# Every rational function of $f \in k(x_1,x_2,\dots,x_n)$ is transcendental over $k$.

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 $k$.

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If it is homework, please tag as such and show effort. Your question is almost trivial. –  Lierre Feb 12 '12 at 18:40

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, beware that in the OP's notation $f$ is just a non constant rational function, so that if you want to write the first sentence you should change the notation. –  Georges Elencwajg Feb 13 '12 at 21:56
Dear @Georges: Thank you! I hope it's ok now. Your formulation is much better, in particular because it's more self-contained. [In French I'd have written "ne serait-ce que parce que...", but I was unable to find an English equivalent. What could I have written?] Your answer only uses unique factorization (which seems unavoidable). –  Pierre-Yves Gaillard Feb 13 '12 at 22:36
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... –  Georges Elencwajg Feb 13 '12 at 23:44

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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Dear Lierre: $+1$! Very nice! I hadn't thought of that. I think one should define $\overline k$ as an algebraic closure of $k$ containing $f$. But then it's simpler to say that $f$ is constant in $k[f](x_1,\dots,x_n)$. (As $k[f]$ is a field, $k[f](x_1,\dots,x_n)$ is defined in the usual way, and is a field.) - I agree that the argument is very concise, but it would be exaggerated (I think) to say it's trivial. –  Pierre-Yves Gaillard Feb 13 '12 at 17:39
Dear Lierre: I wonder if one shouldn't justify more precisely the following step. If $f$ is in $k(x_1,\dots,x_n)$, if $K/k$ is an extension, and if $f$ is constant in $K(x_1,\dots,x_n)$, then $f$ is constant in $k(x_1,\dots,x_n)$. This is clearly false if $f$ is transcendental over $k$ and $K=k(f)$. What is the exact argument that makes this work if $K/k$ is algebraic? –  Pierre-Yves Gaillard Feb 13 '12 at 18:27
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 $\bar k\otimes_k k(x_1,...,x_n)$]. –  Georges Elencwajg Feb 13 '12 at 22:19

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? –  Mohan Feb 14 '12 at 20:00
Dear @user774025, I have changed my former formulation : thanks for forcing me to be clearer! –  Georges Elencwajg Feb 14 '12 at 21:50

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 ! –  Lierre Feb 13 '12 at 21:04