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Here is the proposition I want to prove:

There is a one-to-one order preserving correspondence between the ideals $\mathfrak{b}$ of $A$ which contain $\mathfrak{a},$ and the ideals $\bar{\mathfrak{b}}$ of $A/\mathfrak{a},$ given by $\phi^{-1}(\bar{\mathfrak{b}}) = {\mathfrak{b}}.$

But, I am not quite sure:

1- How can I prove the "one-to-one order preserving correspondence", what does this statement means? Should I find a bijection between $\mathfrak{b}$ and $\bar{\mathfrak{b}}$? how can I do this?

2- What does it mean "order preserving"?

Could someone help me please in this?

Intuition
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    The ideals are (partially) ordered by inclusion. Order preserving means that if $I\leq J$, then the image of $I$ under the correspondence is less than or equal to the image of $J$. That said, given your confusion of this, I am sorry to say, very basic result that is covered in any introductory course on ring theory, are you sure you should be reading this book? Sounds to me like you are simply not ready for it. – Arturo Magidin Oct 02 '22 at 02:37
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    @ArturoMagidin I respect my eagerness to learn so I think I will continue to try to learn forever – Intuition Oct 02 '22 at 02:45
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    I did not say "stop learning". I said you do not seem to be ready for that particular book. You should instead learn the basics needed to understand that book, because right now it is apparent you do not have them. No shame in that, but a lot of silly stubbornness and wasted effort in persisting in trying to read stuff that is too far beyond your current level. – Arturo Magidin Oct 02 '22 at 03:31
  • Along the lines of what @Arturo Magidin is saying, let me suggest Fraleigh's "A First Course in Abstract Algebra". I enjoyed it. Then you could work your way up to Commutative Algebra. – suckling pig Oct 02 '22 at 04:03

1 Answers1

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Hint Use the "canonical" projection $\rho:A\twoheadrightarrow A/\mathfrak a$, which sends each $x\in A$ to the coset $x+\mathfrak a$ in $A/\mathfrak a$.

The bijection is between the sets $\{\mathfrak b:\mathfrak a\subset \mathfrak b\subset A \,\text {is an ideal}\}$ and $\{\bar{\mathfrak b}:\bar {\mathfrak b}\subset A/\mathfrak a\ \text{is an ideal}\}$.

suckling pig
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  • Are every ideals in the world partially ordered by inclusion? does this means that every ideal is contained in a chain? has an upper bound? – Intuition Oct 02 '22 at 16:28
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    The left, right or two-sided ideals of a ring form a poset. Maximal ideals are important. See also ACC, "ascending chain condition". They were important in the work of Hilbert, Noether and Artin. – suckling pig Oct 02 '22 at 17:26
  • Do we need to prove that $\rho$ and $\rho^{-1}$ are the inverse of one another? If so, why? the statement of the question is just saying $1-1$ correspondence. – Intuition Oct 03 '22 at 18:07
  • I am asking the question in my previous comment because here https://math.stackexchange.com/questions/69578/bijection-between-ideals-of-r-i-and-ideals-containing-i?rq=1 they proved bijection .... But I believe it is not required, am I correct? – Intuition Oct 03 '22 at 18:10
  • Oh, I am sorry, I think I just forgot the definition of 1-1 correspondance which means bijection – Intuition Oct 03 '22 at 19:42