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I am trying to solve Ex. 5.4.M in Vakil's notes. Quoting the text:

Suppose $A$ is a $k$-algebra, and $l/k$ is a finite extension of fields. (Most likely your proof will not use finiteness; this hypothesis is included to avoid distraction by infinite-dimensional vector spaces.) Show that if $A\otimes_k l$ is a normal integral domain, then $A$ is a normal integral domain as well.

At a certain point the hint suggests to show that $K(A)\otimes_k l$ is a field ($K(A)$ is the fraction field of $A$).

I showed that $K(A)\otimes_k l$ is an integral domain and that a $K(A)$-basis of it is given by $\{1\otimes b_i\}_{i \in I}$, where $\{b_i\}_{i \in I}$ is a $k$-basis of $l$. In particular, if $l/k$ is a finite extension then $K(A)\otimes_k l$ is a field.

How to proceed in the infinite case?

user26857
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Pgatti
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  • How did you conclude that $K(A)⊗_kl$ is a field and where exactly used that the field extension is finite? – user26857 Jul 20 '15 at 15:04
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    If $l/k$ is finite than $K(A)\otimes_k l$ is a finite dimensional $K(A)$-vector space. Finite dimensional vector spaces that are domains are fields: multiplication by a non-zero element is an injective linear endomorphism, hence surjectve. – Pgatti Jul 20 '15 at 15:04
  • I'm not sure if Vakil suggests that the same proof works for infinite dimensional field extensions. Moreover, the product of two fields extensions is a field under special circumstances as one can see here (which I'm afraid are not fulfilled by your case). – user26857 Jul 20 '15 at 15:10
  • I saw that post and I shared your opinion, but I was not totally convinced, so I asked this question. In particular the fact that $A \otimes_k l$ is normal may be a sufficient condition for the tensor to be a field. At least I could not come up with a counterexample. – Pgatti Jul 20 '15 at 15:16
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    Btw, the claim holds for algebraic fields extensions $l/k$. – user26857 Jul 21 '15 at 21:56
  • @user26857 What is the argument for the algebraic extension case? If you could add it to your last answer that would be great. – Pgatti Jul 22 '15 at 07:11
  • A bit late, but $K(A) \otimes_k l$ is a field because it is a tensor product of fields. In particular the inverse of $a \otimes b$ is $\frac{1}{a} \otimes \frac{1}{b}$. – user477805 Apr 23 '18 at 03:47
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    @user477805 The tensor product of fields is not a field in general https://en.wikipedia.org/wiki/Tensor_product_of_fields#Examples Note that you need to find an inverse for elements of the form $\sum_{i} a_i\otimes b_i$. – Pgatti Apr 24 '18 at 13:39

2 Answers2

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This answer doesn't address directly to the OP's question, but to the exercise in Vakil's textbook.

Let $A\subset B$ be a faithfully flat extension of integral domains, and $B$ is integrally closed. Then $A$ is also integrally closed.

This follows immediately from $A=B\cap K(A)$.

user26857
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Let $A=K$ be a field extension of $k$, and $l=k(x)$ a purely transcendental extension of $k$. Then $A\otimes_kl\simeq S^{-1}K[x]$, where $S=k[x]\setminus\{0\}$. This is clearly a normal integral domain which is not necessarily a field.

However, if the extension $k\subset l$ is algebraic, and $K\otimes_kl$ is an integral domain, then it is a field since the ring extension $K\subset K\otimes_kl$ is generated by algebraic elements ($1\otimes a$, $a\in l$), so it is algebraic.

user26857
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