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This question is from the book Atiyah Macdonald- Commutative algebra.

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Firstly I know that $g$ is an $A$-module homomorphism and $p$ is both an $A$-module and $B$-module homomorphism. Also $pg = id_N$. Now for any $b \otimes_A n \in N_B$,

$gp(b \otimes_A n) - b \otimes_A n$ $\in Ker(p) $

So, $b \otimes_A n = (gp(b \otimes_A n)-b \otimes_A n)+(b \otimes_A n+b \otimes_A n-gp(b \otimes_A n))$

So, I am left to show that $b \otimes_A n \in Im(g)$.

$b \otimes_A n = b(1 \otimes_A n)$ [ since $N_B$ is a $B$-module] $= b g(n) $

But I cannot show this is equal to $g(bn)$. How do I proceed?

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  • $\begingroup$ I'm trying to break $b \otimes n$ into two terms.One of them is in ker p so I'm trying to show the other is in Im g. $\endgroup$ – timotheechalamet Sep 18 '19 at 7:52
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Notice that $g$ may not be an $B$-module homomorphism.

Instead of trying to show $bg(n)=g(bn)$, you could use your result $p \circ g =id_N$.

Let $b \otimes_A n \in N_B$ be given. Then $p(b \otimes_A n)=bn=(p \circ g)(bn)=p(1 \otimes_A bn)$

Thus $$b \otimes_A n = (b \otimes_A n - 1 \otimes_A bn) + 1 \otimes_A bn $$ where $ (b \otimes_A n - 1 \otimes_A bn) \in \ker(p)$ and $1 \otimes_A bn = g(bn) \in \text{im}(g)$.

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