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