Let $k$ be a field and $G$ be a totally ordered abelian group. Let $k((G))$ the set of all formal power series in the indeterminate t:
$f=\underset{\gamma \in G} \sum a_{\gamma}t^{\gamma}$
with the coefficents $a_{\gamma}\in k$ and the support $supp(f):=\{ \gamma \in G : a_{\gamma} \neq 0\}$ is a well ordered subset of $G$.
The sum is defined as follow:
$(\underset{\gamma \in G} \sum a_{\gamma}t^{\gamma})+\underset{\gamma \in G} \sum b_{\gamma}t^{\gamma}:=\underset{\gamma \in G} \sum (a_{\gamma}+b_{\gamma})t^{\gamma}$;
The multiplication is defined as follow:
$(\underset{\gamma \in G} \sum a_{\gamma}t^{\gamma}) \times \underset{\gamma \in G} \sum b_{\gamma}t^{\gamma}:=\underset{\gamma=\alpha+\beta} \sum (a_{\alpha}+b_{\beta})t^{\gamma}$.
$(k((G)),+, \times)$ is a field.
MY QUESTION IS: What is the transcendence degree of $k((\Gamma))$ over $k$? Could we get that it is greater or equal than the cardinality of $\Gamma$?
I think it is yes, but there's one complicated case that I cannot solve which that of $\Gamma=\mathbb{Z}$.
If $G$ is an additively denoted group and $B$ is a set, then I write $G^{(B)}$ for the set of functions $f:B \rightarrow G$ with finite support $\{b \in B \ : \ f(b)\neq 0\}$.
Consider the notion of $\mathbb{Q}$-independant basis of $\Gamma$. This is a family $(\beta_i)_{i \in I} \in \Gamma^I$ such that for each $\gamma \in \Gamma$, there is a unique family family $(k_i)_{i \in I} \in \mathbb{Z}^{(I)}$ and $r \in \mathbb{Z}$ with $\exists j \in I, r\wedge k_i=1$ and $r.\gamma=\sum \limits_{i \in I} k_i.\beta_i$. Such a basis exists in any torsion-free group, and thus in any orderable group. It is obtained by considering a $\mathbb{Q}$-basis of the divisible hull $\widehat{\Gamma}$ of $\Gamma$ and mutiplying each element of the basis by an integer so that the result lie in $\Gamma$.
Now for $\gamma_1,...,\gamma_n \in \Gamma$ and $P \in k[X_1,...,X_n]$, the relation $P(t^{\gamma_1},...,t^{\gamma_n})=0$ entails a relation $k_1.\gamma_1 + ... + k_n.\gamma_n = 0$ for some $n$-uple of integers $(k_1,...,k_n) \neq (0,...,0)$. If $\gamma_1,...,\gamma_n$ lie in a $\mathbb{Q}$-independant basis, then there are no such relations. Thus the transcendence degree of $k((\Gamma))$ over $k$ is greater or equal to the cardinal of any such basis (which is the dimension of $\widehat{\Gamma}$).
Since the set $\mathbb{Z}^{(I)} \times \mathbb{Z}$ surjects onto $\Gamma$, we have $|\Gamma|=|I|=\dim_{\mathbb{Q}}(\widehat{\Gamma})$ if $|I| \geq \aleph_0$, whereas $|\Gamma| = \aleph_0$ if $I$ is finite and non-empty. So this yields the result for $\Gamma$ with $\dim_{\mathbb{Q}}(\widehat{\Gamma})\geq \aleph_0$. In the remaining non trivial case $\dim_{\mathbb{Q}}(\widehat{\Gamma})\in \mathbb{N}^{>0}$, it is enough to prove that the transendence degree of $k((\mathbb{Z}))$ over $k$ is $\geq \aleph_0$ since $\Gamma$ contains a copy of $\mathbb{Z}$.
So can you do this case? This probably exists in the literature.