Show the existence of a valuation with a given residue field

I do not know what to do with the following exercise:

For every totally ordered group $$(G,+,\le)$$ and field $$F$$ there exists a valuation (on a field $$K$$) with valuation group $$(G,+,\le)$$ and residue field $$F$$.

As far as I understand we have to find a valuation $$v$$ such that $$F \cong R_v / M_v$$. However, I do not see any way on how to do this. Could you help me or tell me a source where this construction is presented?

Below you can find all the relevant definitions and information I know about valuations thus far:

In lecture I learned the following definition of valuation:
Let $$(K,+,\cdot)$$ be a field and let $$(G,+,\le)$$ be a totally ordered group. A map $$v: K \longrightarrow G\cup\{\infty\}$$ is a valuation if the following properties hold:

• $$v(ab) = v(a)+v(b)$$
• $$v(a+b) \ge \min\{v(a),v(b)\}$$
• $$v(a) = \infty \iff a = 0$$

Then we proved the basic properties:

1. $$v(1) = 0$$
2. $$v(a^{-1}) = -v(a)$$
3. $$v(-a) = v(a)$$
4. $$v(a - b) \ge \min\{v(a), v(b)\}$$
5. $$\text{If }v(a) \ne v(b), \text{ then } v(a+b) = \min\{v(a), v(b)\}$$

And finally we defined $$R_v := \{k \in K \mid v(k) \ge 0\}$$, which is a local ring with maximal ideal $$M_v = \{k ∈ K \mid v(k) > 0\}$$.

The residue field of a valuation $$v$$ is the residue field of the maximal ideal of the valuation ring, i.e.: $$R_v/M_v$$.

• To be clear, you are trying to find a field $K$ and a valuation on $v: K \longrightarrow G \cup \{\infty\}$? If there are no other restrictions then you can take $K = F$ and the trivial valuation. May 16, 2021 at 23:54
• @paulblartmathcop: I would assume the condition "with value group $G$" means that said $v$ is supposed to be surjective. May 17, 2021 at 14:47
• OP: The Wikipedia article on valuation rings used to contain a construction. Now it just contains a link to en.wikipedia.org/wiki/Hahn_series, but that still does the job. May 17, 2021 at 15:09
• If you need a book reference: Bourbaki, Commutative Algebra, Chapter VI, § 3.4, Example (6) gives a construction. Jun 24, 2021 at 11:44

A simple construction is as follows. Consider the domain $$F[X^G]$$ of polynomials with coefficients in $$F$$ and indeterminates $$X^g,g \in G$$, with the pointwise sum and the Cauchy product which extends $$\forall g,h \in G,X^g X^h=X^{g+h}$$. For $$P \in F[X^G]$$, define $$v P$$ as the least $$g \in G$$ such that $$X^g$$ occurs in $$P$$ (with a non-zero coefficient). Extend $$v$$ to the fraction field $$F(X^G)$$ of $$F[X^G]$$ by setting $$v \frac{P}{Q}= v P - v Q$$.
Then $$(F(X^G),v)$$ is a valued field with value group $$G$$ and residue field $$F$$. It need not embed in every such valued field, but it does in certain conditions.