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How to see formally/algebraically that the field of rational functions $K(x)$ embeds into the ring of formal power series $K[[x]]$?

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    perhaps to $K[[x]][x^{-1}]$?2011-04-05

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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 $K[[x]]$.

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HINT $\ $ Embed $\rm\:K(x)\:$ into $\rm\:K\:,\:$ e.g. consider $\rm\: K\ =\ \mathbb Q(x_1,x_2,x_3,\ldots)\:$

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    Sorry agai$n$ a$n$d tha$n$ks for your a$n$swers! I see now that it cannot be done. What is possible is to embed K(x) into K((x)) (the laurent series) and so we can view K((x)) as a vector space over K(x) but we cannot view K[[x]] as a vector space over K(x) (at least i dont see it, but first i thought there was an embedding). Thanks again.2011-04-05
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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.

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    @Matt: Fair enough; and you can invert any polynomial with nonzero constant in $K[[x]]$, which gives you $\frac{1}{p(x)}$ as well. The only issue is inverting $x$, which is trivial.2011-04-06