homomorphisms from fields are either injective or trivial

Suppose $F$ is a field, $R$ is a ring, and $\varphi :F\to R$ is a homomorphism of rings. Then $\varphi$ is either trivial or injective.

Proof.

We use the fact that kernels of ring homomorphism are ideals. Since $F$ is a field, by the above result, we have that the kernel of $\varphi$ is an ideal of the field $F$ and hence either empty or all of $F$. If the kernel is empty, then since a ring homomorphism is injective iff the kernel is trivial, we get that $\varphi$ is injective. If the kernel is all of $F$, then $\varphi$ is the zero map from $F$ to $R$.∎

Finally, it is clear that both of these possibilities are in fact achieved:

• The map $\varphi :\mathbb{Q}\to \mathbb{Q}$ given by $\varphi (n)=0$ is trivial (has all of $\mathbb{Q}$ as a kernel)

• The inclusion $\mathbb{Q}\to \mathbb{Q}[x]$ is injective (i.e. the kernel is trivial).