Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 27 (Topics in Number Theory)

Every nonzero P-adic number has a unique decomposition $z=p^{m}u$ with $m\in \mathbb{Z}$ and $u\in {\mathbb{Z}}_p^{\times }$. The group homomorphism kernel of reduction from the unit group onto ${\mathbb{F}}_p^{\times }$ is $1+p{\mathbb{Z}}_p$, the group of principal units. Each nonzero residue class has a unique lift $\omega$ satisfying ${\omega }^{p-1}=1$, by the simple-root Hensel lemma proved in question 3. Uniqueness makes these lifts multiplicative. They form the cyclic group ${\mu }_{p-1}$, and every unit has a unique factorizationThus ${\mathbb{Q}}_p^{\times }=p^{\mathbb{Z}}\times {\mu }_{p-1}\times (1+p{\mathbb{Z}}_p)$.

For odd $p$, the p-adic logarithm and p-adic exponential function give inverse group isomorphisms $1+p{\mathbb{Z}}_p\leftrightarrow p{\mathbb{Z}}_p$. To verify the convergence domain, for $t\in p{\mathbb{Z}}_p$ the p-adic logarithm terms have valuations $n{v}_p(t)-v_p(n)\to \infty$, and the exponential terms have valuations $n{v}_p(t)-v_p(n!)\to \infty$, since $v_p(n!)\le (n-1)/(p-1)$ and $p>2$. In both series every term after the linear one has strictly larger valuation than the linear term. Consequently $\log (1+t)\in p{\mathbb{Z}}_p$ and $\exp (t)\in 1+p{\mathbb{Z}}_p$. The formal identities $\log (vw)=\log v+\log w$ and $\exp (\log v)=v$, $\log (\exp t)=t$ hold on these convergent domains, as follows by multiplying the convergent series or passing to their formal identities termwise.

Choose a generator $\zeta$ of ${\mu }_{p-1}$. An explicit isomorphism isThe inverse reads off the valuation, the root-of-unity unit factor, and $p^{-1}\log (v)$. Both directions are continuous with the standard product topology, so this is also a topological group isomorphism. The odd-prime condition is essential for using all of $1+p{\mathbb{Z}}_p$ as the p-adic logarithm-isomorphism domain.

For a finite extension of fields complete for a Non-Archimedean absolute value, use the unique extended field absolute value. The extension is unramified when its residue extension is a separable field extension andEquivalently, it has ramification index one, a separable field extension of residue fields, and no defect. For general nondiscrete valued fields, merely saying that the ramification index is one is insufficient; the displayed degree equality is part of the unramified condition.

Put $n=[L:K]=[k_L:k_K]$. Since $\bar{x}$ generates the residue extension, $1,\bar{x},\dots ,{\bar{x}}^{n-1}$ are linearly independent over $k_K$. Their lifts $1,x,\dots ,x^{n-1}$ are linearly independent over $K$: a nonzero relation could be divided by a coefficient of largest field absolute value, giving an integral relation whose reduction has at least one nonzero coefficient, contrary to residue independence. They therefore form a $K$-basis of $L$.

For any $y={\sum }_{j=0}^{n-1}{a}_j{x}^j$, choose $a_r$ of largest field absolute value. All coefficients of $y/a_r$ belong to ${\mathcal{O}}_K$, and at least one is a unit. Its reduction is a nonzero linear combination of the residue basis, so $y/a_r$ is a unit of ${\mathcal{O}}_L$. This proves the residue-basis norm formula for an unramified extensionHence $y\in {\mathcal{O}}_L$ exactly when every $a_j\in {\mathcal{O}}_K$. We concludeThe last equality also uses the fact that every polynomial in the integral element $x$ is integral. The proof requires neither a uniformizer nor discrete valuation, and applies to every residue generator specified in the question.

The Mahler theorem states that a function $f:{\mathbb{Z}}_p\to {\mathbb{Q}}_p$ is continuous if and only if it has a unique expansionand the expansion converges uniformly. Moreover $\Vert f{\Vert }_{\infty }={\sup }_n|c_n{|}_p$; in particular $f$ takes values in ${\mathbb{Z}}_p$ exactly when all $c_n$ belong to ${\mathbb{Z}}_p$. Here $\left(\genfrac{}{}{0ex}{}{x}{0}\right)=1$ and $\left(\genfrac{}{}{0ex}{}{x}{n}\right)=x(x-1)\cdots (x-n+1)/n!$. These binomial polynomials are continuous and integer-valued on ${\mathbb{Z}}_p$, since they are integer-valued on the dense nonnegative integers. The coefficient condition $c_n\to 0$ therefore makes the series uniformly convergent.

