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I have problem to understand and solve the exercise 1.2.14 on Qing Liu's book "Algebraic Geometry and Arithmetic Curves". It goes as follows:

Let $A\to B$ be a ring homomorphism, and let $J$ be an ideal of $B$ such that $B/J$ is flat over $A$. Show that for any ideal $I$ of $A$, we have that $(IB)\cap J=IJ$ (tensor the injection $I\to A$ by $B/J$).

Where do we need the homomorphism $A\to B$ as Liu does not use it explicitly in the problem? Could someone elaborate why tensoring the injection helps to prove the theorem?

Jaakko Seppälä
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1 Answers1

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I don't get Liu's hint, but the following it's useful and not hard to prove:

Let $M$ be a flat $A$-module. Then if tensor a short exact sequence of $A$-modules $0\to M_1\to M_2\to M\to 0$ by any $A$-module it remains exact.

Now consider $0\to J\to B\to B/J\to 0$ a short exact sequence of $A$-modules. By tensoring this with $A/I$ we get $0\to J\otimes_AA/I\to B\otimes_AA/I\to B/J\otimes_AA/I\to 0$. We know that $J\otimes_AA/I\simeq J/IJ$, $B\otimes_AA/I\simeq B/IB$, and $B/J\otimes_AA/I\simeq B/(IB+J)$. (I've used here that $M\otimes_AA/I\simeq M/IM$.) This shows that $(IB+J)/IB\simeq J/(IB\cap J)=J/IJ$, and we are done.

user26857
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  • A reference for the highlighted result: http://www.math.purdue.edu/~sbasu/teaching/fall08/557/Week6.pdf (Proposition 14). – user26857 Jan 27 '15 at 15:55
  • I am a little confused about what is meant by $IJ$ here as $I$ and $J$ are ideals of different ring. We have the following result: If $R/I$ is a flat $R$ module, then for any ideal $J$ of $R$, $I\cap J=IJ$. Using this result in the above situation we get $J\cap IB=JIB$. THis could perhaps resolve your concern why the homomorphism $f$ is mentioned, $IB$ is the ideal generated by $f(I)$. –  May 18 '16 at 07:53
  • @user114539 Reading your comment up to the end, I don't think you are confused at all about the meaning of $IJ$. Anyway, one can use the result you mentioned as an "alternative" approach, but, of course, this should be proved, and what I did is more or less a proof for such a result. – user26857 May 18 '16 at 09:36
  • You are correct, what you have proved more or less proves what I have used. As you notice, it is just a remark and not an answer. –  May 18 '16 at 09:52