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I am working on a problem and after much work I have gotten out an answer $2\mathbb{Z}_8\otimes \mathbb{Z}_4$ which I decided to take on a detour and work with in my mind, mostly to see if it is isomorphic to $\mathbb{Z}_4\otimes \mathbb{Z}_4$ which struck me as the most "evident" thing. We are working with modules here over the ring $R=\mathbb{Z}_8$. We have that

$$2\mathbb{Z}_8=\{0,2,4,6\}$$ so a $R$-homomorphism of $\varphi(a\otimes b)=\frac{a}{2}\otimes b$ should suffice. It is a monomorphism because we have $\varphi(a\otimes b)=0$ we have that either $b=0$ or $\frac{a}{2}=0$, the latter implying that $a=0$ which means that $\ker \varphi = 0$ and ergo monomorphism. I am a bit uncertain about $0$'s in tensor products so I may have done wrong here.

for epimorphism I think it is self-evident, or I may be really really wrong here. Is my reasoning right? Or is the gut feeling right but reasoning wrong or am I again completely wrong?

Zelos Malum
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1 Answers1

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Simply use that $2\mathbf Z/8\mathbf Z\simeq\mathbf Z/4\mathbf Z$ as $\mathbf Z$-modules, and that for any $R/I$-modules $M$ and $N$, $M\otimes_RN\simeq M\otimes_{R/I}N$.

Bernard
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