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Reading over an editing my dissertation "Elementary functions" and i am having trouble with my definition of a rational functions in n variables, this is what i have written but its missing one part:

Definition Let F be a differential field $f_{1},f_{2},....,f_{n}$ $\in$ Dont know what to put here,then we define $F(f_{1},f_{2},....,f_{n})$ = $\lbrace\dfrac{h(f_{1},f_{2},....,f_{n})}{g(f_{1},f_{2},....,f_{n})} : h,g \in F[X_{1},X_{2},...,X_{n}], g\neq0\rbrace$ that is to say the field of rational functions in n variables $f_{1},f_{2},....,f_{n}$

Eseentially my problem is what are $f_{1},...,f_{n}$ i wanna use something like meromorphic functions on an open connected subset of $\mathbb{C}$ however i dont want to really interpret the elements of these differential fields as functions until the next section of my project where i talk about the interpretation of these elements as actual functions.

How would a decent definition of rational funtions in n variables read?

ENAFMTH
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    Why does one need to interpret them as anything, in particular? In fact, why not simply use $X_1,...,X_n,$ instead of $f_1,...,f_n$? If one wants to choose particular $X_1,...,X_n$ later (instead of just indeterminates), then that's fine. You'll just need to justify that the notions of field operations in $F(X_1,...,X_n)$ agree with what we'd like them to be when we replace the $X_k$ with something that has operations associated with them already (such as your meromorphic functions)--that is, you'll need to show that the sets of operations are compatible. – Cameron Buie Apr 04 '15 at 16:29
  • Ok yeah, i completely agree with that! I have the lemma following it which says that if we let $F$ be a field then $F(X_{1},..,X_{n})$ is a differential field, with operations addition and multiplication of functions (and the unary differential operator), does this lemma actually work because i see if $X_{1},...,X_{n}$ are just variables then the product rule and sum of functions rule would hold (although how would i formally prove this) but if say we interpreted these variables as other functions will we always be closed under differentiation? – ENAFMTH Apr 04 '15 at 16:38
  • Well, offhand, I think that you need only look at $F[X_1,...,X_n],$ with operations (say) $\oplus$ and $\odot.$ If you have some differentiable functions $f_1,...,f_n:F\to F,$ then you probably have a definition for what addition and multiplication of these functions looks like, yes? You probably even use the operation symbols that you use for the elements of $F.$ Let $P$ be the intersection of all sets of functions $F\to F$ that are closed under $+$ and $\cdot,$ and which contain $f_1,...,f_n.$ Now, it can be shown that $f_1,...,f_n\in P,$ that $P$ is closed under $+$ and $\cdot,$ and... – Cameron Buie Apr 04 '15 at 16:48
  • (cont'd) in fact that $\langle P,+,\cdot\rangle$ is a ring into which $F$ can be naturally included. Then, one can show that, in fact, $\langle P,+,\cdot\rangle$ is isomorphic to $\bigl\langle F[X_1,...,X_n],\oplus,\odot\bigr\rangle,$ and that the isomorphism takes elements of $F$ to themselves, and takes $f_k\mapsto X_k$ for $k=1,...,n.$ In that sense, we can consider $P$ to consist of the elements of $F[X_1,...,X_n],$ evaluated at $X_1=f_1,...,X_n=f_n,$ and we instead refer to $P$ as $F[f_1,...,f_n],$ without equivocation. That should extend readily to $F(f_1,...,f_n).$ – Cameron Buie Apr 04 '15 at 16:54
  • Once you've proved that compatibility, you can start using $+$ and $\cdot$ again (instead of $\oplus$ and $\odot$), at which point you can start proving the various laws of differentiability. – Cameron Buie Apr 04 '15 at 16:55

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