# Ring Homomorphism from Field is Monomorphism or Zero Homomorphism

## Theorem

Let $\struct {F, +_F, \circ}$ be a field whose zero is $0_F$.

Let $\struct {S, +_S, *}$ be a ring whose zero is $0_S$.

Let $\phi: F \to S$ be a ring homomorphism.

Then either:

$(1): \quad \phi$ is a monomorphism (that is, $\phi$ is injective)

or

$(2): \quad \phi$ is the zero homomorphism (that is, $\forall a \in F: \map \phi a = 0_S$).

## Proof 1

We have by definition that a field is a division ring.

The result can be seen to be an application of Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism.

$\blacksquare$

## Proof 2

Let $\phi: F \to S$ be a ring homomorphism.

Suppose $\phi$ is not a monomorphism.

By definition, $\phi$ is not an injection.

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

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

Then:

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

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

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

 $\ds \map \phi x$ $=$ $\ds \map \phi {k \circ \paren {k^{-1} \circ x} }$ $\ds$ $=$ $\ds \map \phi k * \map \phi {k^{-1} \circ x}$ $\ds$ $=$ $\ds 0_S * \map \phi {k^{-1} \circ x}$ $\ds$ $=$ $\ds 0_S$

So if $\phi$ is not a monomorphism, it is the zero homomorphism.

$\blacksquare$