How to prove that field of rational functions is a *proper* subset of field of formal Laurent series?

Question

Now, if $F$ is a field, I can prove easily that $F(x)\subseteq F((x))$ but I'm having problems to show this is a proper inclusion.

If for example $F=\Bbb R$ or $\Bbb C$, I can take a well-known function, say $\cos x$ and use its power series to show proper inclusing, because if

$$\left(F((x))\ni\right)\;\cos x=1-\frac{x^2}2+\frac{x^4}{24}-\cdots=\frac{f(x)}{g(x)}\in F(x)$$

then, since $f(x),g(x)\in F[x]$ (polynomials), we'd get that $\cos x$ has a finite number of zeros, which is absurd, and thus $\cos x\in F((x))\setminus F(x)$.

My problem now is: what to do if the field $F$ is not one of the usual, infinite ones? For example, if $F$ has positive characteristic?

Any input will be duly appreciated.

Answer 1

We prove a very weak version of the result. Let $F$ be a finite field, and let $\frac{f(x)}{g(x)}$ be a rational function with coefficients in $F$.

Do a formal division of $f$ by $g$. There are only finitely many polynomials of degree less than the degree of $g$ with coefficients in $F$. Thus the formal division yields a terminating or ultimately periodic formal Laurent series.

Now let $q(x)$ be a "sparse" Laurent series, say with the coefficient of $x^n$ equal to $1$ if $n$ is a perfect square, and $0$ otherwise. This is not ultimately periodic.

Answer 2

An easy argument : $1-x$ has a square root in $F((x))$ but not in $F(x)$.

Indeed, in $F((x))$, take the usual Taylor expansion of $\sqrt{1-x}$ at $0$. But in $F(x)$, looking at decomposition in irreducible factors easily shows that such a square root can't exist.

If the characteristic of $F$ is $2$, this won't work, you will have to take a cubic root.

Maybe I can comment a little on the argument : what I'm using is the fact that $F((x))$ is henselian, whereas $F(x)$ is not. This is a quite different argument from the one by @user26857 : they use the fact that some elements of $F((x))$ are not solutions of algebraic equations (over $F(x)$), I use the fact that some elements of $F((x))$ are solutions of such equations. This is really a different argument because there are henselian extensions of $F(x)$ which are algebraic over $F$ (namely the henselianization).