Quotient Ring of Noetherian Ring is Noetherian

Theorem
Let $A$ be a Noetherian ring.

Let $\mathfrak a \subseteq A$ be an ideal.

Let $A / \mathfrak a$ be the quotient ring of $A$ by $\mathfrak a$.

Then $A / \mathfrak a$ is a Noetherian ring.

Proof
Observe that:
 * $0 \longrightarrow \mathfrak a \longrightarrow A \longrightarrow A / \mathfrak a \longrightarrow 0$

is a short exact sequence of $A$-modules.

By Short Exact Sequence Condition of Noetherian Modules, $A / \mathfrak a$ is a Noetherian $A$-module.

As $A / \mathfrak a$ is an $A / \mathfrak a$-module, $A / \mathfrak a$ is also a Noetherian $A / \mathfrak a$-module.

By, $A / \mathfrak a$ is a Noetherian ring.