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Let $k$ be a field and let $x_1,x_2,\dots,x_n$ be indeterminates.

How do I show that every non-constant rational function $f \in k(x_1,x_2,\dots,x_n)$ is transcendental over $k$.

4 Answers 4

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Let $f$ be a non-constant element of $k[x_1,\dots,x_n]$.

We claim that $f$ is transcendental over $k$.

If $f$ was algebraic over $k$, the domain $k[x_1,\dots,x_n]$, being integrally closed, would contain $f$ and $1/f$, and $f$ would be constant.

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    Dear Pierre-Yves, I would simply translate as "if only because...", which is quite close to what you wrote. But that does not mean that I agree with you on the better answer...2012-02-13
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Since I've been a little rough in my comment, here is my trivial answer to make it up.

You can consider $f$ as a rational function in $\bar k(x_1,\dotsc,x_n)$, where $\bar k$ is an algebraic closure, right ? And $f$ is a constant if and only if it is a constant in this new field. But if $f$ is algebraic over $k$, it is certainly in $\bar k$, i.e. it is a constant.

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    Pierre-Yves makes a valid and extremely subtle point. The field $\bar k$ is not uniquely defined and even if you have chosen such an algebraic closure, $\bar k(x_1,...,x_n)$ is not uniquely defined either *if you already have defined* $k(x_1,...,x_n)$ *before choosing* $\bar k$: you have to choose a composite extension of $\bar k$ and $k(x_1,...,x_n)$.[Fortunately all those composite extensions are isomorphic and isomorphic to $\bar k\otimes_k k(x_1,...,x_n)$].2012-02-13
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Let $\phi=P/Q\in k(X_1,...,X_n)$ be a rational function with $P,Q\in k[X_1,...,X_n]$ relatively prime .
Suppose $\phi$ is algebraic over $k$, say $a(P/Q)^9+b(P/Q)^5+c=0$.
Then $aP^9+bP^5Q^4+cQ^9=0$.
Since $P$ divides $cQ^9$ and is relatively prime to $Q$, it must be a constant: $P \in k$.
Similarly $Q \in k$ and thus $P/Q\in k$.

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    why P divides Q and Q divides P?2012-02-14
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This is a rewriting of Lierre's great answer. The idea being not mine, I'm using the community wiki mode.

We can assume $n=1$.

Recall that $k$ is a field, $x$ an indeterminate, and $f(x)$ an element of $k(x)$ which is algebraic over $k(x)$.

We claim that $f(x)$ is constant.

Let $y$ be another indeterminate.

As $x$ is transcendental over $K:=k[f(x)]=k(f(x))$, there is a $K$-embedding $ \phi:k(x)=K(x)\to K(y)=k(f(x),y) $ mapping $x$ to $y$.

As $\phi$ is a $K$-embedding and $f(x)$ is in $K$, we have $\phi(f(x))=f(x)$.

As $\phi$ is a $k$-embedding and $f(x)$ is in $k(x)$, we have $\phi(f(x)=f(\phi(x))=f(y)$.

This yields $f(x)=f(y)$.

Writing $f(x)=p(x)/q(x)$ with $p(x),q(x)$ relatively prime in $k[x]$, we get $ p(x)\ q(y)=p(y)\ q(x), $ and unique factorization in $k[x,y]$ implies that $f(x)$ is constant.

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    Nice clarification !2012-02-13