Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 123 / 2

Give ${\overline{K}}_{\mathfrak{p}}$ the unique extension of the absolute value of $K_{\mathfrak{p}}$. We use two standard facts: a complete non-Archimedean absolute value extends uniquely to every finite algebraic extension, and any two embeddings of a finite separable extension into an algebraic closure are conjugate by an automorphism of that algebraic closure.

A $K$-embedding $\tau :L\to {\overline{K}}_{\mathfrak{p}}$ pulls back this absolute value to a finite place of a number field on $L$ extending the given place of $K$. By the permitted equivalence of places and primes, it determines a prime ideal ${\mathfrak{q}}_{\tau }$ above $\mathfrak{p}$. Every element of $\mathrm{Gal}({\overline{K}}_{\mathfrak{p}}/K_{\mathfrak{p}})$ preserves the absolute value by uniqueness, so ${\mathfrak{q}}_{\tau }$ is constant on each orbit.

Every prime $\mathfrak{q}$ above $\mathfrak{p}$ occurs: its completion of a number field at a prime ideal $L_{\mathfrak{q}}$ is a finite extension of $K_{\mathfrak{p}}$ and admits a $K_{\mathfrak{p}}$-embedding into ${\overline{K}}_{\mathfrak{p}}$. Composing with $L\hookrightarrow L_{\mathfrak{q}}$ gives an embedding inducing $\mathfrak{q}$.

Conversely, suppose two embeddings induce the same prime. Their pulled-back absolute values have the same normalization on $K$, so they extend continuously to embeddings of the same field $L_{\mathfrak{q}}$. To see that their images are still inside the algebraic closure rather than only its completion, the closure of $\tau (L)$ is exactly the finite complete field $K_{\mathfrak{p}}\tau (L)$. The two resulting embeddings of $L_{\mathfrak{q}}$ are conjugate by the stated separable-extension fact. Restricting the conjugacy to $L$ proves that the original embeddings are in the same orbit. Thus the prime ideals and local embedding orbits are naturally identified:

If $L=K(\alpha )$, an embedding is determined by the image of $\alpha$, which may be any root of its minimal polynomial $f$. The orbits on these roots are exactly the root sets of the monic irreducible factors of $f$ over $K_{\mathfrak{p}}$. HenceIf $f_j$ corresponds to ${\mathfrak{q}}_j$, then $L_{{\mathfrak{q}}_j}\cong K_{\mathfrak{p}}[T]/(f_j)$ and $[L_{{\mathfrak{q}}_j}:K_{\mathfrak{p}}]=\mathrm{deg}{f}_j$. This is the local factorization and extended absolute values correspondence and does not require an integral power basis.

The polynomial is Eisenstein at $2$: every nonleading coefficient is even, and its constant term $126$ is not divisible by $4$. Therefore it is irreducible over $\mathbb{Q}$ andBy part (a), the number of primes above a rational prime is the number of irreducible factors over the corresponding $p$-adic field.

We use the Newton polygon theorem in this form: a segment of slope $-r$ and horizontal length $d$ accounts for $d$ roots of valuation $r$, counted with multiplicity, in an algebraic closure with $v_p(p)=1$. We also use the denominator criterion for Newton polygon irreducibility: if all roots have valuation $a/n$ with $\gcd (a,n)=1$, any monic factor of degree $d$ has constant-term valuation $da/n\in \mathbb{Z}$, hence $n\mid d$.

At $3$, the nonzero coefficient points areThe lower hull is the single segment from $(0,2)$ to $(5,0)$, with slope $-2/5$. All five roots therefore have valuation $2/5$. A proper factor of degree $d$ would require $5\mid d$, which is impossible for $1\le d<5$. Thus $g$ is irreducible over ${\mathbb{Q}}_3$, givingThe completion has ramification index $5$ and residue degree $1$: the value $2/5$ forces its ramification index to be divisible by $5$, and its degree is $5$.

At $7$, the nonzero coefficient points areThe lower hull consists of a length-three segment of slope $-1/3$ and a length-two horizontal segment. Thus three roots have valuation $1/3$, and two have valuation zero. Reduction givesBoth $1$ and $-1$ are simple roots, so the Hensel lemma gives two distinct linear factors over ${\mathbb{Q}}_7$. The remaining cubic accounts for all the valuation-$1/3$ roots and is irreducible by the same denominator argument. Hence the local factor degrees are $1,1,3$, andThe two linear factors give $(e,f)=(1,1)$, and the cubic gives $(e,f)=(3,1)$. These data sum to degree five, as required. This computes the local prime decomposition of T5 plus 6T3 plus 252T2 plus 126. Counting distinct factors merely modulo $3$ would not justify the result, since that reduction is $T^5$.