If $k$ is a finite field, $\bar{k}$ its algebraic closure and $K/k$ a finitely generated field extension of transcendence degree $1$ so that $k$ is algebraically closed in $K$, show that $R=K\otimes_k \bar{k}$ is a finitely generated field extension of $\bar{k}$ of transcendence degree $1$.
I'm heaving a problem with proving this statement. I read a proof that says it suffices to show that R is a domain, but I don't know why and I also didn't quite understand the proof that it is a domain.
Thank you for your help!
Edit:
The proof was like this:
It's sufficient to show that $R^\prime=K\otimes_kk^\prime$ is a domain for every finite extension $k^\prime/k$.
$R^\prime$ is a product of fields and if $R^\prime$ is not a field then the idempotent elements generate a finite $k^\prime$-algebra in $R^\prime$ of the $k^\prime$-dimension >1. With Galois-Descent and Galoisgroup $G$ of $k^\prime/k$ we get a finite $k$-algebra in K with the $k$-dimension >1. This is a contradiction since k is relatively closed in K.
I only know the Galois-Descent as it is shown in S. Bosch, Algebra. I don't understand why the $k$-algebra is an $k$-algebra. I know it is a $k$-vectorspace but I can't see how it follows with Bosch that it is also an algebra.