Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 123 (Algebraic Number Theory)

Use the simple-root form of Hensel lemma: if $F\in {\mathbb{Z}}_p[T]$ and $t_0\in {\mathbb{Z}}_p$ satisfy $F(t_0)\equiv 0(\mathrm{mod}p)$ and $F^{\prime }(t_0)\not\equiv 0(\mathrm{mod}p)$, there is a unique $t\in {\mathbb{Z}}_p$ with $F(t)=0$ and $t\equiv t_0(\mathrm{mod}p)$. More generally the same assertion holds over a complete discrete valuation ring, with its maximal ideal in place of $(p)$.

Every nonzero 3-adic number has a unique expressionIndeed its normalized P-adic valuation fixes $k$, and its unit residue modulo three fixes $ϵ$. Cubing is an automorphism on $\{1,-1\}$, so it remains to compute the cube map on the principal units.

For $t\in {\mathbb{Z}}_3$,Thus $U_1^3\subseteq U_2:=1+9{\mathbb{Z}}_3$. Conversely, given $1+9a\in U_2$, apply Hensel lemma toModulo three it is $T-a$, and $F^{\prime }(T)=1+6T+9T^2$ is always a unit. The residue root $a\mathrm{mod}3$ therefore lifts, giving $t$ with $(1+3t{)}^3=1+9a$. HenceThis is the cube map on 3-adic principal units. Introducing $T$ is essential: applying simple-root lifting directly to $X^3-u$ would fail because its derivative is divisible by three at every unit.

The map $1+3t\mapsto t\mathrm{mod}3$ induces an isomorphism $U_1/U_2\cong (\mathbb{Z}/3\mathbb{Z},+)$, since multiplication adds the first principal-unit digits modulo three. The valuation factor contributes $\mathbb{Z}/3\mathbb{Z}$ independently, so the cube-class group of the 3-adic numbers isExplicit generators are the classes of $3$ and $4=1+3$. Representatives are $3^a{4}^b$ with $0\le a,b<3$. The class of $3$ is detected by valuation modulo three, and that of $4$ by its nonzero first principal-unit digit, so the two generators are independent.

Normalize the discrete valuation on $L$ by $v_L(L^{\times })=\mathbb{Z}$, set $v_L(0)=+\infty$, and write ${\mathcal{O}}_L$ for its valuation ring. For a finite Galois extension of local fields, define $G_{-1}=G=\mathrm{Gal}(L/K)$ and, for integers $i\ge 0$,Equivalently, $G_i$ is the kernel of the action on ${\mathcal{O}}_L/{\mathfrak{m}}_L^{i+1}$. These are the lower ramification numbering of the ramification groups. In particular, $G_0$ is the inertia group, the kernel of the action on the residue field, and $G_1$ is the wild inertia group. They form a decreasing sequence of normal subgroups, eventually trivial.

For a real index $t\ge -1$, extend by $G_t=G_{\lceil t\rceil }$. Define the Herbrand function byIt is continuous, strictly increasing and piecewise linear; denote its inverse by ${\psi }_{L/K}$. The upper ramification numbering isThe definition includes the placement of the groups at break endpoints: the group at a break is the group before the drop. Lower numbering is compatible with subgroups; upper numbering is the one compatible with quotients. In a totally ramified extension, the uniformizer criterion for lower ramification groups permits testing the defining valuation inequality on one uniformizer instead of on every integral element.

Let $a=\sqrt[3]{3}$ and let $\zeta$ be a primitive cube root of unity. The splitting field is $L={\mathbb{Q}}_3(a,\zeta )$. The polynomial $X^3-3$ is an Eisenstein polynomial, so $[{\mathbb{Q}}_3(a):{\mathbb{Q}}_3]=3$. Meanwhile $(1+2\zeta {)}^2=-3$ shows that ${\mathbb{Q}}_3(\zeta )$ is quadratic: $-3$ is not a square because its valuation is odd. A quadratic field cannot lie in the degree-three field. Consequently $[L:{\mathbb{Q}}_3]=6$, and the action on the three roots identifies its Galois group with the symmetric group $S_3$.

A useful uniformizer isSince $(\zeta -1{)}^3=-3\zeta (\zeta -1)$, we obtainThus $\pi$ is a root of the Eisenstein polynomial $X^6+3$. Its degree is six, so $L={\mathbb{Q}}_3(\pi )$, the extension is totally ramified, and $\pi$ is a uniformizer. With $v_L(\pi )=1$,This is the Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3.

