Mapping property of complex fraction field

Question

I recently came across a proof which said that: Suppose $\phi: \mathbb{C}[x]\rightarrow \mathcal{F}$ where $\mathcal{F}$ is a field is a homomorphism. If $ker\phi=0$ then $\phi$ maps isomorphically to subring of $\mathcal{F}$. This can be extended to a injective map $\mathbb{C}(x)\rightarrow \mathcal{F}$. Here $\mathbb{C}(x)$ is a fraction field. My question is why is such an extension injective? I think this has to do with the universal mapping property but I am unable to prove the injectivity.

Answer 1

A more general statement might be easier to prove:

If $\varphi: F \to R$ is a ring homomorphism from a field to a nonzero ring, then $\varphi$ is injective