I have a question about a statement/formulation in Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (see page 122):

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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?


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.

  • $\begingroup$ 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? $\endgroup$ – KarlPeter Apr 2 at 11:42

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