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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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up vote 4 down vote accepted

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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I've stared at this for 20 mins now and I think I am on the right track. Is this correct? This is a UFD, so because the right side contains a multiple of $Q$, the left side must contain a multiple of $Q$. But $P,Q$ are coprime so this means that $a_kP^{k-1} + \ldots a_1Q^{k-1}$ is a multiple of $Q$. So $$mQ = a_kP^{k-1}+a_{k-1}P^{k-2}Q+\ldots+a_1Q^{k-1}$$ for some $m$ which tells me that $$Q(m-(a_{k-1}P^{k-2}+\ldots+a_1Q^{k-2})) = a_kP^{k-1}$$ which says that $P$ must be a multiple of $Q$, contradiction. – nullUser Nov 14 '11 at 22:57
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. – user18119 Nov 14 '11 at 23:12

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