Ring Homomorphism whose Kernel contains Ideal

Theorem
Let $R$ be a ring.

Let $J$ be an ideal of $R$.

Let $\nu: R \to R / J$ be the quotient epimorphism.

Let $\phi: R \to S$ be a ring homomorphism such that:
 * $J \subseteq \map \ker \phi$

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

Then there exists a unique ring homomorphism $\psi: R / J \to S$ such that:
 * $\phi = \psi \circ \nu$

where $\circ$ denotes composition of mappings.


 * CommDiagRingHomKerIdeal.png

Also:
 * $\map \ker \psi = \map \ker \phi / J$

Proof
Suppose $\phi = \psi \circ \nu$.

Let $J + x$ be an arbitrary element of $R / J$.

Then:
 * $(1) \quad \map \psi {J + x} = \psi \circ \map \nu x = \map \phi x$

So there is only one possible way to define $\psi$.

Now suppose $J + x = J + x'$.

Then $x + \paren {-x'} \in J$.

So $x + \paren {-x'} \in \map \ker \phi$ as $J \subseteq \map \ker \phi$.

That is, $\map \phi {x + \paren {-x'} } = 0_S$.

So $\map \phi x = \map \phi {x'}$ and so $\psi$ as defined in $(1)$ is well-defined.

Now suppose $J + x, J + y \in R / J$.

We have:

So $\psi$ preserves ring addition.

Then:

So $\psi$ preserves ring product, and so $\psi$ is a ring homomorphism.

Finally:

So:
 * $\map \ker \psi = \map \ker \phi / J$