I know this is a duplicate question. However, I haven't seen anything that invokes the isomorphism theorem. Here's my idea:
By the isomorphism theorem we have that $M/\ker\varphi \cong \operatorname{im}\varphi = M$ (as $\varphi$ is surjective). Does this imply $\ker\varphi = (0)$?
While typing this I'm beginning suspect the implication does not follow. Can anyone explain why or why not?
Thank you! 🙂
Best Answer
Let $R=\mathbb{Z}$, and consider the $\mathbb{Z}$-module $M=\prod_{i=1}^\infty\mathbb{Z}$ (the direct product of infinitely many copies of $\mathbb{Z}$). Let $\varphi:M\to M$ be the left-shift homomorphism,defined by $$\varphi(a_1,a_2,a_3,\ldots)=(a_2,a_3,a_4,\ldots)$$ Then $\varphi$ is surjective (given any element $a=(a_1,a_2,\ldots)\in M$, we have $\varphi(0,a_1,a_2,\ldots)=a$) but not injective (we have $\varphi(n,0,0,\ldots)=(0,0,0,\ldots)$ for any $n\in\mathbb{Z}$).
However, if $M$ is noetherian $R$-module, we can conclude that $\varphi$ is injective. See my answer here and this thread for more information.