Algebraic closure of a field in the rational functional field is the field itself.

Let $k$ be a field. I need to show that the algebraic closure of $k$ in $k(x_1,x_2,...,x_n)$ is $k$, where the rational function field $k(x_1,x_2,...,x_n)$ is the field of fractions of the polynomial ring $k[x_1,x_2,...,x_n]$. Thank you so much for your help.

• Can you prove the result when $n=1$? – JSchlather Sep 12 '13 at 0:46
• What I did is the following: any non-zero element of the rational function field is of the form f/g, if it is algebraic over k, then we have : (f/g)^n+a_1*(f/g)^(n-1)+..+a_n=0, a_i is in k. Then we get fh=g^n for some h in the ring of polynomials of k. Since the ring of polynomials is a UFD, this implies that f and g have a common factor. Where do I go from here?? – user94396 Sep 12 '13 at 0:53
• Walcher's answer explains the case $n=1$. Do you see how to extend the result? – JSchlather Sep 12 '13 at 4:11

We prove the statement for $n=1$, the rest follows easily by induction.
So consider $k(x)$, we want to show that if an element $a\in k(x)$ is algebraic over $k$, then $a\in k$. Write $a=\frac {f(x)}{g(x)}$ with $f,g\in k[x], (f,g)=1$ and let $p\in k[x]$ be the minimal polynomial of $a$ with degree $n$. Consider $$0=(g(x))^np(a)=(f(x))^n+a_{n-1}(f(x))^{n-1}g(x)+...+a_0(g(x))^n$$ From the above identity we see that any irreducible factor of $f$ must be an irreducible factor of $g$ and vice versa. But $f$ and $g$ are coprime, so they are both constant and hence $a\in k$.
Edit: Obviously $a_0\neq 0$, because $p$ is irreducible.