Suppose that $\pi$ denotes the finite ring of order $2$, $\mathbb Z_+$ the trivial $\pi$-module and $\mathbb Z_-$ the non-trivial $\pi$-module (i.e., the generator acts by multiplication by −1).
I want to compute the product $\mathbb Z_+ \otimes_{\mathbb Z/2\mathbb Z} \mathbb Z_-$.
Can someone give a solution?
So, if I understand correctly, we have $\pi=\Bbb Z/2\Bbb Z=\{{\bf0},{\bf1}\}$ as ring, and we take $\Bbb Z$ as both a $\pi$-module $\Bbb Z_+$ where ${\bf1}\cdot x=x$ and a $\pi$-module $\Bbb Z_-$ where ${\bf1}\bullet x=-x$ (and of course ${\bf0}\cdot x=0={\bf0}\bullet x$).
Then, for any $n\in\Bbb N$. on one hand, only by addition we have $$n\otimes 1=(1\otimes 1)+\dots+(1\otimes 1)=1\otimes n$$ and $$n\otimes 1 \ +\ n\otimes(-1)=n\otimes(1+(-1))=n\otimes0=0$$ On the other hand, using multiplication, $$n\otimes 1=({\bf1}\cdot n)\otimes1=n\otimes({\bf1}\bullet 1 )=n\otimes(-1)=-(n\otimes 1)\,.$$
Using these ideas, try to prove that $\Bbb Z_+\otimes_{\pi}\Bbb Z_-\cong\Bbb Z/2\Bbb Z$.