Ring Homomorphism whose Kernel contains Ideal

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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:

\(\displaystyle \map \psi {\paren {J + x} + \paren {J + y} }\) \(=\) \(\displaystyle \map \psi {J + \paren {x + y} }\) Definition of Quotient Ring
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {x + y}\) Definition of $\psi$ in $(1)$ above
\(\displaystyle \) \(=\) \(\displaystyle \map \phi x + \map \phi y\) Definition of Ring Homomorphism
\(\displaystyle \) \(=\) \(\displaystyle \map \psi {J + x} + \map \psi {J + y}\)

So $\psi$ preserves ring addition.


Then:

\(\displaystyle \map \psi {\paren {J + x} \paren {J + y} }\) \(=\) \(\displaystyle \map \psi {J + \paren {x y} }\) Definition of Quotient Ring
\(\displaystyle \) \(=\) \(\displaystyle \map \phi {x y}\) Definition of $\psi$ in $(1)$ above
\(\displaystyle \) \(=\) \(\displaystyle \map \phi x \map \phi y\) Definition of Ring Homomorphism
\(\displaystyle \) \(=\) \(\displaystyle \map \psi {J + x} \map \psi {J + y}\)

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


Finally:

\(\displaystyle \map \psi {J + x}\) \(=\) \(\displaystyle 0_S\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \map \phi x\) \(=\) \(\displaystyle 0_S\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x\) \(\in\) \(\displaystyle \map \ker \phi\)

So:

$\map \ker \psi = \map \ker \phi / J$

$\blacksquare$


Sources

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