Embed $K(x)$ into $K[[x]]$ How to see formally/algebraically that the field of rational functions $K(x)$ embeds into the ring of formal power series $K[[x]]$?
 A: You can embed a field $F$ into $K[[x]]$ if and only if $K$ contains a subfield isomorphic to $F$; in particular, you can embed $K(x)$ into $K[[x]]$ if and only if $K$ contains a subfield isomorphic to $K(x)$  (as in Bill's example).
The ring of formal power series $K[[x]]$ is a local ring with maximal ideal $\mathfrak{m}=(x)$, since an element is a unit if and only if it has nonzero constant term. So any embedding of a field into $K[[x]]$ will go through to the residue field $K[[x]]/\mathfrak{m}\cong K$, since the image of $F$ must intersect $\mathfrak{m}$ trivially. Therefore, any embedding of a field $F$ into $K[[x]]$ will induce an embedding of $F$ into $K$. Of course, any embedding of $F$ into $K$ gives an embedding into $K[[x]]$. 
A: HINT $\ $ Embed $\rm\:K(x)\:$ into $\rm\:K\:,\:$ e.g. consider $\rm\: K\ =\ \mathbb Q(x_1,x_2,x_3,\ldots)\:$
A: I'll assume you meant an embedding into $K((x))$ (the field of formal Laurent series). Once you verify that this is indeed a field, this follows from the fact that any embedding of an integral domain into a field extends to an embedding of the field of fractions into the field, and embedding $K[x]$ into $K((x))$. 
To be totally explicit, send $\frac{1}{1 - ax}$ to $\sum_{n \ge 0} a^n x^n$ and $\frac{1}{x}$ to $\frac{1}{x}$, and extend linearly and multiplicatively. 
