Can u help me prove that if $N$ is a free Module and $0 \rightarrow M_1 \rightarrow M_2 \rightarrow M_3 \rightarrow 0 $ is a short exact sequence, with $f_1 : M_1 \rightarrow M_2$ , in this case is gonna be injective, then $ 0 \rightarrow M_1\bigotimes_AN \rightarrow M_2\bigotimes_AN \rightarrow M_3\bigotimes_AN \rightarrow 0$ is also a short exact sequence. My ony problem is proving that $f_1\bigotimes id$ is injective. I tried using the fact that if N is free then $N\cong \bigoplus_{i \in I} A$ but im not sure if it is the correct way to think about it, Thanks in advance.
Short exact sequences of tensor products
abstract-algebramodulestensor-products
Best Answer
Consider $0\to M\to N$ short exact sequence of modules over ring $A$. Set $F=\oplus_{i\in I}A$ where $I$ is abstract index. For any $A$ module $K$, $K\otimes_AF\cong\oplus_{i\in I} K$. This is a natural morphism. Compare the following sequence with previous sequence.
$0\to M\to N$
$0\to F\otimes M\to F\otimes N$
By naturality of $F\otimes_A(-)\to Id$, one deduces inducing isomorphism vertically between 2 short exact sequences. Note that the diagram is commutative. If upper map is injective, the lower map $F\otimes M\to F\otimes N$ will be injective by commutativity and isomorphism. Take $x\in Ker(F\otimes M\to F\otimes N)$. Now $x$ lifts to an element $x'$ of $M$ by isomorphism and this element has image in $N$ by $M\to N$ map. However $N\cong F\otimes N$ and commutativity of diagram forces the image in $N$ being $0$. Hence by $M\to N$ injective $x'=0$.