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I've been having some trouble proving the following in Vakil's FOAG recently:

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. (Although we won’t need this, a version of the converse is true if $l/k$ is separable, [Gr-EGA, IV2.6.14.2].) Hint: fix a $k$-basis for $l$, $b_1 = 1, \dots, b_d$. Explain why $1\otimes b_1, \dots, 1 \otimes b_d$ forms a free $A$-basis for $A \otimes_k l$. Explain why we have injections $$ \require{AMScd} \begin{CD} A @>>> K(A) \\ @VVV @VVV \\ A \otimes_k l @>>> K(A) \otimes_k l \end{CD} $$ Show that $K(A) \otimes_k l = K(A \otimes_k l) $. (Idea: $ A \otimes_k l \subset K(A) \otimes_k l \subset K(A \otimes_k l) $. Why is $K(A) \otimes_k l $ a field?) Show that $ (A \otimes_k l) \cap K(A) = A$. Now assume $P(T) \in A[T]$ is monic and has a root $\alpha \in K(A)$, and proceed from there.

I actually never wind up showing that $1 \otimes b_1, \dots, 1 \otimes b_d$ is an explicit basis for $A \otimes_k l$ but instead use the fact that field extensions are (faithfully) flat so that the map $A \to A \otimes_k l$ is an injection

1. Does this logic hold for $K(A) \to K(A) \otimes_k l$ injective as well?

Assuming everything works up to here, the author points out in this post $K(A) \otimes_k l$ must be a field as it is a finite-dimensional $K(A)$ vector space which is a domain (c.f. Vakil's exercise 3.2.G) . Then by the universal property of the field of fractions, $K(A \otimes_k l)$ is the smallest field containing $A \otimes_k l$ so that $K(A \otimes_k l) \subset K(A) \otimes_k l$. By the hint provided we get the reverse inequality.

I'm a bit more concerned with what comes next (namely $(A \otimes_k l) \cap K(A) = A$). One direction should be obvious since $A$ is clearly contained in both. For the reverse direction, I'm not really sure how to make sense of notation — my idea is that if $t = \sum_i a_i \otimes s_i \in (A \otimes_k l)$ is also an element of $K(A)$, then we could multiply by some $b \in A$ so that $b\cdot t = \sum_i ba_i \otimes s_i \in A$ which I want to say is equivalent to something of the form $ba^\prime \otimes 1_l$ which would allow us to conclude by left cancellation that $t \in A$.

2. Is this idea correct ?

Thanks in advance for any clarification.

user26857
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cdsb
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  • I don't understand your try. In my answer to the linked question I pointed out that the equality holds for a faithfully flat extension. If you admit that $A\to A\otimes_kl$ is faithfully flat, then there nothing to show. – user26857 May 19 '22 at 12:43
  • @user26857 from your answer though one would first have to prove $A = B \cap K(A)$? Maybe I misunderstand the wording. But the theorem you claim is essentially just a more general version of the question, so of course there would be nothing to show. – cdsb May 20 '22 at 00:32
  • If I understand well do you have a problem with the proof of $A=B\cap K(A)$? – user26857 May 20 '22 at 07:02
  • @user26857 yes i suppose that is the one part of my proof above that I am unsure of; obviously $A \subset B \cap K(A)$ since $A$ is contained in both, but in my proof above I was asking if the way to show this would be by saying that if we let $t \in B \cap K(A)$, then by virtue of $t \in K(A)$ we can multiply by some $c \in A$ so that $ct \in B \cap A = A$; then somehow this implies $t \in A$ to begin with – cdsb May 20 '22 at 20:54
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    I suppose that you know that $IB\cap A=I$ for every ideal $I$ of $A$. Then try $I=(c)$. – user26857 May 20 '22 at 22:15
  • @user26857 ah i see the logic thank you! – cdsb May 21 '22 at 20:06

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