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 poset $$\left({\mathbb{L} \left({R / J}\right); \subseteq}\right)$$ 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 canonical epimorphism from $$a$$ to $$a / J$$ as defined in Quotient Ring.

Then $$\Phi_J$$ is an isomorphism.

Prof

 * Let $$b \in \mathbb{L}_J$$.

From the way $$\mathbb{L}_J$$ is defined, $$J \subseteq b$$.

Thus $$q_J^{-1} \left({q_J \left({b}\right)}\right) = b + J = b$$ by Ring Epimorphism Composite with Inverse of Subring.

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

Then, by Ring Epimorphism Composite Of Subring with Inverse, $$q_J \left({q_J^{-1} \left({c}\right)}\right) = c$$.

Thus by Bijection iff Left and Right Inverse, $$\Phi_J$$ is a bijection, and $$\forall c \in \mathbb{L} \left({R / J}\right): q_J^{-1} \left({\Phi_J}\right) c = q_J^{-1} \left({c}\right)$$.


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

Let $$b_1, b_2 \in \mathbb{L}_J$$.

If $$b_1 \subseteq b_2$$, 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, $$b_1 = q_a^{-1} \left({q_a \left({b_1}\right)}\right) \subseteq q_a^{-1} \left({q_a \left({b_2}\right)}\right) = b_2$$.

Thus $$\Phi_J$$ is an isomorphism.