# Ideals Containing Ideal Isomorphic to Quotient Ring

## Theorem

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

Let $\mathbb L_J$ be the set of all ideals of $R$ which contain $J$.

Let the ordered set $\left({\mathbb L \left({R / J}\right), \subseteq}\right)$ be the set of all ideals of $R / J$.

Let the mapping $\Phi_J: \left({\mathbb L_J, \subseteq}\right) \to \left({\mathbb L \left({R / J}\right), \subseteq}\right)$ be defined as:

$\forall a \in \mathbb L_J: \Phi_J \left({a}\right) = q_J \left({a}\right)$

where $q_J: a \to a / J$ is the quotient epimorphism from $a$ to $a / J$ from the definition of quotient ring.

Then $\Phi_J$ is an isomorphism.

## Proof

Let $b \in \mathbb L_J$.

From the way $\mathbb L_J$ is defined:

$J \subseteq b$
$q_J^{-1} \left({q_J \left({b}\right)}\right) = b + J = b$

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

$q_J \left({q_J^{-1} \left({c}\right)}\right) = c$

Thus by Bijection iff Left and Right Inverse, $\Phi_J$ is a bijection.

Hence:

$\forall c \in \mathbb L \left({R / J}\right): q_J^{-1} \left({\Phi_J}\right) c = q_J^{-1} \left({c}\right)$

Now to show that $\Phi_J$ is an isomorphism.

Let $b_1, b_2 \in \mathbb L_J$.

Let $b_1 \subseteq b_2$.

Then from Subset Maps to Subset:

$q_J \left({b_1}\right) \subseteq q_J \left({b_2}\right)$

Conversely, suppose $q_J \left({b_1}\right) \subseteq q_J \left({b_2}\right)$.

By what we have just proved:

$b_1 = q_J^{-1} \left({q_J \left({b_1}\right)}\right) \subseteq q_J^{-1} \left({q_J \left({b_2}\right)}\right) = b_2$

Thus $\Phi_J$ is an isomorphism.

$\blacksquare$