Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism

Theorem
Let $$\left({R_1, +_R, \cdot}\right)$$ and $$\left({S, +_S, *}\right)$$ be rings.

Let $$\varphi: R \to S$$ a ring homomorphism.

Then if $$R$$ is a field, then $$\varphi$$ is injective or trivial (that is, $$\forall ~a \in R: \varphi \left({a}\right) = 0_S$$).

Proof
As the kernel of a homomorphism is an ideal of $R_1$, and the only ideals of a division ring are trivial, we have $$\operatorname{ker} \left({\varphi}\right) = \left\{{0_R}\right\}$$ or $$R$$.

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

If $$\operatorname{ker}\left({\varphi}\right) = R$$, $$\varphi$$ is trivial.

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

As $$\varphi'$$ 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.