Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 24 / 2

An extending absolute value on a field is non-Archimedean by Question 1(a). We prove finite-dimensional non-Archimedean norm equivalence over a complete field, using completeness rather than local compactness.

Fix a basis $e_1,\dots ,e_d$ of a finite-dimensional vector space over $K$, and let $\Vert \cdot \Vert$ be a norm homogeneous for the given absolute value on a field and satisfying the ultrametric inequality. Write $\Vert \sum a_i{e}_i{\Vert }_0={\max }_i|a_i|$. The upper boundis immediate. For the lower bound use induction on $d$. The one-dimensional case is homogeneity. For the induction step, the span $W$ of $e_1,\dots ,e_{d-1}$ is complete for its restricted norm by the induction comparison with the coordinate norm and completeness of $K$. A complete vector subspace of a metric space is closed, soFor $v=w+a{e}_d$ with $a\ne 0$, scaling by $a$ gives $\Vert v\Vert \ge |a|\delta$; the same coefficient bound is trivial when $a=0$. AlsoThe induction bound controls every coordinate of $w$ by a constant times $\Vert w\Vert$. Together with the bound on $a$, this proves $\Vert v{\Vert }_0\le C^{\prime }\Vert v\Vert$.

Two extending absolute values on a field on $L$ are norms of this kind on its finite-dimensional vector space over $K$. Hence there is $B>0$ with $|x{|}_2\le B|x{|}_1$ for every $x\in L$. Apply this to $x^N$ and use multiplicativity:Letting $N\to \infty$ gives $|x{|}_2\le |x{|}_1$. Interchanging the two norms gives equality. ThusThe proof covers the trivial base absolute value on a field too. It proves uniqueness, without assuming an existence theorem or compactness of the coordinate sphere $\{v:\Vert v{\Vert }_0=1\}$.

False. This is an example of integral trace need not imply integrality. Take $K={\mathbb{Q}}_p(\theta )$ with ${\theta }^2=p$, the splitting field of $X^2-p$. Its two conjugates are $\theta ,-\theta$, and its normalized discrete valuation satisfies $v_K(\theta )=1$, $v_K(p)=2$, by the Eisenstein polynomial criterion. Forthe field trace is zero, but $v_K(x)=-1$. Therefore $x$ is not in the valuation ring, despite having integral field trace. Cancellation in a sum of conjugates explains why a field trace test fails.

True. Each element of the Galois group preserves the extending absolute value on a field, because composing it with that absolute value on a field gives another extension, equal to the original by part (a). Thus all conjugates of $x$ have the same absolute value on a field. Writing $d=[K:{\mathbb{Q}}_p]$, the field norm givesA unit field norm has absolute value on a field one, forcing $|x|=1$. This means both $x$ and $x^{-1}$ are in the valuation ring, so $x$ is a unit.

Equivalently, with normalized discrete valuations and residue degree $f$,This is unit norm detects units in a local extension.

True. Put $H=\mathrm{Gal}(K/{\mathbb{Q}}_p(\beta ))$. Every $\sigma \in H$ fixes $\beta$, permutes the polynomial's polynomial roots, and is an isometry by the uniqueness proved in part (a). ThereforeThe strict inequalities single out ${\alpha }_1$ as the unique closest polynomial root, so $\sigma ({\alpha }_1)={\alpha }_1$ for every $\sigma \in H$. The Fundamental theorem of Galois theory then givesThis closest-root descent in a Galois extension is the relevant form of Krasner's lemma. If polynomial roots repeat, the strict hypothesis excludes a second occurrence of ${\alpha }_1$; otherwise the same argument applies unchanged.

False. Use the polynomial $X(X^2-p)$, with splitting field $K={\mathbb{Q}}_p(\theta )$, ${\theta }^2=p$, and label the polynomial roots $0,\theta ,-\theta$. Take $\beta =p\theta$. ThenBoth $p-1$ and $p+1$ are p-adic units, also when $p=2$. Hence zero is the uniquely closest polynomial root, but $\beta \notin {\mathbb{Q}}_p(0)={\mathbb{Q}}_p$, because $\theta \notin {\mathbb{Q}}_p$. The polynomial need not be irreducible in the given setup. The inclusion established by Krasner's lemma goes in the direction in part (iii), not its converse.