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 $k$.
 A: 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.
A: 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 $P/Q\in k$. 
A: 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.
A: 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.
