolute value, by the preceding classification. Completing $\mathbb{Q}$ embeds the P-adic number field ${\mathbb{Q}}_p$ into the complete field $K$.

Now $p\mathcal{O}$ is open, so $\mathcal{O}/p\mathcal{O}$ is finite. It is an ${\mathbb{F}}_p$-vector space; choose lifts $e_1,\dots ,e_d\in \mathcal{O}$ of a basis. Successively reducing modulo $p\mathcal{O}$ gives, for every $x\in \mathcal{O}$,Convergence in $K$ follows from $|p|<1$. Every element of $K$ becomes integral after multiplication by a power of $p$, so $K={\sum }_j{\mathbb{Q}}_p{e}_j$. Consequently $\boxed{[K:{\mathbb{Q}}_p]<\infty }$: $K$ is a p-adic field. This finite-module proof does not assume discreteness of the original value group in advance.

Take the formal Laurent series field $K=\mathbb{C}((T))$, with $v_T$ the least exponent of a nonzero term and $|x|=2^{-v_T(x)}$. Its value group is $\mathbb{Z}$, so this is a discrete valuation. A Cauchy sequence has an eventually stable coefficient at every exponent, and its tail has a common lower bound on its exponents. The stabilized coefficients therefore give a formal Laurent series field element to which the sequence converges. Hence $K$ is complete.

Its valuation ring is $\mathbb{C}[[T]]$, and its residue field is the infinite field $\mathbb{C}$. If $K$ were locally compact, the previous compact-ball argument would make $\mathbb{C}[[T]]$ compact. Its quotient by the open ideal $T\mathbb{C}[[T]]$ would then be compact and discrete, hence finite, contradicting its identification with $\mathbb{C}$. Thus $\boxed{\mathbb{C}((T))\text{ is complete and discretely valued but not locally compact}}$.