Proposition 3.7 in Atiyah-Macdonald (Tensor product of fractions is fraction of tensor product)

Question

Let $M, N$ be two $A$ modules. The proposition states that $S^{-1}M \otimes_{S^{-1}A} S^{-1}N \cong S^{-1}(M \otimes_A N)$.

Here is my attempted proof:

We know $S^{-1}(M \otimes_A N) \cong S^{-1}A \otimes_A (M \otimes_A N)\cong (S^{-1}A \otimes_A M) \otimes_A N \cong S^{-1}M \otimes_A N$. How do I show that $S^{-1}M \otimes_A N \cong S^{-1}M \otimes_{S^{-1}A} S^{-1}N$?

Answer 1

We have \begin{align} S^{-1}M\otimes_{S^{-1}A}S^{-1}N &\cong (S^{-1}A\otimes _A M)\otimes_{S^{-1}A}(S^{-1}A\otimes_A N)&&\text{by Proposition 3.5}\\ &\cong (M\otimes_A S^{-1}A)\otimes_{S^{-1}A}(S^{-1}A\otimes_A N)&&\text{by Proposition 2.14(i)}\\ &\cong M\otimes_A (S^{-1}A\otimes_{S^{-1}A}(S^{-1}A\otimes_A N))&&\text{by Exercise 2.15}\\ &\cong M\otimes_A ((S^{-1}A\otimes_{S^{-1}A}S^{-1}A)\otimes_A N)&&\text{by Exercise 2.15}\\ &\cong M\otimes_A (N\otimes_A(S^{-1}A\otimes_{S^{-1}A}S^{-1}A))&&\text{by Proposition 2.14(i)}\\ &\cong (M\otimes_A N)\otimes_A(S^{-1}A\otimes_{S^{-1}A}S^{-1}A)&&\text{by Proposition 2.14(ii)}\\ &\cong (S^{-1}A\otimes_{S^{-1}A}S^{-1}A)\otimes_A(M\otimes_A N)&&\text{by Proposition 2.14(i)}\\ &\cong S^{-1}A\otimes_A(M\otimes_A N)&&\text{by Proposition 2.14(iv)}\\ &\cong S^{-1}(M\otimes_A N)&&\text{by Proposition 3.5} \end{align}

Answer 2

An alternative method: consider the homomorphism induced by the bilinear map, and prove it is an isomorphism.

$(m/s, n/t)\mapsto (m\otimes n)/st$ is bilinear. It induces $f$. It suffices to prove $f$ injective.

If $\sum_{i=1}^{n}(m_i\otimes n_i)/s_it_i=0$,(we only need to consider elements of that form, because $a(m\otimes n)=(am)\otimes n$.) then $\exists s'\in S$ s.t. $s'(\sum_i(S_im_i\otimes T_in_i))=0$, where $S_i=\Pi_{j\ne i}s_i$, $T_i=\Pi_{j\ne i} t_i$. Let $\pi: M\otimes_A N\rightarrow S^{-1}M\otimes_{S^{-1}A} S^{-1}N$ be the canonical homomorphism given by $m\otimes n\mapsto m/1\otimes n/1$. Then $(s'/1)(\sum_i (S_im_i/1\otimes T_in_i/1))=\pi(s)\pi(\sum_i(S_im_i\otimes T_in_i))=\pi(s'(\sum_i(S_im_i\otimes T_in_i)))=\pi(0)=0$. Hence $(s'/(s'ST))(\sum_i (S_im_i/1\otimes T_in_i/1))=(1/s'ST)(s'/1)(\sum_i (S_im_i/1\otimes T_in_i/1))=0$ (where $S=\Pi s_i,\ T_i=\Pi t_i$) $\Rightarrow (1/(ST))(\sum_i (S_im_i/1\otimes T_in_i/1))$, $\Rightarrow$ $\sum_i (m_i/s_i)\otimes(n_i/t_i)=0$ i.e. $\ker f=\{0\}$