3.2.7

Let $$\mathbb{F}, \mathbb{F}' $$ be two fields, and let $$\phi:\mathbb{F} \rightarrow \mathbb{F}'$$ be a mapping between them. We say that $$\phi $$ is a homomorphism of fields if for all $$f,g \in \mathbb{F} $$, $$\phi $$ satisfies $$\phi(f+g) = \phi(f) + \phi(g) $$ and $$ \phi(f*g) = \phi(f)*\phi(g)$$. That is, $$\phi $$ is compatible with both laws of composition. (Note that in the last two equations, the composition is performed in $$\mathbb{F}$$ on the left hand side, and in $$\mathbb{F}'$$ on the right hand side).

We claim that every homomorphism of fields is injective. Let $$\phi $$ be such a homomorphism, and then since $$\mathbb{F} $$ and $$\mathbb{F}' $$ form abelian groups under addition, and $$ \phi$$ is a group homomorphism compatible with addition, $$\phi $$ maps the additive identity to the additive identity, so that $$\phi(0) = 0' $$, where 0 and 0' are the additive identities for $$\mathbb{F} $$and $$ \mathbb{F}'$$ respectively. Similarly $$\phi(1) = 1' $$, since $$\mathbb{F}/\{0\} $$ and $$\mathbb{F}'/\{0'\} $$ form abelian groups under multiplication and $$\phi $$ is also compatible with multiplication.

Now suppose that for some f and g in $$\mathbb{F} $$, we have $$\phi(f) = \phi(g) $$. Then, $$ \phi(f-g) = \phi(f) - \phi(g) = 0'$$, where 0' is the additive identity of $$\mathbb{F}' $$. Letting $$q = f-g \in \mathbb{F}$$, we have $$\phi(q) = 0' $$. We claim that q = 0. For contradiction, suppose that q was not 0, then it would have a multiplicative inverse $$ q^{-1}$$, so that $$\phi(q*q^{-1} = \phi(1) = 1' $$, but also $$\phi(q*q^{-1}) = \phi(q)*\phi(q^{-1}) = 0' * \phi(q^{-1}) = 1' $$, implying that 0' has a multiplicative inverse $$ \phi(q^{-1})$$, which is a contradiction. Thus q = 0, so f=g, showing that $$\phi $$ is injective.