Rational power series If we let $R=\mathbb{Z}[x]$ and $D=\mathbb{Z}[[x]]$. We say that $z\in D$ is rational if there is  $g\in R$, $g\ne 0$ such that $zg\in R$. Let $S$ be the set of all rational elements in $D$. Then $S$ is a subring of $D$.
So, in this case, we may assume that $zg=f\in R$ with $f,g$ have no common factors. By some computations, we get that $g(0)=\pm 1$.
So, is it still true for the finitely many variables case and also, infinitely many variables?
In particular, if we let $P$ be the kernel of the map from $R$ to $\mathbb{Z}$ mapping all the variables to 0, fixing the constant term, then $P=(x_1,x_2,\cdots)$ is a prime ideal in $R$. Let $S^*$ be the set of invertible elements in $S$, so, may we have 
$S^*\cap R=\pm 1 + P$ ???
For the case $n=1$ variable above, I can do since $\mathbb{Q}[x]$ is PID. But in general, I dont know.
Would some one give me some ideas or suggested sources for reading?
Thanks in advance.
 A: For any domain $R$ an element of $R[[X]]$ is invertible if and only if the constant term is invertible in $R$. 
Applying this repeatedly, one gets that an element of $\mathbb{Z}[[X_1,\dots,X_n]]$ is invertible (in this domain) if and only if its constant term is invertible in $\mathbb{Z}$ thar is it is in $\pm 1 + (X_1,\dots,X_n)$. 
So, regarding your 'in particular' with finitely many variables, you have for $S$ the intersection of $\mathbb{Q}(X_1,\dots,X_n)$ and $\mathbb{Z}[[X_1,\dots,X_n]]$ that is the elements are a quotient of two rational (or equivalently integral) polynomials that are an integral power series, as in the question, that $S^{\ast}$ is contained in the set of power series with constant coeffiecient $\pm 1$ as$S^{\ast} \subset \mathbb{Z}[[X_1,\dots,X_n]]^{\ast}$. Yet, not each power series with constant coefficient $\pm 1$ is an element of $S^{\ast}$ for example as $\mathbb{Q}(X_1,\dots,X_n)$ is countable while there are uncountably many power series with constant coefficient $\pm 1$.   
Thus, using the notation of the question, $S^{\ast} \subset \pm 1 + (X_1,\dots,X_n)$ yet the inclusion is strict, and this is not an equality.
The case of infinitely many variables: since you consider quotients of polynomials one can reduce to only considering the substructure where only the (finitely many) variables occuring in the polynomials are present (for each quotient individually).
Alternatively, we mainly need that the constant term of an invertible power series (finitely or infinitely many variable) is invertible in the base domain. This follows just by noting that the constant term of the product is the product of the constant terms, so if the product is $1$ it/they have to be invertible. Thus also showing the inclusion. That it is strict follows by restricting to a substructure with finitely many variables (or a suitable adaption of the cardinality argument).  
One thing I am now somehow not completely sure about, though I think so, (but in any case it is not needed here) is whether in infinitely many variables also the invertability of the constant term is sufficient to imply that the power series is invertible. 
A: Edit: As pointed out by unknown (google) and Andreas Blass it's not true that the power series ring is in general a direct limit of it's subrings in finitely many variables. So what I have written below only holds in the subring $\varinjlim_J R[[x_j|j \in J]]$ where $J$ runs through the finite subsets of $I$.

Let me work out the case of an infinite index set $I$. Let $R$ be a ring with unit. Then 
$$U := R[[x_i| i \in I]]$$ 
can be seen as the union of the "finite" $R[[x_{i_1}, \dots, x_{i_n}]]$ with $i_1, \dots, i_n \in I$. 
Now suppose $f\in U$ is invertible. So there is $g \in U$ such that $fg=1$. Moreover we can find a large enough $n$ such that $f,g \in R[[x_{i_1}, \dots, x_{i_n}]]$. From the finite case it follows that 
$f \in R^{\times} + (x_{i_1}, \dots, x_{i_n}) \le R^{\times} + (x_i | i \in I) =: P$. 
Conversely suppose $f \in P$. Then $f \in R^{\times} + (x_{i_1}, \dots, x_{i_n})$ for some $n$. Thus there is $g \in R[[x_{i_1}, \dots, x_{i_n}]] \le U$ such that $fg=1$. So $f$ is invertible in $U$, showing $U^{\times} = P$. 
Most properties of $U$ can be derived in this way. 
