Kernel is Trivial iff Monomorphism

Kernel of Group Monomorphism
Let $$\phi: \left({S, \circ}\right) \to \left({T, *}\right)$$ be a group homomorphism.

Then $$\phi$$ is a monomorphism iff $$\mathrm{ker} \left({\phi}\right)$$ is trivial.

Kernel of Ring Monomorphism
Let $$\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$$ be a ring homomorphism.

Then $$\phi$$ is a ring monomorphism iff $$\mathrm{ker} \left({\phi}\right) = 0_{R_1}$$.

Proof

 * Let $$\phi: \left({S, \circ}\right) \to \left({T, *}\right)$$ be a group monomorphism.

By Homomorphism to Group Preserves Identity and Inverses, $$e_S \in \mathrm{ker} \left({\phi}\right)$$.

If $$\mathrm{ker} \left({\phi}\right)$$ contained another element $$s \ne e_S$$, then $$\phi \left({s}\right) = \phi \left({e_S}\right) = e_T$$ and $$\phi$$ would not be injective, thus not be a group monomorphism.

So $$\mathrm{ker} \left({\phi}\right)$$ can contain only one element, and that must be $$e_S$$, which is therefore the trivial subgroup of $$S$$.


 * Now suppose $$\mathrm{ker} \left({\phi}\right) = \left\{{e_S}\right\}$$.

Then, for $$x, y \in S$$:

Thus $$\phi$$ is injective, and therefore a group monomorphism.

The proof for the ring monomorphism follows directly from Ring Homomorphism of Addition is Group Homomorphism and the above result for the group monomorphism.