Evaluate at a nonnegative integer $m$. Terms with $n>m$ vanish, giving the finite triangular relation $f(m)={\sum }_{n=0}^m\left(\genfrac{}{}{0ex}{}{m}{n}\right)c_n$. Binomial inversion gives the Mahler coefficientsTo obtain the generating function, multiply the last finite identity by $T^n/n!$ and compare coefficients. Setting $n=j+k$ yieldsThe Mahler coefficient exponential generating function is an identity in ${\mathbb{Q}}_p[[T]]$: each coefficient is a finite sum. It does not assert convergence for every $p$-adic substitution for $T$.

Let $c_n$ be the Mahler coefficients of $f$. Define its discrete antiderivative byBecause $c_n\to 0$ and the binomial polynomials have field absolute value at most one on ${\mathbb{Z}}_p$, this series converges uniformly to a continuous ${\mathbb{Z}}_p$-valued function. Every summand vanishes at zero. Pascal's identity gives $\left(\genfrac{}{}{0ex}{}{x+1}{n+1}\right)-\left(\genfrac{}{}{0ex}{}{x}{n+1}\right)=\left(\genfrac{}{}{0ex}{}{x}{n}\right)$, so taking differences through the uniformly convergent series proves $g(x+1)-g(x)=f(x)$. For nonnegative integers, induction from $g(0)=0$ gives precisely $g(m)={\sum }_{j=0}^{m-1}f(j)$. Thus this is the required continuous extension, unique because the nonnegative integers are dense in ${\mathbb{Z}}_p$.

The Mahler expansion of $g$ has coefficient zero in degree zero and coefficient $c_{m-1}$ in degree $m\ge 1$. This discrete antidifferentiation on the p-adic integers is the mechanism behind the next part.

Take an arbitrary $f\in C({\mathbb{Z}}_p,{\mathbb{Q}}_p)$. Compactness of ${\mathbb{Z}}_p$ bounds its field absolute value, so some $p^{m}f$ takes values in ${\mathbb{Z}}_p$. Apply part (ii) to obtain a continuous $h$ with $h(x+1)-h(x)=p^{m}f(x)$. Then $g=p^{-m}h$ belongs to the same continuous-function space and satisfies $f=g(\cdot +1)-g$.

Translation invariance by one alone now gives $L(f)=L(g(\cdot +1))-L(g)=0$. Since $f$ was arbitrary,This proves translation-invariant linear forms on p-adic continuous functions vanish without assuming that the linear form itself is continuous. A proof using only density of polynomials would not suffice for an arbitrary, potentially discontinuous linear form; surjectivity of the forward difference operator on the entire function space is what is needed.

We first prove the simple-root Hensel lemma directly. Put $x_1=x$. Suppose $x_n\equiv x(\mathrm{mod}\pi )$ and $f(x_n)\equiv 0(\mathrm{mod}{\pi }^n)$. The derivative remains a unit since its reduction agrees with $f^{\prime }(x)$. Choose $t_n\in {\mathcal{O}}_K$ whose residue solvesand set $x_{n+1}=x_n+{\pi }^n{t}_n$. Polynomial expansion gives $f(x_{n+1})=f(x_n)+{\pi }^n{t}_n{f}^{\prime }(x_n)+{\pi }^{2n}B$ with $B\in {\mathcal{O}}_K$. Since $2n\ge n+1$, the new value vanishes modulo ${\pi }^{n+1}$. The sequence is Cauchy because $v(x_{n+1}-x_n)\ge n$, and completeness supplies $y\in {\mathcal{O}}_K$ with $y\equiv x(\mathrm{mod}\pi )$. Continuity gives $f(y)=0$.

For uniqueness, if $y,z$ are two roots in that residue class, thenThe factor in parentheses is a unit, since $f^{\prime }(z)$ is a unit and $y-z\in \pi {\mathcal{O}}_K$. Thus $y=z$. This proves existence and uniqueness of the specified lifted root.

Now suppose the residue field has $q=p^r$ elements and put $e=v(p)$. Every element of $k_K^{\times }$ is a simple root of $T^{q-1}-1$. The Hensel lemma therefore lifts its $q-1$ roots uniquely to a subgroup of the roots of unity in $K$. These are the Teichmuller representatives and give all the prime-to-$p$ torsion. To verify the latter assertion, if a root of unity $u\equiv 1$ has order $m$ prime to $p$, write $u=1+t$. In $u^m-1=mt+\left(\genfrac{}{}{0ex}{}{m}{2}\right)t^2+\cdots$, the first term has valuation $v(t)$ and all later terms have greater valuation, so the sum cannot vanish unless $t=0$. Reduction is therefore injective on prime-to-$p$ torsion, whose order divides $q-1$.

