Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism

Theorem
Let $\left({R, +_R, \circ}\right)$ and $\left({S, +_S, *}\right)$ be rings whose zeros are $0_R$ and $0_S$ respectively.

Let $\phi: R \to S$ be a ring homomorphism.

If $R$ is a division ring, then either:
 * $(1): \quad \phi$ is a monomorphism (that is, \phi$ is injective)
 * $(2): \quad \phi$ is the zero homomorphism (that is, $\forall a \in R: \phi \left({a}\right) = 0_S$).

Proof
We have that:
 * The kernel of a homomorphism is an ideal of $R$
 * the only ideals of a division ring are trivial

So $\ker \left({\phi}\right) = \left\{{0_R}\right\}$ or $R$.

If $\ker \left({\phi}\right) = \left\{{0_R}\right\}$, then $\phi$ is injective by Kernel of Monomorphism is Trivial.

If $\ker \left({\phi}\right) = R$, $\phi$ is the zero homomorphism by definition.

Alternative Proof
From Surjection by Restriction of Codomain, we can restrict the codomain of $\phi$ and consider the mapping $\phi': R \to \operatorname {Im} \left({R}\right)$

As $\phi'$ is now a surjective homomorphism, it is by definition an epimorphism.

Then we invoke the result that an epimorphism from a division ring to a ring is either null or an isomorphism.

As an isomorphism is by definition injective, the result follows.