Ring Homomorphism from Ring with Unity to Integral Domain Preserves Unity

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Theorem

Let $\struct {R, +_R, \circ_R}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $\struct {D, +_D, \circ_D}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.

Let $\phi: R \to D$ be a ring homomorphism such that:

$\map \ker \phi \ne R$

where $\map \ker \phi$ denotes the kernel of $\phi$.


Then $\map \phi {1_R} = 1_D$.


Proof

Aiming for a contradiction, suppose $\map \phi {1_R} = 0_D$.

Let $x \in R$ be arbitrary.

Then:

\(\ds \map \phi x\) \(=\) \(\ds \map \phi {x \circ_R 1_R}\) Definition of Unity of Ring
\(\ds \) \(=\) \(\ds \map \phi x \circ_D \map \phi {1_R}\) Definition of Ring Homomorphism
\(\ds \) \(=\) \(\ds \map \phi x \circ_D 0_D\) by hypothesis
\(\ds \) \(=\) \(\ds 0_D\) Definition of Ring Zero

But this contradicts the assertion that $\map \ker \phi \ne R$.

It follows that $\map \phi {1_R} \ne 0_D$.


Then we have:

\(\ds 1_D \circ_D \map \phi {1_R}\) \(=\) \(\ds \map \phi {1_R}\) Definition of Unity of Ring
\(\ds \) \(=\) \(\ds \map \phi {1_R \circ_R 1_R}\) Definition of Unity of Ring
\(\ds \) \(=\) \(\ds \map \phi {1_R} \circ_D \map \phi {1_R}\) Definition of Ring Homomorphism
\(\ds \leadsto \ \ \) \(\ds 1_D\) \(=\) \(\ds \map \phi {1_R}\) Cancellation Law for Ring Product of Integral Domain

Hence the result.

$\blacksquare$


Sources

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