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.: $R_v/M_v$.

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    $\begingroup$ 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. $\endgroup$ May 16, 2021 at 23:54
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    $\begingroup$ @paulblartmathcop: I would assume the condition "with value group $G$" means that said $v$ is supposed to be surjective. $\endgroup$ May 17, 2021 at 14:47
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    $\begingroup$ 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. $\endgroup$ May 17, 2021 at 15:09
  • $\begingroup$ If you need a book reference: Bourbaki, Commutative Algebra, Chapter VI, § 3.4, Example (6) gives a construction. $\endgroup$
    – Daniel W.
    Jun 24, 2021 at 11:44

1 Answer 1


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.


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