I really have to know a few things though.... with that surjection map from ideals in a ring to ideals in a quotient ring by an ideal i.e. ( $\mathbf{R} \; \mapsto \; \dfrac{ \mathbf{R}}{ \mathbf{I}}$ ) that is a surjective homomorphism, why cant I do the simplest example of restricting the map to ideals containing $\mathbf{I}$ so the homomorphism is then bijective?
Here is an example of the process: $\mathbb{I} \; = \; \mathbf{3Z} \; and \; \mathbf{5Z} \; = \; \{\; 0 \; , \; 3 \, , \; 5 \; , \; 6 \; , \; 8 \; , \; 9 \; , \; 10 \; , \; …. \; \} \; \mapsto \; the \; unit \; ideal \; \in \; \dfrac{\mathbf{Z}}{\mathbf{3Z}}$. $\mathbb{I}$ contains $\mathbf{3Z}$ and so does the unit ideal in $\mathbf{Z} \; which \; also \; maps \; to \; the \; unit \; ideal \; in \dfrac{\mathbf{Z}}{\mathbf{3Z}}$. Do I really just make every single ideal one to one and onto the unit ideal?? Because well initially it doesn't even seem like a bijection when two ideals map to the unit ideal. The tutorials could all just tell that restricted to each ideal containing $\mathbf{I}$, the map is bijective. ; !
The ring $\mathbb{Z}/3\mathbb{Z}$ is a field and therefore only has two ideals: The zero ideal and the field itself. Not every ideal is sent to the entire field, as for example $3\mathbb{Z}$ is mapped to the zero ideal. If you restrict the map to the ideals containing $3\mathbb{Z}$, you will only have two ideals left in your domain as $3\mathbb{Z} \subset \mathbb{Z}$ is maximal, namely the ideal $3\mathbb{Z}$ itself and the integers $\mathbb{Z}$. As I said before the ideal will get send to the zero ideal and the integers will get sent to the entire quotient and thus it is a bijection as you wanted.