[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:

• $f$ is the constant $0$ map: for every $x \in F$, we have $f(x) = 0_G$.
• $f$ is injective.

Proof

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.