# Atiyah Macdonald- Commutative algebra Chapter 2 Question 13

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

• 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. – timotheechalamet Sep 18 '19 at 7:52

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 $$1 \otimes_A bn = g(bn) \in \text{im}(g)$$.