Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 28 (Local Fields)

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.

A simple-root version of Hensel lemma is: if $R$ is a complete discrete valuation ring, $f\in R[T]$, $f(a_1)\equiv 0(\mathrm{mod}\pi )$, and $f^{\prime }(a_1)$ is a unit, then there is a unique root $a\in R$ congruent to $a_1$ modulo $\pi$.

Suppose $f(a_n)\in {\pi }^{n}R$. The derivative remains a unit, so choose a residue $c_n$ satisfyingand put $a_{n+1}=a_n+c_n{\pi }^n$. Taylor expansion gives $f(a_{n+1})\equiv f(a_n)+c_n{\pi }^n{f}^{\prime }(a_n)(\mathrm{mod}{\pi }^{n+1})$, because $2n\ge n+1$. Thus $f(a_{n+1})\in {\pi }^{n+1}R$. The sequence is a Cauchy sequence and converges by completeness; continuity of polynomial evaluation makes its limit a root.

If roots $a,b$ have the prescribed residue, factor $f(a)-f(b)=(a-b)Q(a,b)$. Modulo $\pi$, the second factor is $f^{\prime }(a_1)$, so it is a unit. Therefore $a=b$. This is the simple-root Hensel lifting by successive residues.

For odd $p$, eighth roots of unity have order prime to the residue characteristic. The total ramification criterion for local cyclotomic extensions shows that they generate an unramified extension of degree $f={\mathrm{ord}}_8(p)$. More directly, a primitive eighth root in ${\mathbb{F}}_{p^f}$ lifts uniquely by Hensel lemma, since the derivative of $T^8-1$ is a unit at each nonzero root. Its residue has exact order eight, and the Frobenius automorphism acts by ${\zeta }_8\mapsto {\zeta }_8^p$. Consequently $f=1$ when $p\equiv 1(\mathrm{mod}8)$ and $f=2$ for the other three odd residue classes.

At $p=2$, the translated cyclotomic polynomial isIt is Eisenstein at two, so the extension has degree four. All primitive eighth roots are odd powers of ${\zeta }_8$ and already lie in the field; the extension is Galois, and its automorphisms give the full group $(\mathbb{Z}/8\mathbb{Z}{)}^{\times }\cong C_2\times C_2$. Thus

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.

Expand the translated cyclotomic polynomial:Every nonleading coefficient is divisible by $p$, and the constant term is not divisible by $p^2$. The polynomial is Eisenstein, hence irreducible. Its root $\lambda ={\zeta }_p-1$ is a uniformizer of $E={\mathbb{Q}}_p({\zeta }_p)$, and the Eisenstein polynomial valuation criterion givesThus this local cyclotomic extension is totally ramified.

Put $\alpha =\sqrt[p]{p}$ and $\lambda ={\zeta }_p-1$. The polynomial $T^p-p$ is Eisenstein, so $[{\mathbb{Q}}_p(\alpha ):{\mathbb{Q}}_p]=p$, while the preceding part gives degree $p-1$ for $E$. Both numbers divide $[K:{\mathbb{Q}}_p]$ by the tower law, and the compositum degree is at most their product. Coprimality forces $[K:{\mathbb{Q}}_p]=p(p-1)$. Since $K$ contains all the roots ${\zeta }_p^a\alpha$ of $T^p-p$, it is the splitting field of a separable polynomial and is Galois.

Extend $v_p$ with $v_p(p)=1$. Then $v_p(\alpha )=1/p$, $v_p(\lambda )=1/(p-1)$, andThe value group therefore forces the ramification index to be at least $p(p-1)$; it cannot exceed the extension degree. Thus $K/{\mathbb{Q}}_p$ is totally ramified and $\varpi$ is a uniformizer for the normalized valuation $v_K=p(p-1)v_p$.

Write its automorphisms as ${\sigma }_{a,b}(\alpha )={\zeta }_p^a\alpha$ and ${\sigma }_{a,b}({\zeta }_p)={\zeta }_p^b$, with $a\in {\mathbb{F}}_p$, $b\in {\mathbb{F}}_p^{\times }$. The degree count makes all $p(p-1)$ pairs occur, giving $G\cong {\mathbb{F}}_p⋊{\mathbb{F}}_p^{\times }$. Their action on the uniformizer isIf $b\ne 1$, the residue of this ratio is $b$, so subtracting one gives a unit and $v_K({\sigma }_{a,b}(\varpi )-\varpi )=1$. If $b=1$ and $a\ne 0$, the ratio is ${\zeta }_p^{-a}$, giving displacement valuation $1+v_K({\zeta }_p^{-a}-1)=p+1$. The identity has infinite displacement valuation.

By the uniformizer criterion for lower ramification groups, $\sigma \in G_i$ precisely when that displacement valuation is at least $i+1$. Hence the ramification groups of the splitting field of Xp minus p over the p-adic numbers have sizesFor $1\le i\le p$ the group itself is $\{{\sigma }_{a,1}:a\in {\mathbb{F}}_p\}\cong C_p$. The lower wild ramification break is $p$.