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.
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:
\(\ds \map \psi {\paren {J + x} + \paren {J + y} }\) | \(=\) | \(\ds \map \psi {J + \paren {x + y} }\) | Definition of Quotient Ring | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {x + y}\) | Definition of $\psi$ in $(1)$ above | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi x + \map \phi y\) | Definition of Ring Homomorphism | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \psi {J + x} + \map \psi {J + y}\) |
So $\psi$ preserves ring addition.
Then:
\(\ds \map \psi {\paren {J + x} \paren {J + y} }\) | \(=\) | \(\ds \map \psi {J + \paren {x y} }\) | Definition of Quotient Ring | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {x y}\) | Definition of $\psi$ in $(1)$ above | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi x \map \phi y\) | Definition of Ring Homomorphism | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \psi {J + x} \map \psi {J + y}\) |
So $\psi$ preserves ring product, and so $\psi$ is a ring homomorphism.
Finally:
\(\ds \map \psi {J + x}\) | \(=\) | \(\ds 0_S\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \map \phi x\) | \(=\) | \(\ds 0_S\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(\in\) | \(\ds \map \ker \phi\) |
So:
- $\map \ker \psi = \map \ker \phi / J$
$\blacksquare$
Sources
- 1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): $\S 2.2$: Homomorphisms: Theorem $2.8$