Theorem. Let $A$ be an algebra and $I\subseteq A$ be an ideal. There is a one-to-one correspondence between the ideals of $A$ containing $I$ and the ideals of $A/I$.
Take quotient map $\pi: A \to A/I$. It's straightforward to check that
- The image of an ideal (containing $I$) in $A$ under $\pi$ is an ideal in $A/I$.
- The preimage of an ideal in $A/I$ under $\pi$ is an ideal in $A$.
So $\pi$ allows us to obtain a surjective map $\pi':\{\text{Ideals in } A \text{ that contain } I\}\to \{\text{Ideals in } A/I\}$.
Main question. Why is $\pi'$ a bijection? The main obstruction I have in mind is that for an ideal $K'\in A/I$, the preimage $K:=\pi^{-1}(K')$ is an ideal -- but what is preventing the existence of a subideal $J\subseteq K$ satisfying $\pi(J)=K'$?
Minor question. Is the specification that ideals $K$ in $A$ contain $I$ just to guarantee that $\mathbf{0}\in\pi(K)$ in the construction, hence ensuring that $\pi(K)$ is a subvector space -- or is it achieving something extra?