Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 28 / 3

Any extending absolute value is also non-Archimedean. Indeed its values on integers are the given base-field values, bounded by one, so the integer criterion for a non-Archimedean absolute value applies. For the criterion's sufficiency, the binomial theorem gives $|x+y{|}^n\le (n+1)\max (|x|,|y|{)}^n$; taking $n$th roots yields the ultrametric inequality.

Fix a $K$-basis of $L$ and its coordinate maximum norm. Every extending absolute value is a vector-space norm satisfying $\Vert av\Vert =|a|\Vert v\Vert$. The finite-dimensional non-Archimedean norm equivalence over a complete field makes it equivalent to that coordinate norm. Here is the relevant proof: induct on dimension. The span $W$ of all but the last basis vector $e$ is complete by the induction hypothesis, hence closed in the given norm. Therefore $\delta ={\inf }_{w\in W}\Vert e-w\Vert >0$. For $v=w+ae$, one has $|a|\le \Vert v\Vert /\delta$ and $\Vert w\Vert \le \max (\Vert v\Vert ,|a|\Vert e\Vert )$. The induction bound on $W$ bounds all its coordinates by a constant times $\Vert v\Vert$. The reverse bound follows from the triangle inequality. This proof applies to every complete base field here, without a local compactness assumption.

Thus two extending absolute values satisfy $c|x{|}_1\le |x{|}_2\le C|x{|}_1$ for fixed positive constants. Apply this to $x^n$ and take $n$th roots:Letting $n\to \infty$ proves equality. The bounded comparison forces equality of multiplicative absolute values argument gives the required unique extension of an absolute value to a finite extension, without proving existence.

For a nonzero Laurent series let ${\mathrm{ord}}_t$ be its least exponent. Choose $|x|=p^{-{\mathrm{ord}}_t(x)}$ and $|0|=0$. This is a multiplicative Non-Archimedean absolute value, and its valuation ring is ${\mathbb{F}}_p[[t]]$.

TakeIf ${\gamma }_1,\dots ,{\gamma }_f$ is an ${\mathbb{F}}_p$-basis of ${\mathbb{F}}_{p^f}$, the elements ${\gamma }_i{u}^j$, $1\le i\le f$, $0\le j<e$, form a $K$-basis of $L$. To see this, group the exponents of each Laurent series modulo $e$ and then expand its coefficients in the constant-field basis. Uniqueness of those two expansions proves linear independence. Thus $[L:K]=ef$.

Its valuation ring is ${\mathbb{F}}_{p^f}[[u]]$, with maximal ideal $(u)$ and residue field ${\mathbb{F}}_{p^f}$. Moreover ${\mathrm{ord}}_u(t)=e$, so the ramification index is $e$ and the residue degree is $f$. The unique extending absolute value is explicitlyThis restricts to the chosen absolute value because ${\mathrm{ord}}_u(x)=e{\mathrm{ord}}_t(x)$ on $K$. The prescribed ramification and residue degree for a Laurent series field construction works even when $p\mid e$; separability is not required in the question.