Is this a finite dimensional $L(t)$-algebra?

Question

I recently read this argument in a book:

$k$ is a field, $L|k$ is a field extension, $K$ is a finite extension of $k(t)$. This implies that $K\otimes_kL$ is a finite dimensional $L(t)$-algebra.

But this doesn't seem right. If $L|k$ is algebraic, then I think we would have $k(t)\otimes_kL=L(t)$, but does this hold in general? It feels like this should be obvious but my field theory is a bit rusty...

Answer 1

There's a natural map $\phi:k(t)\otimes_k L\to L(t)$. As you say, it is an isomorphism when $L/k$ is algebraic, but not in general. Take, say $k=\Bbb Q$ and $L=\Bbb R$. Then $1/(t-\pi)$ is not in the image of $\phi$,