How to prove there exists a bijection between the ideals of $A/a$ and the ideals of $A$ containing $a$

Question

original proposition is

there is a one-to-one order-presserving correspondence between the ideals of $A$ which contain $a$ and the ideals of $A/a$.

I think one-to-one correspondence mean bijection and assume the order on the ideal is partially order.

define the order by "contained"

then bijection can imply order-presserving.

so the proposition we need prove is

there is a bijection between the ideals b of A which contain a and the ideal c of A/a.

I stuck here because I cant find a expression to describe the element of c.

Answer 1

You don't need to describe the element $c$, instead, you need to note, and use some well-known properties below:

1. Note that, you have the canonical epimorphism: $\rho: A \to A/\mathfrak{a}$.

2. If $f: A \to B$ is a ring homomorphism, and if $\mathfrak{b}$ is an ideal of $B$, then $f^{-1}(\mathfrak{b})$ is also an ideal of $A$.

3. If $f: A \to B$ is a ring epimorphism, then if $\mathfrak{a}$ is an ideal of $A$, then $f(\mathfrak{a})$ is also an ideal of $B$.

Please not that 3. does not hold when $f$ is not an epimorphism. However, it still holds in a weaker version, though: If $\mathfrak{a}$ is a subring of $A$, then $f(\mathfrak{a})$ is also a subring of $B$.

Using 1, 2, 3 above, I think you should be above to prove your statement. Just give it a try.

Regards,