Lemma 10.12.10. Let
\begin{align*} M_1\xrightarrow {f} M_2\xrightarrow {g} M_3 \to 0 \end{align*}
be an exact sequence of $R$-modules and homomorphisms, and let $N$ be any $R$-module. Then the sequence
10.12.10.1
\begin{equation} \label{algebra-equation-2ndex} M_1\otimes N\xrightarrow {f \otimes 1} M_2\otimes N \xrightarrow {g \otimes 1} M_3\otimes N \to 0 \end{equation}
is exact. In other words, the functor $- \otimes _ R N$ is right exact, in the sense that tensoring each term in the original right exact sequence preserves the exactness.
Proof.
We apply the functor $\mathop{\mathrm{Hom}}\nolimits (-, \mathop{\mathrm{Hom}}\nolimits (N, P))$ to the first exact sequence. We obtain
\[ 0 \to \mathop{\mathrm{Hom}}\nolimits (M_3, \mathop{\mathrm{Hom}}\nolimits (N, P)) \to \mathop{\mathrm{Hom}}\nolimits (M_2, \mathop{\mathrm{Hom}}\nolimits (N, P)) \to \mathop{\mathrm{Hom}}\nolimits (M_1, \mathop{\mathrm{Hom}}\nolimits (N, P)) \]
By Lemma 10.12.8, we have
\[ 0 \to \mathop{\mathrm{Hom}}\nolimits (M_3 \otimes N, P) \to \mathop{\mathrm{Hom}}\nolimits (M_2 \otimes N, P) \to \mathop{\mathrm{Hom}}\nolimits (M_1 \otimes N, P) \]
Using the pullback property again, we arrive at the desired exact sequence.
$\square$
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