Ring Homomorphism from Ring with Unity to Integral Domain Preserves Unity

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
$\map \phi {1_R} = 0_D$.

Let $x \in R$ be arbitrary.

Then:

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

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

Then we have:

Hence the result.