Hahn series

The ring of **Hahn series** with value group? $G$, denoted $k[x^G]$, is the ring of functions $f\colon G \to k$ such that $\{x \in G : f(x) \neq 0\}$ is well-ordered when considered as a subset of the opposite poset $G^{op}$. Addition is defined pointwise, and multiplication is defined by the convolution product:

$(f \cdot g)(x) = \sum_{y+z = x \in G} f(y)g(z)$

The multiplicative valuation $v(f)$ is the least $x \in G^{op}$ for which $f(x) \neq 0$.

We obtain a valuation ring from this construction since a valuation ring determines and is determined by a valuation on a field.

The ring $k[x^G]$ is a field. If $k$ is algebraically closed, then $k[x^G]$ is algebraically closed provided that $G$ is divisible.

As a corollary, if $G$ is divisible, $k[x^G]$ is real closed if $k$ is real closed. This is because the adjunction of a square root of $-1$ would make $k[x^G]$ algebraically closed, since this gives the same result as constructing the Hahn series over the algebraically closed field $k[\sqrt{-1}]$.

Revised on February 5, 2012 03:27:18
by Todd Trimble
(74.88.146.52)