Tensor Product over Algebraically Closed Field

I have a question about a statement/formulation 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 my interest is the resulting tensor product $$K \otimes L$$. We know that $$K \otimes L$$ is finite dimensional $$L(t)$$-algebra.

Consider following formulation:

"... the assumption on $$K ⊗_k L$$ is satisfied when $$L|k$$ is a separable algebraic extension, or when $$k$$ is algebraically closed. In the latter case $$K ⊗_k L$$ is in fact a field for all $$L ⊃ k$$ ..."

I'm a bit irritated about this formulation since the "or" suggests that if $$k$$ is algebraically closed that we don't need the other assumption separate algebraic for the extension $$L|k$$ to obtain that $$K ⊗_k L$$ is a field. And this seems to be highly wrong. For example take $$K=L=k(t)$$ and $$k= \mathbb{C}$$. Then $$k$$ is alg closed but $$\mathbb{C}(t) \otimes \mathbb{C}(t)$$ is not a field.

What does the author has here in mind?

That if $$L|k$$ is a separable algebraic extension and $$k$$ is algebraically closed then $$K ⊗_k L$$ is a field?

Or did I misunderstood him?

1 Answer

I think he means that $$L/k$$ is an algebraic field extension. The clue is in the earlier statement: to get that $$K\otimes_kL$$ is a finite dimensional $$L(t)$$-algebra, he seems to be assuming that $$k(t)\otimes_kL\cong L(t)$$, which happens provided $$L/k$$ is algebraic, but not in general, as your example shows.

Update: If $$L/k$$ is algebraic, then $$k(t)\otimes_kL\cong L(t)$$.

For, we know that $$k[t]\otimes_kL\cong L[t]$$, so the left hand side is the localisation of $$L[t]$$ using the multiplicatively closed set $$S=k[t]-\{0\}$$. We therefore need to show that the saturation of $$S$$ is $$L[t]-\{0\}$$; in other words, if $$0\neq f\in L[t]$$, then there exists $$0\neq g\in L[t]$$ such that $$fg\in k[t]$$. We can do this by taking $$g$$ to be the product over all possible polynomials given by exchanging the coefficients of $$f$$ by their conjugates.

• yes then the statement becomes indeed trivial since $k$ alg closed would imply $L=k$. Another aspect: could you give a reference/ sketch of the proof that if $L/k$ is algebraic then $K\otimes_kL$ is a finite dimensional $L(t)$-algebra? – KarlPeter Apr 2 at 11:42