Is $k[x]_{(x)}$ never isomorphic with $k[[x]]$? Let $k$ be a field . I can show that there is an injective ring homomorphism from the localization $k[x]_{(x)}$ into $k[[x]]$ . My question is : Can $k[x]_{(x)}$ be isomorphic to $k[[x]]$ ? If $k$ is algebraically closed , then is it true that $k[x]_{(x)}$ cannot be isomorphic with $k[[x]]$ ?
 A: I assume your question is whether the natural map $k[x]_{(x)} \to k[[x]]$ is an isomorphism. I claim it is never an isomorphism; to prove this it suffices to construct, for every field $k$, a power series which does not lie in the image. In fact it's possible to write down a single power series which works for every $k$: I claim that
$$f(x) = \sum_{n \ge 0} x^{n^2}$$
never lies in the image of $k[x]_{(x)}$. This is equivalent to saying that there is no polynomial $p(x) \in k[x]$ with nonzero constant term such that $p(x) f(x)$ is a polynomial, or said another way, that $f$ is not rational. 
To see this, suppose that $\deg p(x) = d$. Then for $n$ greater than somewhere around $\frac{d}{2}$, the distances between consecutive terms $x^{n^2}$ becomes greater than or equal to $d$; beyond that point there is no longer any cancellation when computing $p(x) f(x)$, so the coefficients of $f(x)$ become the coefficients of $p(x)$, spaced out further and further apart. In particular, they are never eventually zero. 
(A more complicated version of this argument establishes that $f(x)$ is not only not rational, it is not even algebraic.)
There are also various other arguments for different fields $k$: in all cases the point is to show that there exists an irrational power series and there are just many ways to do this.


*

*If $k$ is at most countable there is a cardinality argument: $k[x]_{(x)}$ is countable but $k[[x]]$ is, as a set, the product of countably many copies of $k$, and hence is uncountable. Explicitly this boils down to diagonalizing over all rational power series to write down an irrational one.

*If $k$ is a subfield of $\mathbb{C}$ then growth arguments become possible: coefficients of rational power series grow at most exponentially and so e.g. $f(x) = \sum n! x^n$ cannot be rational. A more analytic way to say this is that rational power series have nonzero radius of convergence so any nonzero power series with zero radius of convergence is irrational. Many, many other constructions and arguments are possible here; for example $f(x) = e^x$ is irrational because, as a holomorphic function on $\mathbb{C}$, it has no poles but grows faster than a polynomial. 

*If $k$ is infinitely generated, observe that the coefficients of a rational power series over $k$ necessarily lie in a finitely generated subfield of $k$. So any $f(x) = \sum f_n x^n$ with the property that the coefficients generate a subfield of $k$ not contained in a finitely generated subfield is irrational. But now observe that any finitely generated field is countable, and so argument 1 applies in that case; so this argument together with argument 1 furnishes an alternate proof in general. 
