Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 21 / 1

An additive valuation on a field is a map $v:K^{\times }\to \Gamma$, where $\Gamma$ is a totally ordered abelian group, with $v(xy)=v(x)+v(y)$ and $v(x+y)\ge \min (v(x),v(y))$ whenever $x+y\ne 0$. Set $v(0)=\infty$. We consider nontrivial valuations; if the trivial valuation is allowed, it must be listed separately from the asserted classification. Two valuations are equivalent when their valuation rings agree, or equivalently their ordered value-group images identify in a way compatible with the maps. Real-valued equivalent nontrivial valuations differ by positive scaling.

For a valuation on $\mathbb{Q}$, every integer has nonnegative value, because it is a sum of copies of $1$ or its negative. If all prime numbers had value zero, unique factorization would make the valuation trivial. Thus some prime $p$ has $v(p)>0$. There cannot be two such primes: for distinct $p,q$, Bézout's identity gives $ap+bq=1$ with integers $a,b$, and the valuation inequality would give $v(1)>0$. All other prime values therefore vanish. Factoring the numerator and denominator of a rational number givesThe image is the cyclic ordered group generated by $v(p)>0$, proving equivalence to the P-adic valuation. This is the classification of nontrivial valuations on the rational numbers; an Archimedean absolute value is not an additive valuation satisfying this ultrametric inequality.

The simple-root form of Hensel's lemma says that for a complete discrete valuation ring $\mathcal{O}$ with maximal ideal $\mathfrak{p}$, a polynomial $f\in \mathcal{O}[X]$ and $a\in \mathcal{O}$ with $f(a)\equiv 0(\mathrm{mod}\mathfrak{p})$ and $f^{\prime }(a)\not\equiv 0(\mathrm{mod}\mathfrak{p})$ have a unique root in $a+\mathfrak{p}$.

For $X^3+1$ over ${\mathbb{Q}}_7$, all roots are integral: a negative valuation would make the leading term the unique term of lowest valuation. Modulo $7$, the roots are $3,5,6$, and $3X^2$ is nonzero at each. Hensel's lemma gives three distinct lifts, and a cubic has no further roots. The number is $3$.

For $X^2+2X+4$ over ${\mathbb{Q}}_2$, a root is again integral and reduces to zero modulo $2$. Write it as $2y$ with $y\in {\mathbb{Z}}_2$. The equation becomes $y^2+y+1=0$, impossible modulo $2$. The number is $0$.

For $3X^3+X+3$ over ${\mathbb{Q}}_3$, a putative root of valuation $k<0$ gives term valuations $1+3k,k,1$. The first is uniquely minimal, a contradiction. Thus all roots are integral. Reduction modulo $3$ is $X$, whose root zero is simple since the derivative $9X^2+1$ is a unit everywhere. Hensel's lemma gives exactly one lift. The number is $1$.