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Let $R(x)$ denote the rational functions over the reals and $\overline{R(x)}$ its algebraic closure. Also let $P(X,Y)$ denote the set of partial functions from $X$ to $Y$, where partial functions means in simple terms a "function" for which not every term in the domain is mapped.

Lastly let $h$ be the canonical homomorphism $h:R(x)\rightarrow P(R,R)$ that is described by simply evaluating the rational function over the $R$.

My question is whether there exists a homomorphism $f:\overline{R(x)}\rightarrow P(R,C)$ such that $f|_{R(x)}=h$ and $f(\sqrt[n]{q})=\sqrt[n]{f(q)}$

It is clear or any element of $\overline{R(x)}$ that can be expressed through the field operations of the rational functions and roots we have an obvious corresponding partial function associated with it (simply evaluate the expression as you would in an grade school algebra class). However, there are many elements of $\overline{R(x)}$ that are not expressible through roots and arithmetic operations and so this questions boils down to is there meaningful and consistent way of "evaluating" them over the reals so that the outputs lie entirely within $C$ whilst agreeing the the above mentioned canonical mapping of rational functions.

A pure existence proof for this correspondence would be nice but a construction for a method of evaluation for any $q\in \overline{R(x)}$

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What do you mean by "the set of functions over C"? –  Tom Goodwillie Mar 9 '11 at 2:12
I am not sure if that's what you are asking but, no, $\sqrt x$ is not in $\mathbb{C}(x)$. –  Felipe Voloch Mar 9 '11 at 2:23
Thinking of subsets alone may probably be the wrong intuition here. Try to consider maps between the rings of such functions, and see what goes wrong. –  David Roberts Mar 9 '11 at 5:38
Sorry for the ambiguity, I have made what I meant much more precise now. This kinda of problem seems like that kind that has a clear answer and a method of construction as well. –  Christian Bueno Mar 9 '11 at 19:57
There's not much point in editing a question once it has been closed, since no one can post an answer, unless you also open a thread on meta asking people to look at the new wording and consider reopening (in which case you post a link here to the discussion on meta). Alternatively, one can just post a new question (with a link to the old one). –  Gerry Myerson Mar 9 '11 at 22:59
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closed as too localized by Andres Caicedo, Pete L. Clark, Chandan Singh Dalawat, George Lowther, Todd Trimble Mar 9 '11 at 20:13

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