Field homomorphism is trivial or injective

by Patrick Stevens Dec 31 2016

Field homomorphisms preserve a *lot* of structure; they preserve so much structure that they are always either injective or totally boring.

[summary(Technical): Let $~$F$~$ and $~$G$~$ be fields, and let $~$f: F \to G$~$ be a [-field_homomorphism]. Then $~$f$~$ either is the constant $~$0$~$ map, or it is injective.]

[summary: A structure-preserving map between two fields turns out either to be totally trivial (sending every element to $~$0$~$) or it preserves so much structure that its image is an embedded copy of the domain. More succinctly, if $~$f$~$ is a field homomorphism, then $~$f$~$ is either the constant $~$0$~$ map, or it is injective.]

Let $~$F$~$ and $~$G$~$ be fields, and let $~$f: F \to G$~$ be a [-field_homomorphism]. Then one of the following is the case:


Let $~$f: F \to G$~$ be non-constant. We need to show that $~$f$~$ is injective; equivalently, for any pair $~$x,y$~$ of elements with $~$f(x) = f(y)$~$, we need to show that $~$x = y$~$.

Suppose $~$f(x) = f(y)$~$. Then we have $~$f(x)-f(y) = 0_G$~$; so $~$f(x-y) = 0_G$~$ because $~$f$~$ is a field homomorphism and so respects the "subtraction" operation. Hence in fact it is enough to show the following sub-result:

Suppose $~$f$~$ is non-constant. If $~$f(z) = 0_G$~$, then $~$z = 0_F$~$.

Once we have done this, we simply let $~$z = x-y$~$.

Proof of sub-result

Suppose $~$f(z) = 0_G$~$ but that $~$z$~$ is not $~$0_F$~$, so we may find its multiplicative inverse $~$z^{-1}$~$.

Then $~$f(z^{-1}) f(z) = f(z^{-1}) \times 0_G = 0_G$~$; but $~$f$~$ is a homomorphism, so $~$f(z^{-1} \times z) = 0_G$~$, and so $~$f(1_F) = 0_G$~$.

But this contradicts that the Image of the identity under a group homomorphism is the identity, because we may consider $~$f$~$ to be a Group homomorphism between the multiplicative groups $~$F \setminus \{ 0_F \}$~$ and $~$G \setminus \{0_G\}$~$, whereupon $~$1_F$~$ is the identity of $~$F \setminus \{0_F\}$~$, and $~$1_G$~$ is the identity of $~$F \setminus \{0_G\}$~$.

Our assumption on $~$z$~$ was that $~$z \not = 0_F$~$, so the contradiction means that if $~$f(z) = 0_G$~$ then $~$z = 0_F$~$. This proves the sub-result and hence the main theorem.