Ring Homomorphism from Field is Monomorphism or Zero Homomorphism/Proof 2

Theorem
Let $\left({F, +_F, \circ}\right)$ be a field whose zero is $0_F$.

Let $\left({S, +_S, *}\right)$ be rings whose zero is $0_S$.

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

Then either:
 * $(1): \quad \phi$ is a monomorphism (that is, $\phi$ is injective)

or
 * $(2): \quad \phi$ is the zero homomorphism (that is, $\forall a \in F: \phi \left({a}\right) = 0_S$).

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

Suppose $\phi$ is not a monomorphism.

By definition, $\phi$ is not an injection.

So there must exist $a, b \in F: \phi \left({a}\right) = \phi \left({b}\right)$.

Let $k = a +_F \left({-b}\right)$.

Then:

As $a \ne b$ then $k \ne 0$ and so has a product inverse $\exists k^{-1} \in F$.

So for any $x \in F$ we can write $x = k \circ \left({k^{-1} \circ x}\right)$ and so:

So if $\phi$ is not a monomorphism, it is the zero homomorphism.