Let $f:A\to B$ be a ring homomorphism, and let $N$ be a $B$-module. Regarding $N$ as an $A$-module by restriction of scalars, form the $B$-module $N_B = B\otimes_A N$. Show that the homomorphism $g:N\to N_B$ which maps $y$ to $1\otimes y$ is injective and that $g(N)$ is a direct summand of $N_B$. [Define $p:N_B\to N$ by $p(b\otimes y) = by$ and show that $N_B = \text{Im}(g)\oplus\text{Ker}(p)$.]
First, as a $A$-module homomorphism, $p\circ g =1_{N}$ so $g$ is injective.
I want to prove the summand part in two ways first by finding isomorphism between $N_B$ and $\text{Im}(g)\oplus\text{Ker}(p)$ and next by splitting lemma.
1). How can I define a $A$-module isomorphism between $N_B$ and $\text{Im}(g)\oplus\text{Ker}(p)$?
I first tried this by defining $b\otimes n\mapsto (1\otimes p(b\otimes n),b\otimes n-(g\circ p)(b\otimes n))$ as $b\otimes n = 1\otimes p(b\otimes n)+b\otimes n-(g\circ p)(b\otimes n)$. But I think this map is not surjective.
2). This is a verification question. For splitting lemma, I considered the exact sequence $0\to\text{Ker}(p)\to N_B\xrightarrow{p} N\to 0$. Since $p\circ g = 1_N$, by splitting lemma, $N_B\simeq N\oplus \text{Ker}(p)\simeq \text{Im}(g)\oplus\text{Ker}(p)$ as $g$ is injective. Is this correct?
I saw some related post to this question(defining explicit isomorphism) but I think the defined map in the post is wrong. Anyway, any comment will be appreciated.
Edit: Both $A,B$ are commutative rings with unity
Best Answer
1)
Take an $x\in N_B$, and decompose it as $$ \Big(g\circ p(x),x-g\circ p(x)\Big) $$ Clearly the first component is in the image of $g$. And since $g\circ p$ (which maps $b\otimes_A n$ to $1\otimes_A bn$) is idempotent, the second component is clearly in the kernel of $g\circ p$, which by injectivity of $g$ means the kernel of $p$.
As for mapping in the other direction, we have simply $(x,y)\mapsto x+y$.
2)
Yes, that looks good.