Any root of unity has a unique decomposition into a $p$-power-order factor and a prime-to-$p$ factor, by splitting its finite order using coprime powers. For a primitive $p^a$th root $\zeta$, $a\ge 1$, reduction in characteristic $p$ gives $\bar{\zeta }=1$. The cyclotomic identityshows that all factors have the same valuation: $({\zeta }^j-1)/(\zeta -1)=1+\zeta +\cdots +{\zeta }^{j-1}$ reduces to $j\ne 0$. Consequently the valuation of a primitive p-power root of unity isThe left side is a positive integer. Hence $p^{a-1}(p-1)\le e$, so the possible $p$-power orders are bounded. Combining this bound with the prime-to-$p$ result bounds all root orders by one integer, proving finiteness. A finite multiplicative subgroup of a field is cyclic, and it contains exactly the full prime-to-$p$ subgroup of order $q-1$. ThusIf $e<p-1$, even a primitive $p$th root would violate the valuation formula, so $s=0$.

At the boundary $e=p-1$, assume first that $p$ is odd. Take $K_{+}={\mathbb{Q}}_p({\zeta }_p)$. The polynomial ${\Phi }_p(1+T)=p+\left(\genfrac{}{}{0ex}{}{p}{2}\right)T+\cdots +T^{p-1}$ is Eisenstein, so $K_{+}/{\mathbb{Q}}_p$ is totally ramified of degree $p-1$ and $v_{K_{+}}(p)=p-1$. It contains a primitive $p$th root, while the valuation formula excludes primitive $p^2$th roots. Thus $s=1$.

For the other example, take $K_0={\mathbb{Q}}_p(\pi )$ with ${\pi }^{p-1}=p$. The polynomial $T^{p-1}-p$ is Eisenstein, giving the same ramification index and residue field ${\mathbb{F}}_p$. If a primitive $p$th root belonged to $K_0$, the valuation formula would give ${\zeta }_p-1=u\pi$ with $u$ a unit. Substitute this into ${\Phi }_p(1+({\zeta }_p-1))=0$ and divide by $p$. All intermediate terms reduce to zero, leaving $1+{\bar{u}}^{p-1}=0$ in ${\mathbb{F}}_p$. But ${\bar{u}}^{p-1}=1$, contradicting $2\ne 0$. Thus $s=0$ in $K_0$. This proves both cases of roots of unity at the tame ramification boundary for odd primes.

The printed boundary claim needs the qualification $p>2$. When $p=2$, every characteristic-zero field already contains the root $-1$ of order two, so $s=0$ is impossible. If $v(2)=1$, the valuation formula excludes order four and higher, forcing $s=1$. For example $K={\mathbb{Q}}_2$ has exactly this behavior. This is a genuine exception to the final request as printed, rather than an omitted construction.

The valuation ring is $\mathcal{O}=\{x\in K:|x|\le 1\}$, and $\mathfrak{m}=\{x\in K:|x|<1\}$ is an ideal by the ultrametric inequality. Its complement inside $\mathcal{O}$ consists exactly of elements of field absolute value one, whose inverses also lie in $\mathcal{O}$. Every proper ideal must avoid units and hence lie in $\mathfrak{m}$. Therefore $\mathfrak{m}$ is the unique maximal ideal: $\mathcal{O}$ is a local ring. Also its fraction field is $K$, since every nonzero element outside the ring has its inverse inside it.

If $z\in K$ is integral over $\mathcal{O}$, it satisfies a monic equation $z^n+a_{n-1}{z}^{n-1}+\cdots +a_0=0$ with $|a_i|\le 1$. Were $|z|>1$, the leading term would have strictly larger field absolute value than every other term, so the ultrametric inequality would prevent cancellation to zero. Thus $z\in \mathcal{O}$, proving the valuation ring is integrally closed.

For the ideal criterion, assume the valuation is nontrivial and write $v=-\log |\cdot |$. If its value group is discrete, normalize it to $\mathbb{Z}$ and choose a uniformizer $\pi$ of valuation one. In any nonzero ideal $I$, the nonnegative integer valuations have a least value $m$. Choose $a\in I$ of that value. For every $b\in I$, $v(b/a)\ge 0$, so $b\in a\mathcal{O}$. Hence $I=(a)=({\pi }^m)$ and $\mathcal{O}$ is a principal ideal domain.

Conversely, if $\mathcal{O}$ is a principal ideal domain, its nonzero maximal ideal is $(\pi )$. For every element $x$ of positive valuation, $x\in \mathfrak{m}$ implies $x=\pi y$ with $y\in \mathcal{O}$, so $v(x)\ge v(\pi )>0$. Thus $\gamma =v(\pi )$ is the least positive value. For any value $t$, subtract $\lfloor t/\gamma \rfloor \gamma$. The remainder belongs to the value group and lies in $[0,\gamma )$, so is zero. The value group is therefore $\gamma \mathbb{Z}$, and the valuation is discrete. We have proved the principal ideal criterion for a nontrivial rank-one valuation ring.

