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Exercise
Let $I$ be an ideal on $R$ and $M$ an $R$-module. Prove that
\[M\otimes_R \frac{R}{I}\cong \frac{M}{IM}.\]In particular: $M\otimes_R R\cong R\otimes_R M\cong M$.
Proof
Define the function
\[\begin{align*} \varphi: M \times \frac{R}{I} &\to \frac{M}{IM}\\[3mm] (m, \overline{r}) &\mapsto \overline{rm}. \end{align*}\]This function is well-defined because if $ \overline{r} = \overline{s} $ in $ R/I $, then $ r - s \in I $, which implies that
\[rm - sm = (r - s)m \in IM,\]so that $ \overline{rm} = \overline{sm} $ in $ M/IM $.
It is easy to check the bilinearity of $ \varphi $. By [1, Lemma 10.12.2], $ \varphi $ induces a well-defined $ R $-module homomorphism given by
\[\begin{align*} \tilde{\varphi}: M \otimes_R \frac{R}{I} &\to \frac{M}{IM}\\[3mm] m \otimes \overline{r} &\mapsto \overline{rm}. \end{align*}\]It is also easy to verify that $ \tilde{\varphi} $ is surjective, since for every $ \overline{m} \in M/IM $, we have $ \tilde{\varphi}(m \otimes \overline{1}) = \overline{m}$.
We only need to verify that $\tilde{\varphi}$ is injective, for which we will prove that its kernel is trivial. Let $\sum_{i=1}^n m_i\otimes \overline{r_i}\in \ker(\tilde{\varphi})$, which means that
\[0_{M/IM}=\tilde{\varphi}\left(\sum_{i=1}^n m_i\otimes \overline{r_i}\right)=\sum_{i=1}^n \overline{r_i m_i}\]and this happens if and only if
\[\sum_{i=1}^n r_i m_i=\sum_{j=1}^N a_j m_j'\]with $ a_j\in I $ and $ m_j’\in M $. By bilinearity, we have
\[\begin{align*} \sum_{i=1}^n m_i\otimes \overline{r_i} &=\left(\sum_{i=1}^n r_i m_i\right)\otimes \overline{1}\\ &=\left(\sum_{j=1}^N a_j m_j'\right)\otimes \overline{1}\\ &=\sum_{j=1}^N m_j'\otimes \overline{a_j}\\ &=\sum_{j=1}^N m_j'\otimes 0_{R/I}\\ &=0_{M\otimes R/I}. \end{align*}\]Thus, $\ker(\tilde{\varphi})$ is trivial, and $\tilde{\varphi}$ is an isomorphism. $\blacksquare$
References
- The Stacks Project Authors. (n.d.). Tensor products (Section 10.12, Tag 00CV). The Stacks Project. Retrieved February 13, 2025, from https://stacks.math.columbia.edu/tag/00CV