Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 28 / 1

The discrete valuation identities $v(xy)=v(x)+v(y)$ and $v(x+y)\ge \min (v(x),v(y))$ make $R$ closed under products and sums; it contains $0,1$ and additive inverses. As a subring of the field $K$, it is an integral domain.

Its units are exactly the elements of valuation zero: $x^{-1}\in R$ if and only if $v(x)=0$. The remaining elements form the ideal $\mathfrak{m}=\{x:v(x)>0\}$, and every proper ideal is contained in it, since an ideal containing a unit is the whole ring. Hence $R$ is a local ring with this unique maximal ideal.

For a nonzero ideal $I$, the set of valuations of its nonzero elements is a nonempty subset of ${\mathbb{Z}}_{\ge 0}$ and has a least value. Choose $x\in I$ attaining it. Every $y\in I$ satisfies $v(y/x)\ge 0$, so $y/x\in R$ and $I=(x)$. The zero ideal is also principal. This proves that $R$ is a discrete valuation ring, in particular a principal ideal domain.

Choose a uniformizer $\pi$, with $v(\pi )=1$. For $z\in K$, some $n\ge 0$ makes ${\pi }^{n}z\in R$, and $z=({\pi }^{n}z)/{\pi }^n$ is a quotient of elements of $R$. Thus $K$ is the field of fractions of $R$.

Let $n_0=v(x)$ and $y_0={\pi }^{-n_0}x\in R^{\times }$. Choose $a_{n_0}$ to represent the residue of $y_0$, and set $y_1=(y_0-a_{n_0})/\pi \in R$. Continue by choosing $a_{n_0+j}$ representing $y_j$ and putting $y_{j+1}=(y_j-a_{n_0+j})/\pi$. The first digit is not in $\mathfrak{m}$, because $y_0$ is a unit. After $r$ steps,The error has valuation at least $n_0+r$, so the partial sums converge to the existing element $x$. This proves the digit expansion in a discretely valued field without any completeness assumption.

The leading exponent is forced to be $v(x)$. Reduction of ${\pi }^{-n_0}x$ modulo $\mathfrak{m}$ forces the first digit, and the recursive quotient then forces every later digit. Thus the expansion is unique for the chosen representatives. No assumption that the representative of the zero residue is literally zero is needed.

Arbitrary series of this form need not converge in $K$. Their partial sums are Cauchy, so they do converge if $K$ is complete. For a counterexample take $K=\mathbb{Q}$ with the three-adic valuation, $\pi =3$, and $A=\{0,1,2\}$. Hensel lemma applied to $T^2-10$ at the simple residue root $1$ gives a square root of $10$ in ${\mathbb{Q}}_3$. Its three-adic digits form such a series, but its sums cannot converge to an element of $\mathbb{Q}$, since that would give a rational square root of $10$. Convergence to an already given element and convergence of every possible digit sequence are different assertions.