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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 $$f(x) = fracsum_i=0^na_ix^isum_j=0^mb_jx^j;,$$ and I would like to show that I can write $f$ in the form $sum_k=r^inftyc_kx^k$. 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!

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 $h(x)=sum_kge 0b_kx^k$, it’s clear that you want $b_0=a_0^-1$. Then the first degree term in $g(x)h(x)$ must be $$(a_0b_1+a_1b_0)x=(a_0b_1+a_0^-1a_1)x;,$$

so $a_0b_1+a_0^-1a_1=0$, 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.

In case $R$ is finite or countable, the rational-function field is countable, while the Laurent-series field is uncountable.

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 $fin R[x]$, a series $uin R((x))$ such that $fcdot u=1$.
First consider the case that $f$ has constant term $1$. Then we can define $u_i$, $iinBbb N$, recursively such that for all $n$
$$tag1 f(x)cdot sum_i=0^nu_ix^iin 1+x^nR[x]$$
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 $f(x)cdotsum_i=0^n-1u_ix^i$.
We obtain a power series $u(x)$ with $f(x)u(x)=1$ as desired.

Now consider general $fne 0$. Then it can be written as $acdot x^kcdot hat f$ where $ain Rsetminus 0$, $kin Bbb N_0$, $hat f$ is a polynomial with constant term $1$.
As just seen, there is a power series $hat u$ with $hat fhat u=1$. Then $u:=a^-1x^-khat u$ 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 $gne 0$, say $g(x)=sum_j=0^d a_jx^j$ with $a_dne 0$.
Pick $mge maxd,1$.
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 $x^m^2+d+1mid x^kg$ for $k>m$. Hence $ug$ has infinitely many nonzero coefficients and is not a polynomial.

Hint $rmdisplaystylequad 1: =: (a-xf)(b-xg) Rightarrow ab=1$
$$Rightarrow displaystylermfrac1b-xf = fraca1-axf = a:(1+axf+(axf)^2+(axf)^3+:cdots:)$$

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 $bigcup_ninmathbf N R^n$, 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 $2^aleph_0$.