11
$\begingroup$

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!

  • 4
    @Hagen: You might flesh that counterexample out to an answer.2012-11-04

5 Answers 5

5

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.

  • 0
    @Alex: Yes, it is. And **Hagen** has now even added an answer that deals with that aspect. That’s fine, and doubtless useful for someone coming upon this page in the future. It has, however, no bearing on **my** answer, which, as I said, was for the specific question asked by the OP.2016-04-09
5

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

3

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) for $k and $x^{m^2+d+1}\mid x^kg$ for $k>m$. Hence $ug$ has infinitely many nonzero coefficients and is not a polynomial.

  • 0
    Don't you want "because \deg(x^{k^2}g) for k and $x^{m^2+d+1} | x^{k^2}g$ for k >m" in the penultimate sentence?2017-12-19
0

Hint $\rm\displaystyle\quad 1\: =\: (a-xf)(b-xg)\ \Rightarrow\ ab=1$ $\Rightarrow\ \ \displaystyle\rm\frac{1}{b-xf}\ =\ \frac{a}{1-axf}\ =\ a\:(1+axf+(axf)^2+(axf)^3+\:\cdots\:)$

  • 2
    Could you explain why this implication/equality follows? I'm afraid that I don't see it =(2012-11-03