Consider a $R$-module homomorphism $\varphi\colon M\to N$. It is well-known that $\ker\varphi=\{0\}$ iff $\varphi$ is injective (equivalently is monic). However, suppose we only have $\ker\varphi\cong0$ (where $0$ is the trivial module). Does it still hold that $\varphi$ has to be injective?
I think the main reason why I struggle to accept/dismiss this is due to the fact that the equivalence given is elementwise while the second characterisation is structurewise. Intuitively, 'lifting' $\{0\}$ to the trivial module $0$ should not be problematic at all as there is only one sensible way of making $\{0\}$ into a module. Also, as $0$ is a zero object in $R$–${\rm Mod}$ it is unique up to unique isomorphism, so if $\ker\varphi\cong0$ the kernel technically already is the zero module and only has one element: $0$.
To sum up: does it hold that $\ker\varphi\cong0$ iff $\varphi$ is injective? If not, what does $\ker\varphi\cong0$ imply instead?
Thanks in advance!
Best Answer
If $\ker \varphi \cong 0$, then $\ker \varphi = \{0\}$.
To see this, suppose the former condition, then $\ker \varphi = \{a\}$ for some $a \in M$. Since the kernel is a submodule of $M$, $ra =a$ for all $r \in R$, and take $r=0$.