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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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    Not as an extension of the obvious embedding $K[x]\to K[[x]]$, at any rate, since $x$ has an inverse in $K(x)$, but not in $K[[x]]$...2011-04-05
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    the image of $K(x)$ would have to be closed under addition, so no two series with the same constant term could be in the image... but then you couldnt scale by elements of $K$... i dont think its possible?2011-04-05
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    Certainly not under a K-algebra homomorphism.2011-04-05
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    I don't see it. The embedding restricts to an embedding of $K[x]$; it must send $1$ to a nonzero idempotent of $K[[x]]$. But since $K[[x]]$ is an integral domain, the only idempotents are $0$ and $1$, so the embedding must send $1$ to $1$. In particular, the embedding must send the prime field of $K$ identically to the prime field of $K$ inside $K[[x]]$. If $K=\mathbb{Q}$, and if the image of $x$ has constant term $a_0$, then $x-a_0$ maps to a noninvertible element, contradicting that we have an embedding of $K(x)$ into $K[[x]]$.2011-04-05
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    perhaps to $K[[x]][x^{-1}]$?2011-04-05

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