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 $f\left(x\right)=0_G$.
• $f$ 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 $f\left(x\right)=f\left(y\right)$, we need to show that $x=y$.

Suppose $f\left(x\right)=f\left(y\right)$. Then we have $f\left(x\right)-f\left(y\right)=0_G$; 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 $f\left(z\right)=0_G$, then $z=0_F$.

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

Proof of sub-result

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

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

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 $F\setminus \left\{0_F\right\}$ and $G\setminus \left\{0_G\right\}$, whereupon $1_F$ is the identity of $F\setminus \left\{0_F\right\}$, and $1_G$ is the identity of $F\setminus \left\{0_G\right\}$.

Our assumption on $z$ was that $z\ne 0_F$, so the contradiction means that if $f\left(z\right)=0_G$ then $z=0_F$. This proves the sub-result and hence the main theorem.