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Let $f: A \to B$ a faithful flat map between rings. Let $a \in A$.

I want to know how to prove that

$$aA = aB \cap A$$ holds.

Ideas:

Consider the exact sequence

(1) $$0 \to aB \cap A\to A \to A/(aB \cap A)\to 0$$

Tensoring by $B$ gives exact

(2) $$0 \to (aB \cap A) \otimes_A B \to B \to A/(aB \cap A) \otimes_A B \to 0$$

Does here $(aB \cap A) \otimes_A B= aB$ hold?

My idea is to show that $aA = \operatorname{Ker}(A\to A \to A/(aB \cap A)) =aB \cap A$

by going back from (2) to (1). Since $aB = aA \otimes B$ then by faithful flatness that suffice.

But I need $(aB \cap A) \otimes_A B= aB$.

Or is this generally wrong and I should try another idea?

Bernard
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user267839
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    How do you mean $aB\cap A$? It's presumably $f(a)B\cap f(A)$, and is a subring of $B$. – Berci Dec 06 '18 at 21:36
  • See https://math.stackexchange.com/a/2449611/660 – Pierre-Yves Gaillard Dec 06 '18 at 21:41
  • Yes, exactly. Implicitely I mean the images of $A$ wrt $f$. – user267839 Dec 06 '18 at 21:41
  • @Berci faithfully flat ring extension are necessarily injective, so $A$ is a subring of $B$ and writing out $f$ is a bit excessive – Badam Baplan Dec 07 '18 at 05:05
  • @KarlPeter An important property of faithfully flat ring extension $A \rightarrow B$ is that $IB \cap A = I$ for all ideals $I$ of $R$. I think you'd be best off reading through some introductory notes rather than asking here. These ones are short and sweet http://www.math.lsa.umich.edu/~hochster/615W14/fthflat.pdf – Badam Baplan Dec 07 '18 at 05:11
  • @BadamBaplan - Write $\otimes$ for $\otimes_A$. It suffices to show $$B\otimes aA=(B\otimes aB)\cap(B\otimes A).\quad()$$ There is a sub-$B$-module $M$ of $B\otimes B$ such that $$B\otimes B=(B\otimes A)\oplus M.$$ We get $$B\otimes aB=a(B\otimes B)=(B\otimes aA)\oplus aM.$$ Intersecting with $B\otimes A$ gives $()$. – Pierre-Yves Gaillard Dec 07 '18 at 11:12
  • @Pierre-YvesGaillard: Could you explain why the intersection of $B\otimes aB=a(B\otimes B)=(B\otimes aA)\oplus aM$ with $B \otimes A$ gives (*)? It's generally not clear to me why intersection should "commute" with tensor products and direct sums – user267839 Dec 08 '18 at 23:54
  • @KarlPeter - Recall that we have $$B\otimes B=(B\otimes A)\oplus M.$$ It's enough to show $$\Big((B\otimes A)\oplus0\Big)\cap\Big((B\otimes aA)\oplus aM\Big)\subset(B\otimes aA)\oplus0.$$ (The other inclusion is obvious.) If $z$ is in the LHS, then $z=(y,0)=(x,w)$ with $$y\in B\otimes A,\ x\in B\otimes aA,\ w\in aM.$$ This gives $y=x,0=w$, and thus $z=(x,0)$ is in the RHS. (Don't hesitate to tell me if this is unclear or false!) – Pierre-Yves Gaillard Dec 09 '18 at 01:23

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