Field Homomorphism is either Trivial or Injection/Proof 2

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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.


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


\(\ds \map \phi k\) \(=\) \(\ds \map \phi {a +_E \paren {-b} }\)
\(\ds \) \(=\) \(\ds \map \phi a +_F \map \phi {-b}\)
\(\ds \) \(=\) \(\ds \map \phi a +_F \paren {-\map \phi b}\)
\(\ds \) \(=\) \(\ds 0_F\) 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.