Tensor Product of Fields is a Field

I have two questions about a construction introduced in Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (see page 122):

We fix an integral proper normal curve $$X$$ over a field $$k$$. We consider it's function field $$K$$ which is a finite extension of $$k(t)$$ and take an arbitrary field extension $$L \vert k$$.

The point of interest is the resulting tensor product $$K \otimes L$$. We know that $$K \otimes L$$ is finite dimensional $$L(t)$$-algebra.

Following two questions:

1. Assume $$K \otimes L$$ is a finite direct product of fields $$L_i$$. Why these fields are finitely generated (as $$L$$-modules)?

2. Assume non $$k$$ is algebraically closed. Why is $$K \otimes L$$ then a field?

3. 1.
• If $L/k$ is separable then the primitive element gives $L = k[x]/(f)$ and $K \otimes_k L = K[x]/(f) = K[x]/(\prod_j f_j) = \prod_j K[x]/(f_j)$ since $f$ separable implies the $(f_j)$ are comaximal. If $k$ is algebraically closed and $L/k$ is a tower of purely transcendental and algebraic extensions then rename the transcendental elements so they are not in $\overline{K}$ thus the algebraic extensions keep the same degree with $k$ or $K$ as the basefield – reuns Apr 1 at 1:42
• I'm quite not sure if it is possible to reduce the problem 1. to the case when $L \vert k$ is separable. Or why do you implicitely assumed that? – KarlPeter Apr 1 at 1:54
• If $K$ is a finite extension of $k(t)$ then $K = k(t)[Y_1,\ldots,Y_m]/I$, finitely generated as field, not as an algebra or module, and if $K \otimes_k L = L(t)[Y_1,\ldots,Y_m]/I= \prod_j L_j$ then each $L_j$ is a quotient of $L(t)[Y_1,\ldots,Y_m]/I$, finitely generated as a field over $L$. – reuns Apr 1 at 1:56
• @reuns: yes, so the 1. question is solved. Concerning the second one I'm a bit confused. Assume $k$ is algebraically closed. $K$ is finite extension of $k(t)$ and therefore has transcendence degree $1$ over $k$. On the other hand $L \vert k$ might be an arbitrary extension. – KarlPeter Apr 1 at 12:52
• Then Grothendieck-Sharp tells as in mathoverflow.net/questions/82083/… that $\dim_{\mathrm{Krull}}(K\otimes_k L) = \min(\operatorname{trdeg}_k(K),\operatorname{trdeg}_k(L))$. Then if we take as $L$ for example a transcendent extension $k(t)$ of $k$ we obtain $\dim_{\mathrm{Krull}}(K\otimes_k L)=1$ so the tensor product can't be a field. But this contradicts the statement 2. The author didn't explicitely siad that $L \vert k$ should be a finite extension – KarlPeter Apr 1 at 12:52