Showing that $R(x)$ is a proper subset of $R((x))$ if $R$ is a field 
I would like to show that if $R$ is a field, then $R(x)$ is a proper subset of $R((x))$, where $R(x)$ is the ring of rational functions, and $R((x))$ is the ring of formal Laurent series.

If $f \in R(x)$, then $f(x) = f_1(x)f_2^{-1}(x)$, where $f_1(x), f_2(x) \in R[x]$. So I wrote this as $$f(x) = \frac{\sum_{i=0}^{n}a_ix^i}{\sum_{j=0}^{m}b_jx^j}\;,$$ and I would like to show that I can write $f$ in the form $\sum_{k=r}^{\infty}c_kx^k$. However, I am unsure how to manipulate $f$ in order to show this. What I was thinking was to find some formal power series expansion for $f_2^{-1}(x)$, multiply out the summation with $f_1(x)$, then rearrange the coefficients and terms to obtain the desired form. However, I can't seem to derive a formula for the inverse of a polynomial in general that I could use for this. How can I go about manipulating $f_2^{-1}(x)$ to show this? Any suggestions?
Thanks!
 A: In case $R$ is finite or countable, the rational-function field is countable, while the Laurent-series field is uncountable.
A: HINT: Write $f_2(x)$ in the form $x^rg(x)$, where $g$ has a non-zero constant term. Then $g(x)$ has an inverse in $R[[x]]$.
An easy induction shows that its coefficients can be calculated recursively: just start calculating! For instance, if $g(x)=a_0+a_1x+\ldots+a_mx^m$, and the inverse is to be $h(x)=\sum_{k\ge 0}b_kx^k$, it’s clear that you want $b_0=a_0^{-1}$. Then the first degree term in $g(x)h(x)$ must be $$(a_0b_1+a_1b_0)x=(a_0b_1+a_0^{-1}a_1)x\;,$$
so $a_0b_1+a_0^{-1}a_1=0$, and you can solve for $b_1$. It’s easy to prove that this can be continued recursively.
And from there you’re pretty much home free.
A: To show that $R(x)$ is a proper subset of $R((x))$, we first need to ignore the "is". Using more precision, I'd prefer to say that $R(x)$ is canonically isomorphic to a proper subring of $R((X))$.
First part: subset
We have a canonical and straightforward map from the ring $R[x]$ of polynomial to the ring $R((X))$ fo formal Laurent series (this does not even require $R$ to be a field) and accordingly identify polynomials with their corresponding power series.
To extend this map to $R(x)$ we need to find, for every non-zero polynomial $f\in R[x]$, a series $u\in R((x))$ such that $f\cdot u=1$.
First consider the case that $f$ has constant term $1$. Then we can define $u_i$, $i\in\Bbb N$, recursively such that for all $n$
$$\tag1 f(x)\cdot \sum_{i=0}^nu_ix^i\in 1+x^nR[x]$$
Indeed, we can just let $u_0=1$ and then recursively let $u_n$ be $-1$ times the coefficient of $x^n$ in the polynomial $f(x)\cdot\sum_{i=0}^{n-1}u_ix^i$.
We obtain a power series $u(x)$ with $f(x)u(x)=1$ as desired.
Now consider general $f\ne 0$. Then it can be written as $a\cdot x^k\cdot \hat f$ where $a\in R\setminus \{0\}$, $k\in \Bbb N_0$, $\hat f$ is a polynomial with constant term $1$.
As just seen, there is a power series $\hat u$ with $\hat f\hat u=1$. Then $u:=a^{-1}x^{-k}\hat u$ is a Laurent series with $fu=1$, as desired. (Here is the only place where we use that $R$ is a field: We need to find $a^{-1}$).
Remark: Actually, it suffices to know that $R((x))$ is itself a field; which by itself can be proved by finding a multiplicative inverse recursively (almost) precisely as above.
Second part: proper
It suffices to exhibit a single formal Laurent series that cannot be written as quotient of polynomials.
Consider
$$ u(x)=\sum_{k=0}^\infty x^{k^2} $$
and assume that $u=\frac fg$ with $g\ne 0$, say $g(x)=\sum_{j=0}^d a_jx^j$ with $a_d\ne 0$.
Pick $m\ge \max\{d,1\}$. 
Then in multiplying $u(x)g(x)$ we see that the coefficient of $x^{m^2+d}$ equals $a_d$ because $\deg(x^kg)<m^2+d$ for $k<m$ and $x^{m^2+d+1}\mid x^kg$ for $k>m$. Hence $ug$ has infinitely many nonzero coefficients and is not a polynomial.
A: Hint  $\rm\displaystyle\quad 1\: =\: (a-xf)(b-xg)\ \Rightarrow\ ab=1$ 
$$\Rightarrow\ \ \displaystyle\rm\frac{1}{b-xf}\ =\ \frac{a}{1-axf}\ =\ a\:(1+axf+(axf)^2+(axf)^3+\:\cdots\:)$$
A: If your field $R$ is countable, if I am not mistaken, another possible argument is a cardinality argument, probbably (?) under the Axiom of Choice though :
$R[x]$ can be viewed as $\bigcup_{n\in\mathbf N} R^n$, a countable union of countable sets, so $R[x]$ is countable. Its fraction field $R(x)$ is a quotient of $R[x]\times R[x]$, hence countable.
On the other hand $R((x))$ contains "$\{0,1\}((x))$", hence has cardinality at least $2^{\aleph_0}$.