Let $\tau (a)=\zeta a$, $\tau (\zeta )=\zeta$, and let $s(a)=a$, $s(\zeta )={\zeta }^{-1}$. Then $\tau$ has order three, $s$ has order two, and $s\tau s={\tau }^{-1}$. Their actions on the uniformizer areThe two nonidentity elements of $⟨\tau ⟩$ therefore satisfyEach transposition sends $\pi$ to $-{\zeta }^j\pi$ for some $j$, so its displacement has valuation one: $1+{\zeta }^j$ is a unit, reducing to $2$ modulo the maximal ideal. The uniformizer criterion for lower ramification groups now givesFor real indices, this is $S_3$ on $[-1,0]$, $C_3$ on $(0,3]$, and $1$ for $t>3$. In particular the lower breaks are $0$ and $3$.

On $(0,3]$ the index $[G_0:G_t]$ is two; beyond three it is six. Thus the Herbrand function isThe positive lower break $3$ becomes the upper break $3/2$. ThereforeThe fractional upper break is allowed because the full extension is nonabelian. As a check, the different exponent from ramification groups is $5+3\cdot 2=11$. The derivative of the minimal polynomial $X^6+3$ gives the same result: $v_L(6{\pi }^5)=6+5=11$. This is the $p=3$ case of ramification groups of the splitting field of Xp minus p over the p-adic numbers.

Normalize $v_K$ by $v_K({\pi }_K)=1$ and put $v_K(0)=+\infty$. Write $f(X)={\sum }_{j=0}^d{a}_j{X}^j$, with $a_d=1$ and $a_0\ne 0$. In the usual coefficient-exponent convention, the Newton polygon $N_K(f)$ is the lower boundary of the convex hull of the upward vertical rays starting atIt runs from $(0,v_K(a_0))$ to $(d,0)$, with nondecreasing slopes from left to right. Zero coefficients contribute no finite point. Equivalently it is the largest convex piecewise-linear function lying below all these coefficient points.

The Newton polygon root valuation theorem says that a segment of slope $s$ and horizontal length $r$ accounts for exactly $r$ roots of valuation $-s$, counted with multiplicity, using the extended valuation on an algebraic closure. The nonzero constant coefficient excludes a zero root.

For a short justification, use the weighted Gauss valuation $w_t(P)={\min }_j(v(c_j)+jt)$ for $P=\sum c_j{X}^j$. It is multiplicative: after scaling $X$ by an element of valuation $t$ in a valued extension, the initial nonzero residue polynomials multiply without vanishing. Rational $t$ suffice. Factoring a monic polynomial into linear factors gives $w_t(f)={\sum }_r\min (t,v(r))$. This supporting-line function changes derivative at precisely the root valuations; its derivative drops count their multiplicities. Supporting lines to the lower polygon have slope $-t$, proving the sign and horizontal-length assertion.

Slope convention for part (b). Reflecting the horizontal axis, by plotting $(d-j,v_K(a_j))$, produces the reflected Newton polygon convention; its slopes are the root valuations themselves. Part (b)'s wording with $m$ instead of $-m$ is correct under this positive-root-valuation convention. The proof below gives both forms explicitly, so the result does not depend on silently changing the sign.

The denominator is ${\pi }^{n}X$; dividing by $X$ removes the zero at $X=0$. A uniformizer of a degree-$n$ totally ramified extension generates the field: its valuation over $K$ is $1/n$, so $K(\pi )/K$ already has ramification index at least $n$. Hence $K(\pi )=L$, its minimal polynomial has degree $n$, and its conjugates are the distinct $\sigma (\pi )$ for $\sigma \in G$.

Factor the minimal polynomial over $L$. The reduced ramification polynomial isEvery displayed root $r_{\sigma }=(\sigma (\pi )-\pi )/\pi$ is nonzero and integral, since two uniformizers have a difference of valuation at least one. Thus $g$ is monic of degree $n-1$, has nonzero constant coefficient, and belongs to ${\mathcal{O}}_L[X]$. For $n=1$ it is the constant polynomial $1$, with no slopes and no ramification jumps.

Here is also a justification of the uniformizer criterion for lower ramification groups in this setting. The powers $1,\pi ,\dots ,{\pi }^{n-1}$ are a $K$-basis. In an expansion $x=\sum a_j{\pi }^j$, the nonzero terms have valuations $n{v}_K(a_j)+j$, distinct modulo $n$; therefore there is no cancellation at the least valuation. If $x$ is integral, all those values are nonnegative, which forces $a_j\in {\mathcal{O}}_K$. Thus ${\mathcal{O}}_L={\mathcal{O}}_K[\pi ]$. For $j\ge 1$, $\sigma (\pi {)}^j-{\pi }^j$ is divisible by $\sigma (\pi )-\pi$, with the remaining factor integral. Consequently every $\sigma (x)-x$ has valuation at least $v_L(\sigma (\pi )-\pi )$, and equality is attained at $x=\pi$.

