The Novikov field is a field I have a questions on why the "Novikov field" defined in below is indeed a field: 

Specifically, why every element $\sum_{g\in \Gamma}a_gT^g$ has an inverse? If
$$T^0=\sum_{g\in \Gamma}a_gT^g * \sum_{h\in \Gamma}b_hT^h:=\sum_{l\in \Gamma}(\sum_{g\in \Gamma}a_gb_{l-g})T^l$$
Basically, this means we need to solve the systems:
$\sum_{g\in \Gamma}a_gb_{l-g}=\delta_{0l}$, where $0$ is the additive identity in $\Gamma$. I don't know how to solve this for $b_h$'s.
Although this has a background in symplectic geometry, I hope the solution could be purely algebraic since the definition of this "field" is.
Source
 A: When $\Gamma=\mathbb{Z}\subset \mathbb{R}$, this is the ring of formal Laurent series, and is a field for basically the same reason. Any nonzero formal Laurent series $g(T)$ can be expressed in the form $T^{-n}f(T)$ for $f$ a formal power series in $T$ with $f(0)\neq 0$. Since formal power series are a local ring with maximal ideal $(T)$, we have that $f(T)$ is invertible, and thus the inverse is $T^nf(T)^{-1}$.
So see that any formal power series $f(T)$ with coefficients in $R$ is invertible iff $f(0)\in R^*$, write out the multiplication formula and use induction.
The case of general $\Gamma$ will proceed similarly, since the "support" of a power series always has finitely many elements contained in the ray $(-\infty,a]$, one may perform a similar inductive argument to show that inverses exist.
EDIT- This is a bit trickier that I originally thought.
First multiply by $T^g$ for some $g\in \Gamma$ to get the support of $T^g f(T)$ positive, where we have support of a series being the set of $h\in\Gamma$ with $a_h\neq 0$.
Since $T^g$ is invertible, with inverse $T^{-g}$, we may assume that our power series $f(T)$ has positive support, and multiplying by an element of our field, we may assume that $a_0=1$.(Note that this isn't actually needed, just makes it a bit easier with signs/coefficients).
Then the equation $f(T)*g(T)=1=T^0$ gives a bunch of equations which we will want to set to $0$. The problem is that we don't know a priori what the support of $T$ should be, or if its possible to pick some support with coefficients which allows this. This is definitely trickier than the $\Gamma=\mathbb{Z}$ case.
We can instead construct an inverse inductively by a different method. Let $$f(T)=f_1(T)=1+a_{g_{1,1}}T^{g_{1,1}}+a_{g_{1,2}}T^{g_{1,2}}+...$$
Now let $f_2(T)=(1-(a_{g_{1,1}})^{-1}T^{g_{1,1}})f_1(T)$
Inductively, construct $f_n(T)=(1-(a_{g_{n-1,1}})^{-1}T^{g_{n-1,1}})f_{n-1}(T)$
Where $f_{n-1}(T)=1+a_{g_{n-1,1}}T^{g_{n-1,1}}+...$
We claim first that the sequence $g_{n,1}$ is discrete.
For this, note that if $$f_{n-1}(T)=1+a_{g_{n-1,1}}T^{g_{n-1,1}}+a_{g_{n-1,2}}T^{g_{n-1,2}}...$$
Then smallest coefficient of $f_n(T)$ will either be $T^{2g_{n-1,1}}$ or $T^{g_{n-1,2}}$, and if we had an accumulation point of the $\{g_{n,1}\}$, then this first case couldn't occur, which means that one our our $f_i(T)$ didn't have discrete support, but we know all of them do since the novikov field is a ring.
Then the desired inverse of $f(T)$ is given by $\prod\limits_{i=1}^{\infty}(1-(a_{g_{i,1}})^{-1}T^{g_{i,1}})$. This product is well defined by the discreteness of the sequence $\{g_{i,1}\}$, and by construction, there is nothing except $T^0$ in the support of $f(T)\prod\limits_{i=1}^{\infty}(1-(a_{g_{i,1}})^{-1}T^{g_{i,1}})$, so this is a multiplicative inverse.
A: Let $\Gamma_0:= \Gamma\cap \mathbb{R}_{\geq 0}$ be the non-negative subgroup of $\Gamma$. Then $\Lambda$ is the localisation of the ring
$$\Lambda_0 := \big\{\sum_{g\in\Gamma_0 }a_gT^g:a_g\in \mathcal{K} \big\}  $$
at the unique maximal idea of $\Lambda_0$. Since $\mathcal{K}$ is a field the ring $\Lambda_0$ is an integral domain. And hence the localisation is a field.
