Let \(R\) be a ring with 1, and let \(I\) be a two sided ideal of \(R\). Let \(M\) be a left \(R\)-module. Recall that \(IM\) is the submodule \[IM = \{a_1 m_1 + \cdots + a_n m_n : a_i \in I, m_i \in M\}.\] Prove that \((R/I) \otimes_R M \cong M/IM\) as \(R\)-modules.
Hint
Construct an \(R\)-balanced map from \((R/I)\times M\) to \(M/IM\) and use the universal mapping property of tensor products.
Solution
The isomorphism and its inverse are \[ \begin{aligned} \Phi:(R/I)\otimes_R M&\longrightarrow M/IM, & (a+I)\otimes m&\longmapsto am+IM,\\ \Psi:M/IM&\longrightarrow(R/I)\otimes_R M, &m+IM&\longmapsto(1+I)\otimes m. \end{aligned} \]
1. The tensor product and quotient have the required left\(R\)-module structures.
Because \(I\) is two-sided, \(R/I\) is an \((R,R)\)-bimodule by left and right multiplication. Put \(T=(R/I)\otimes_R M\), using the right \(R\)-action on \(R/I\) and the given left action on \(M\). For \(r\in R\), left multiplication on \(R/I\) is a right-linear map: \(r(ab)=(ra)b\). It therefore induces an additive map on \(T\), with \[ r\big((a+I)\otimes m\big)=(ra+I)\otimes m. \] The module identities follow on elementary tensors from the ring identities, and elementary tensors generate \(T\) additively. Thus this defines a left \(R\)-module structure [@DF04].
The set \(IM\) is an additive subgroup, since sums concatenate the finite expressions and negatives can be absorbed in the coefficients. For \(r\in R\), each \(ra_i\) lies in \(I\), so \(r\sum_i a_im_i=\sum_i(ra_i)m_i\in IM\). Hence \(IM\) is a left submodule, and \(M/IM\) is a left quotient module.
2. The displayed rule defines a left \(R\)-linear map \(\Phi\).
Define \(\beta:(R/I)\times M\to M/IM\) by \(\beta(a+I,m)=am+IM\). If \(a\) is replaced by \(a+i\) with \(i\in I\), the value changes by \(im\in IM\), so it is independent of the representative. It is additive in each variable. For \(r\in R\), \[ \beta((a+I)r,m)=(ar)m+IM =a(rm)+IM=\beta(a+I,rm). \] Thus \(\beta\) is balanced, and the universal property of the tensor product gives an additive map \(\Phi:T\to M/IM\) with the asserted rule [@DF04]. It is left \(R\)-linear since \[ \Phi\big(r((a+I)\otimes m)\big) =(ra)m+IM=r(am+IM). \]
3. The map \(\Psi\) is well defined and left \(R\)-linear,and is inverse to \(\Phi\).
Define \(\eta:M\to T\) by \(\eta(m)=(1+I)\otimes m\). It is additive, and balancing gives \[ \eta(rm)=(1+I)\otimes rm=(r+I)\otimes m =r\eta(m). \] For \(i\in I\) and \(m\in M\), \(\eta(im)=(i+I)\otimes m=0\). It follows by additivity that \(\eta\) vanishes on all of \(IM\). It therefore descends to a left-linear map \(\Psi:M/IM\to T\) with the displayed formula.
For every \(m+IM\), one has \(\Phi\Psi(m+IM)=m+IM\). For every elementary tensor, balancing gives \[ \Psi\Phi((a+I)\otimes m) =(1+I)\otimes am=(a+I)\otimes m. \] Since elementary tensors generate \(T\), these identities prove that both composites are the identity. Thus \(\Phi\) is the claimed left \(R\)-module isomorphism.