Valuations in function fields

Question

"Let $k$ be a field and $K = k(t)$ the function field in one variable. Show that the valuations $v_{\mathfrak p}$ associated to the prime ideals $\mathfrak p = (p(t))$ of $k[t]$, together with the degree valuation $v_{\infty}$, are the only (non-trivial) valuations of $K$, up to equivalence. What are the residue class fields?"

This is an exercise of Neukirch. I was trying to reproduce the proof that the only absolute values of $\mathbb Q$ are the $|\cdot|_p$ with $p$ prime or $p=\infty$, up to equivalence:

Let $v$ be a (non-trivial) valuation of $k(t)$. Every element of $k(t)$ can be written as the product of a finite number of powers of irreducible polynomials. Since $v$ is non-trivial, there is some $p(t)$ irreducible such that $v(p(t))\ne 0$. Let $\mathfrak a=\{f\in k[x]\colon v(f)>0\}$. I wish to prove that $\mathfrak a$ is an ideal of $k[x]$, and then $\mathfrak a=(p(x))$. The problem is that a don't know if every element $f\in k[x]$ satisfies $v(f)\ge 0$, that is, if $k[x]$ is a subset of the valuation ring of $v$.

Answer 1

A complete answer to your question can be found in the post https://math.stackexchange.com/a/2550192/231098

I'll do my own version here, that is closer to the "valuation point of view", rather than the "absolute value point of view".

Let me reformulate the statement of the exercise: we are looking for the valuations $v$ on $k(t)$ which are trivial on $k$. Indeed, there are many more valuations than the ones we are asked to find if we do not suppose $v_{|k}$ to be trivial (you can look up the theory of key polynomials or simply look at this post https://math.stackexchange.com/a/2584924/231098 )

So let us see how can we find them.

Case 1: $v(t)<0$, then $v(t^{-1})>0$, so if we let $s=t^{-1}$, then, by ultrametric property, for every $Q(s)=a_ks^k+a_{k+1}s^{k+1}+\ldots+a_ns^n\in k[s],\: a_k\neq0$ $$v(P)=kv(s).$$

Considering $P(t)=\alpha_0+\alpha_1 t+\ldots+\alpha_rt^r\in k[t],\:\alpha_r\neq0$, then $s^nP(s^{-1})\in k[s]$ and since its constant coefficient is non zero, we get $v(s^nP(s^{-1}))=0$, so $$v(P(t))=v(P(s^{-1})=-nv(s)\geqslant0,$$

so the valuation $v$ is equivalent to the degree in $t$ valuation.

Case 2: $v(t)\geqslant0$, then by ultrametric property $\forall P\in k[t],\:v(P)\geqslant0$. We then consider the prime ideal (you can easily show that it is prime on your own) $$\mathfrak{p}:=\{f\in k[t];\:v(f)>0\}.$$

Subcase 2.1: if $\mathfrak{p}=(0)$, then the valuation $v$ is trivial on $k[t]$, so by the identity $v(f/g)=v(f)-v(g),\:v$ can only extend to the trvial valuation on $k(t)$.

Subcase 2.2: if $\mathfrak{p}=(p(t))$, then we can show that $v$ is the valuation associated to the prime $p(t)$. Indeed, we show that for any other (non-equivalent) irreducible/prime $q(t),\:v(q)=0$. Indeed, by Bézout we have two polynomials $a(t),b(t)\in k[t]$ such that $$ap+bq=1,$$

so $$0\leqslant v(q)\leqslant v(b)+v(q)=v(1-ap).$$

However, since $v(ap)=v(a)+v(p)\geqslant v(p)>0=v(1)$, we get by ultrametric property $v(1-bp)=0$, thus $v(q)=0$. We can conclude by using the UFD property of $k[t]$.

Parting remarks: this type of strategy can work more or less for any Dedekind domain $R$, of which $k[t]$ is a simple example. Indeed, for valuations $v$, that are non-negative on $R$, we set the prime ideal $\mathfrak{p}=\{x\in R;\:v(x)>0\}$ and we localize at this ideal. By the prime factorization property of Dedekind domains, you can then show that $R_\mathfrak{p}$ is a valuation ring and indeed it is the valuation ring associated to the prime ideal $\mathfrak{p}$ (details omitted).

Answer 2

That's because it needs not be true. It's false for the valuation at infinity. So the first thing you might want to do is distinguish the cases where $v(x)\geqslant 0$ or $v(x)<0$.

You will be able to apply your reasoning in the cases where $v(x)\geqslant 0$, and you will have to treat the valuation at infinity separately.