For a nonidentity automorphism, write $i(\sigma )=v_L(\sigma (\pi )-\pi )$. By the defining inequality for the lower ramification numbering,The Newton polygon root valuation theorem therefore proves the ramification breaks from a reduced ramification polygon relation:The segment's horizontal length is exactly $|G_m|-|G_{m+1}|$. Equivalently, in the reflected Newton polygon convention required for a positive-slope formulation,All occurring root valuations are nonnegative integers. For negative indices, extend $G_m=G$; total ramification means that there are no negative-index jumps, so the reflected formulation remains true for all integers.

The sign distinction is substantive. Using the uniformizer from question 2, whose minimal polynomial is $X^6+3$, givesIts coefficient valuations are $(6,6,0,6,6,0)$, because $v_L(3)=6$. The usual lower hull has vertices $(0,6),(2,0),(5,0)$ and slopes $-3,0$, of lengths $2,3$. These correspond exactly to the drops $|G_3|-|G_4|=2$ and $|G_0|-|G_1|=3$. With the reflected axis the slopes are $0,3$, as in the printed positive-$m$ wording. If the coefficient-exponent definition is used throughout, the printed assertion needs the minus sign shown above.

Let $K=\mathbb{Q}(\sqrt{-31})$. Its ring of integers of a quadratic field is $\mathbb{Z}[(1+\sqrt{-31})/2]$, and its fundamental field discriminant is $-31$.

First compute the class number. The ideal-form correspondence for imaginary quadratic fields identifies the ideal classes with proper equivalence classes of primitive positive definite binary quadratic forms of discriminant $-31$. Each has a unique reduced positive definite binary quadratic form representative under the convention $|b|\le a\le c$, with $b\ge 0$ when $|b|=a$ or $a=c$. From $31=4ac-b^2\ge 3a^2$ we get $1\le a\le \sqrt{31/3}<4$. Checking these three possible leading coefficients gives exactlyFor $a=3$, neither odd residue choice $b=\pm 1,\pm 3$ makes $(b^2+31)/12$ integral. The listed forms are primitive and satisfy the reduction convention, soThis computes the ideal class group of Q of square root minus thirty-one.

Now take a root $\theta$ of $F(T)=T^3+T-1$, and let $H$ be its splitting field over $\mathbb{Q}$. There is no rational root: the only candidates are $\pm 1$, and neither vanishes. Thus the cubic is irreducible. Its polynomial discriminant isSince this is nonsquare, the Galois group of an irreducible cubic is $S_3$. Its quadratic subfield is $K$: the product of pairwise root differences squares to $-31$ and changes sign under odd permutations. Thus $H/K$ is cyclic of degree three, and $H=K(\theta )$. Indeed the cubic field and $K$ have coprime degrees, so their compositum already has the full splitting-field degree six. The squarefree polynomial discriminant also shows, by the discriminant-index formula for an integral lattice, that $\mathbb{Z}[\theta ]$ is the full integer ring of the cubic field.

It remains to prove that $H/K$ is unramified everywhere. At any rational prime $\ell \ne 31$, the reduction of $F$ is separable. Over a finite residue extension in which it splits, all residue roots are simple and lift by Hensel lemma to the corresponding unramified extension of ${\mathbb{Q}}_{\ell }$. Therefore the local splitting field is unramified over ${\mathbb{Q}}_{\ell }$, and its extension over a completion of $K$ is unramified as well.

At the only possible ramified prime, direct reduction givesThe simple root lifts to $r\in {\mathbb{Q}}_{31}$. Write $F(T)=(T-r)Q(T)$, with $Q$ monic quadratic over ${\mathbb{Q}}_{31}$. The discriminant product formula givesHere $Q(r)=F^{\prime }(r)$ is a unit, so the quadratic discriminant has square class $-31$. Hence the entire local splitting field isIt is quadratic over ${\mathbb{Q}}_{31}$, but its extension over the completion of $K$ is trivial. Thus the prime over $31$ splits completely in $H/K$; in particular it is not ramified there. This is the local splitting-field argument behind an unramified cyclic cubic extension from a prime-discriminant cubic. There are no real places of the imaginary quadratic field $K$, so no infinite-place ramification remains to check.

The Hilbert class field theorem identifies the maximal everywhere unramified abelian extension with an extension of degree $h_K$. Our cyclic unramified extension already has that degree, so it must be the whole Hilbert class field. ThereforeEquivalently, it is the splitting field of $T^3+T-1$, of degree six over $\mathbb{Q}$ and degree three over $K$. Replacing $\theta$ by ${\theta }^{-1}$ gives the alternative defining cubic $T^3-T^2-1$. This is the Hilbert class field of Q of square root minus thirty-one.