# Field Homomorphism is either Trivial or Injection

## Theorem

Let $\struct {E, +_E, \times_E}$ and $\struct {F, +_F, \times_F}$ be fields.

Let $\phi: E \to F$ be a (field) homomorphism.

Then $\phi$ is either an injection or the trivial homomorphism.

## Proof 1

This is an instance of Ring Homomorphism from Field is Monomorphism or Zero Homomorphism.

$\blacksquare$

## Proof 2

Let $\phi: E \to F$ be a field homomorphism.

Suppose $\phi$ is not an injection.

So there must exist $a, b \in F: \map \phi a = \map \phi b$.

Let $k = a +_E \paren {-b}$.

Then:

 $\ds \map \phi k$ $=$ $\ds \map \phi {a +_E \paren {-b} }$ $\ds$ $=$ $\ds \map \phi a +_F \map \phi {-b}$ $\ds$ $=$ $\ds \map \phi a +_F \paren {-\map \phi b}$ $\ds$ $=$ $\ds 0_F$ as $\map \phi a = \map \phi b$

As $a \ne b$ then $k \ne 0_E$ and so has a product inverse $\exists k^{-1} \in E$.

So for any $x \in E$ we can write $x = k \circ \paren {k^{-1} \circ x}$ and so:

 $\ds \map \phi x$ $=$ $\ds \map \phi {k \times_E \paren {k^{-1} \times_E x} }$ $\ds$ $=$ $\ds \map \phi k \times_F \map \phi {k^{-1} \times_E x}$ $\ds$ $=$ $\ds 0_F \times_F \map \phi {k^{-1} \times_E x}$ $\ds$ $=$ $\ds 0_F$

So if $\phi$ is not an injection, it is the trivial homomorphism.

$\blacksquare$