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

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.