Let $F$ and $K$ be fields.

Proof. Indeed, if $\psi$ is a field homomorphism, in particular it is a ring homomorphism. Note that the kernel of a ring homomorphism is an ideal and a field $F$ only has two ideals, namely $\{0\}, F$ Moreover, by the definition of field homomorphism, $\psi(1)=1$ hence $1$ is not in the kernel of the map, so the kernel must be equal to $\{0\}$

Remark: For this reason the terms ``field homomorphism'' and ``field monomorphism'' are synonymous. Also note that if $\psi$ is a field monomorphism, then $$\psi(F)\cong F, \quad \psi(F)\subseteq K$$ so there is a ``copy'' of $F$ in $K$ In other words, if $$\psi\colon F\to K$$ is a field homomorphism then there exist a subfield $H$ of $K$ such that $H\cong F$ Conversely, suppose there exists $H\subset K$ with $H$ isomorphic to $F$ Then there is an isomorphism $$\chi \colon F \to H$$ and we also have the inclusion homomorphism $$\iota\colon H \hookrightarrow K$$ Thus the composition $$\iota \circ \chi\colon F \to K$$ is a field homomorphism.

Remark: Let $\psi : F \to K$ be a field homomorphism. We claim that the characteristic of $F$ and $K$ must be the same. Indeed, since $\psi(1_F)=1_K$ and $\psi(0_F)=0_K$ then $\psi(n\cdot 1_F)=n\cdot 1_K$ for all natural numbers $n$ If the characteristic of $F$ is $p>0$ then $0=\psi(p\cdot 1)=p\cdot 1$ in $K$ and so the characteristic of $K$ is also $p$ If the characteristic of $F$ is $0$ then the characteristic of $K$ must be $0$ as well. For if $p\cdot 1=0$ in $K$ then $\psi(p\cdot 1)=0$ and since $\psi$ is injective by the lemma, we would have $p\cdot 1=0$ in $F$ as well.