Ring Homomorphism from Field is Monomorphism or Zero Homomorphism

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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.

$\blacksquare$


Also see


Sources

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