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 $F ∖ {0_{F}}$ and $G ∖ {0_{G}}$ , whereupon $1_{F}$ is the identity of $F ∖ {0_{F}}$ , and $1_{G}$ is the identity of $F ∖ {0_{G}}$ .

Our assumption on $z$ was that $z \neq 0_{F}$ , so the contradiction means that if $f (z) = 0_{G}$ then $z = 0_{F}$ . This proves the sub-result and hence the main theorem.