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I am very new to field theory and I am trying to prove that if $F$ is a field and $R \in F(x_1,x_2,\ldots,x_n):=\{ PQ^{-1} : P,Q \in F[x_1,x_2,\ldots,x_n] \}$ is nonconstant, then $R$ is transcendental over $F$.

Suppose not. Then we choose a polynomial $A=a_0+a_1x+\ldots+a_kx^k$ such that $A(R)= 0$. I want to say something along the lines of $A(R)\in F(x_1,x_2,\ldots,x_n)$ and has finite degree, so it can't have too many roots. But then $A(R)=0$ for all $(x_1,\ldots,x_n)$ so it does have too many roots.

If $F$ were infinite I can imagine that this works, though I don't know how to formulate it. But I have no idea what to do if $F$ is finite, so perhaps there is just a better way of doing it?

Any help would be greatly appreciated.

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Suppose $R$ is non-constant and write $R=P/Q$ with $P, Q\in F[x_1,x_2,...,x_n]$, $Q\ne 0$ and $P, Q$ coprime (we have to know that $F[x_1,x_2,...,x_n]$ is a UFD). Then $A(R)=0$ implies
$ a_kP^k+a_{k-1}P^{k-1}Q+\cdots + a_0Q^k=0$ or $(a_kP^{k-1}+a_{k-1}P^{k-2}Q+...+a_1Q^{k-1})P=-a_0Q^k.$ Can you conclude now ?

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    Yes. But $Q$ is allowed to be a constant. So you also need the divisibility $P|Q$ which is true because $P$ divides the lhs of the my second displayed formula.2011-11-14