Exercise 2.13

Question

Let $f\colon A 	o B$ 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 $g\colon N 	o N_B$ which maps $y$ to $1 \otimes y$ is injective and that $g(N)$ is a direct summand of $N_B$.

Amit Awasthi · Accepted Answer

\begin{proof}
  Note that the action of $B$ on $N_B$ is given by $b' \cdot (b \otimes_A n) = (b'b) \otimes_A n$, and $A$ on $N$ by $a \cdot n = f(a) \cdot n$ by definition of restriction of scalars. Define the map $\bar{p}\colon B 	imes N 	o N$ defined by $\bar{p}(b,y) = by$. $\bar{p}$ is $A$-bilinear since letting $c \in A$,
  \begin{align*}
    \bar{p}(b+c \cdot d,y) &= \bar{p}(b+f(c)d,y) = (b + f(c)d)y = by + f(c)dy\
    &= \bar{p}(b,y) + f(c)\bar{p}(d,y) = \bar{p} + c \cdot \bar{p}(d,y),\
    \bar{p}(b,c \cdot x+y) &= \bar{p}(b,f(c)x + y) = b(f(c)x+y) = bf(c)x + by\
    &= f(c)\bar{p}(b,x) + \bar{p}{b,y} = c \cdot \bar{p}(b,x) + \bar{p}{b,y}.
  \end{align*}
  By the universal property of the tensor product (Prop.~2.12) we then get the unique homomorphism $p\colon N_B 	o N, b\otimes_A y \mapsto by$.
  \par Now consider the sequence
\begin{equation*}
\begin{array}{ccccc}
0 & ightarrow & N \stackrel{g}{ightharpoonup} & N_B & \operatorname{coker}(g) & 0 \
 & & & \leftharpoonup_{p} &
\end{array}
\end{equation*}

Since $(p \circ g)(y) = p(1 \otimes_A y) = y$, we see that $p \circ g = \operatorname{id}_N$, and so $g$ is injective; this implies the sequence is short exact. By the splitting lemma, we then see that $N_B \cong g(N) \oplus \operatorname{coker}(g)$ as desired.
\end{proof}

Exercise 2.13

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Exercise 2.13

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