The nontriviality qualification matters under the standard definition of discrete valuation, which requires value group isomorphic to $\mathbb{Z}$. With the trivial field absolute value, $\mathcal{O}=K$ is still a principal ideal domain, but the value group is zero and there is no uniformizer. Thus the literal unrestricted equivalence has a field exception. If “discrete” instead includes the trivial value group as a discrete subgroup of $\mathbb{R}$, that exceptional case satisfies the equivalence too.

Krasner's lemma states: let $K$ be complete for a Non-Archimedean absolute value, let $\alpha$ be a separable algebraic element over $K$, and let $\beta$ be algebraic over $K$, all in a fixed algebraic closure with the extended field absolute value. Ifthen $K(\alpha )\subseteq K(\beta )$. If $\alpha \in K$, the conclusion is automatic and there is no conjugate-distance condition to check.

To prove it, put $F=K(\beta )$ and take a finite splitting field $N/F$ for the minimal polynomial of $\alpha$ over $F$. That is a separable polynomial, so $N/F$ is Galois. The finite extension $F$ is complete, and uniqueness of extensions of an field absolute value from a complete non-Archimedean field makes every $\sigma \in \mathrm{Gal}(N/F)$ an isometry. It fixes $\beta$, soIf $\sigma (\alpha )\ne \alpha$, it is another $K$-conjugate and this inequality contradicts the strict hypothesis. Every such automorphism therefore fixes $\alpha$. The Fundamental theorem of Galois theory gives $\alpha \in F$, provingOnly $\alpha$ needs to be a separable algebraic element: the proof works even when $\beta$ is inseparable, because the splitting field is taken over $K(\beta )$.

Choose compatible primitive roots of unity ${\zeta }_{p^{n+1}}$, with ${\zeta }_{p^{n+2}}^p={\zeta }_{p^{n+1}}$. Use the slightly faster-converging series permitted by “or otherwise”:The terms have field absolute value $p^{-2n}$, so the partial sums are a Cauchy sequence in ${\overline{\mathbb{Q}}}_p$. We show that an algebraic limit would contradict the growth of cyclotomic extension degrees. The faster weights make the strict conjugate-distance estimate work uniformly also at $p=2$.

Put $L_N={\mathbb{Q}}_p({\zeta }_{p^{N+1}})$. Its degree is $p^N(p-1)$: the translated cyclotomic polynomial ${\Phi }_{p^{N+1}}(1+T)$ has constant term $p$ and reduces modulo $p$ to $T^{p^N(p-1)}$, so is Eisenstein. To verify the reduction, write it as ${\sum }_{j=0}^{p-1}(1+T{)}^{j{p}^N}$ and use characteristic $p$ to reduce this to ${\sum }_{j=0}^{p-1}(1+T^{p^N}{)}^j=T^{p^N(p-1)}$. All roots lie in $L_N$, and its Galois automorphisms are ${\sigma }_a(\zeta )={\zeta }^a$ for units $a$ modulo $p^{N+1}$.

For a nonidentity automorphism, put $r=v_p(a-1)$, choosing a representative with $0\le r\le N$. It fixes the terms with $n<r$. For $n\ge r$, the root ${\zeta }_{p^{n+1}}^{a-1}$ has order $p^{n+1-r}$, so the root-of-unity valuation formula, with $v_p(p)=1$, givesThis is strictly increasing as $n$ increases: successive differences are $2-p^{-(n-r+1)}>0$. Thus the first changed term is the unique term of smallest valuation and cannot cancel. ConsequentlyIn particular no nonidentity automorphism fixes $A_N$, so ${\mathbb{Q}}_p(A_N)=L_N$. Also the distance from $A_N$ to every distinct conjugate is at least $p^{-2N-1/(p-1)}$.

Suppose the Cauchy sequence converged to $A\in {\overline{\mathbb{Q}}}_p$. The tail estimate givesApply Krasner's lemma with $\alpha =A_N$, $\beta =A$. It forces $L_N={\mathbb{Q}}_p(A_N)\subseteq {\mathbb{Q}}_p(A)$ for every $N$. The right-hand side has finite degree, while $[L_N:{\mathbb{Q}}_p]=p^N(p-1)$ is unbounded, a contradiction. ThereforeThis proves algebraic closure of the p-adic numbers is not complete by an explicit algebraic Cauchy sequence and a fully quantified conjugate-separation argument.