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 natural epimorphism.

Let $$\phi: R \to S$$ be a ring homomorphism such that:
 * $$J \subseteq \ker \left({\phi}\right)$$

where $$\ker \left({\phi}\right)$$ 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:
 * $$\ker \left({\psi}\right) = \ker \left({\phi}\right) / J$$

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

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

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

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

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

Then $$x + \left({-x'}\right) \in J$$.

So $$x + \left({-x'}\right) \in \ker \left({\phi}\right)$$ as $$J \subseteq \ker \left({\phi}\right)$$.

That is, $$\phi \left({x + \left({-x'}\right)}\right) = 0_S$$.

So $$\phi \left({x}\right) = \phi \left({x'}\right)$$ 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:
 * $$\ker \left({\psi}\right) = \ker \left({\phi}\right) / J$$