field embedding I've come across this problem in Etingof's notes on representation theory (Problem 5.1 on p. 78). It just sounds nice exercise... The question is :

Let $f : k(x_1,\ldots,x_n)\rightarrow k(y_1,\ldots, y_m)$ be a field embedding of field of rational functions. Show that $m\geq  n$.

P.S: His hint was first to show for $f : k[x_1,\ldots,x_n]\rightarrow k(y_1,\ldots, y_m)$
 A: This is false. Let $k=\mathbb{Q}(z_1,z_2,\ldots)$. Then we can embed $k(x_1,\ldots,x_n)$ into $k$ by sending $x_i$ to $z_i$ and sending $z_i$ to $z_{n+i}$. This is actually a field isomorphism between $k(x_1,\ldots,x_n)$ and $k$. Then composing with the canonical embedding of $k$ in $k(y_1,\ldots,y_m)$, we can embed $k(x_1,\ldots,x_n)$ in $k(y_1,\ldots,y_m)$ for any $m$ including $m<n$.
It becomes true with better information on $k$.
For one case, if $k=\mathbb{Q}$ then $k(y_1,\ldots,y_m)$ has transcendence degree $m$ over $\mathbb{Q}$. But if we embed $k(x_1,\ldots,x_n)$ in $k(y_1,\ldots,y_m)$, then the image has transcendence degree $n$ over $\mathbb{Q}$. So $n\leq m$.
But I see that the notes in the link assume $k$ is algebraically closed. If this is the case then when we embed $k(x_1,\ldots,x_n)$ in $k(y_1,\ldots,y_m)$ then $k$ goes to an algebraically closed subfield of $k(y_1,\ldots,y_m)$. Therefore the image of $k$ must be a subfield $k_0$ of $k$. Now we can argue with transcendence degrees again. The image of $k(x_1,\ldots,x_n)$ has transcendence degree $n$ over $k_0$, while $k(y_1,\ldots,y_m)$ has transcendence degree $m$ over $k$ and so transcendence degree at least $m$ over $k_0$. So $m\geq n$.
