How to see formally/algebraically that the field of rational functions $K(x)$ embeds into the ring of formal power series $K[[x]]$?
Embed $K(x)$ into $K[[x]]$
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5Not 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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0the 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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0Certainly not under a K-algebra homomorphism. – 2011-04-05
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0I 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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0perhaps to $K[[x]][x^{-1}]$? – 2011-04-05
3 Answers
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 $K[[x]]$.
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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0-1: I don't understand what this is supposed to mean. This clearly is impossible for many other choices of $K$. Please clarify, and I will remove my downvote. – 2011-04-05
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0My presumption is that this works in the OP's context (which context has been mistakenly omitted from the above statement of the problem). – 2011-04-05
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0That certainly is a plausible explanation for the OP's confusing question. Downvote removed – 2011-04-05
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0Sorry, what i wanted to do is to view K[[x]] as a vector space over K(x) but i dont see how i can do this. – 2011-04-05
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0Sorry again and thanks for your answers! 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
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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0This suffices if $K$ is algebraically closed, so all polynomials factor into linear terms. But do these definitions suffice for, e.g., $\frac{1}{1+x^2}$ if $K=\mathbb{R}$? – 2011-04-06
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1@Arturo: Dear Arturo, You can always extend scalars to the algebraic closure, where this definition makes sense, and then observe that one actually lands in $K[[x]]$. Regards, – 2011-04-06
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1@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