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:

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

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The inclusion $\mathbb{Q}\to \mathbb{Q}[x]$ is injective (i.e. the kernel is trivial).
Title  homomorphisms from fields are either injective or trivial 

Canonical name  HomomorphismsFromFieldsAreEitherInjectiveOrTrivial 
Date of creation  20130322 14:39:07 
Last modified on  20130322 14:39:07 
Owner  mathcam (2727) 
Last modified by  mathcam (2727) 
Numerical id  4 
Author  mathcam (2727) 
Entry type  Corollary 
Classification  msc 12E99 