Kernel is Trivial iff Monomorphism

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

Let $$\ker \left({\phi}\right)$$ be the kernel of $$\phi$$.

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

Let $$\ker \left({\phi}\right)$$ be the kernel of $$\phi$$.

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

Proof for Kernel of Group Monomorphism

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

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

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


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

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

$$ $$ $$ $$ $$ $$ $$

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

Proof for Kernel of Ring 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.