All field
homomorphisms are either trivial or
injective
(This proof is modified from math.stackexchange.com/a/2402008)
Say a field
homomorphism f:A→B is
trivial if it is the constant
function f(x)=0B. A sufficient condition for this is that it maps two distinct values
a1,a2∈A to
0B: say that’s the case, and WLOG assume
a1 is not
0A (since one of
a1,a2 must not be); then for any
a∈A we have
f(a)=f(aa1a1−1)=f(aa1−1)f(a1)=f(aa1−1)0B=0B.
Now take a
homomorphism f:A→B which is not trivial. Since
f is a
homomorphism, we know for all
a,b∈A that
f(b−a)=f(b)−f(a)
Say that
f(b)=f(a). Then
f(b)−f(a) is
0B, so
f(b−a) is
0A (since
f is nontrivial) so
b=a. Hence
f is
injective.