# homomorphisms from fields are either injective or trivial

###### 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 $\phi$ 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 $\phi$ is injective. If the kernel is all of $F$, then $\phi$ is the zero map from $F$ to $R$. ∎

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

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

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

Title homomorphisms from fields are either injective or trivial HomomorphismsFromFieldsAreEitherInjectiveOrTrivial 2013-03-22 14:39:07 2013-03-22 14:39:07 mathcam (2727) mathcam (2727) 4 mathcam (2727) Corollary msc 12E99