Proof. Let $F$ be a field of characteristic $0$ We want to find a one-to-one field homomorphism $\phi:\rat\to F$ For $\frac{m}{n}\in\mathbb{Q}$ , with $m,\,n$ coprime, define the mapping $\phi$ that takes $\frac{m}{n}$ into $\frac{m1_F}{n1_F}\in F$ It is easy to check that $\phi$ is a well-defined function. Furthermore, it is elementary to show

1. additive: for $p,q\in\rat$ $\phi(p+q)=\phi(p)+\phi(q)$
2. multiplicative: for $p,q\in\rat$ $\phi(pq)=\phi(p)\phi(q)$
3. $\phi(1)=1_F$ and
4. $\phi(0)=0_F$

This shows that $\phi$ is a field homomorphism. Finally, if $\phi(p)=0$ and $p\ne 0$ then $1=\phi(1)=\phi(pp^{-1})=\phi(p)\phi(p^{-1})=0\cdot\phi(p^{-1})=0$ a contradiction.

Proof. Let $F$ be a field of characteristic $p$ The idea again is to find an injective field homomorphism, this time, from $\fp$ into $F$ Take $\phi$ to be the function that maps $m\in \fp$ to $m\cdot 1_F$ It is well-defined, for if $m=n$ in $\fp$ then $p\mid (m-n)$ meaning $(m-n)1_F=0$ or that $m\cdot 1_F=n\cdot 1_F$ (showing that one element in $\fp$ does not get ``mapped'' to more than one element in $F$ . Since the above argument is reversible, we see that $\phi$ is one-to-one.

To complete the proof, we next show that $\phi$ is a field homomorphism. That $\phi(1)=1_F$ and $\phi(0)=0_F$ are clear from the definition of $\phi$ Additivity and multiplicativity of $\phi$ are readily verified, as follows:

• $\phi(m+n)=(m+n)\cdot 1_F=m\cdot 1_F + n\cdot 1_F=\phi(m)+\phi(n)$
• $\phi(mn)=mn\cdot 1_F=mn\cdot 1_F\cdot 1_F=(m\cdot 1_F)(n\cdot 1_F)=\phi(m)\phi(n)$

This shows that $\phi$ is a field homomorphism.