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The first line of the proof given in my book says that the ideals of $R/I$ are in bijective correspondence with the ideals of $R$ lying between $I$ and $R$.

What is the bijection?

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    $\begingroup$ google bijection $\endgroup$
    – clark
    Mar 22, 2013 at 10:41
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    $\begingroup$ @clark I think OP is asking about the particular bijection (i.e., the correspondence theorem), and not what a bijection is. $\endgroup$ Mar 22, 2013 at 10:43
  • $\begingroup$ @IttayWeiss you are right, I read "a" instead of "the" $\endgroup$
    – clark
    Mar 22, 2013 at 11:00

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This refers to the correspondence theorem: If $I$ is an ideal of $R$, then there is a bijection between the set of all ideals of $R/I$ and the set of all ideals $I\subseteq J\subseteq R$. The correspondence is given by the inverse map associated to the canonical projection $R\to R/I$.

Thus, if $I$ is a maximal ideal then there are only two intermediate ideals $I\subseteq J\subseteq R$. By the correspondence then, there are only two ideals in $R/I$ ... (and the proof goes on).

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  • $\begingroup$ Weis Your second paragraph puzzles me! If I is maximal then I=J=R, right? $\endgroup$
    – DaveUM
    Mar 24, 2013 at 13:29
  • $\begingroup$ No, J is either I or R. $\endgroup$ Mar 24, 2013 at 20:04
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It is $$\{\mathfrak{a}\trianglelefteq R ~|~ I\subseteq \mathfrak{a}\} \to \{\mathfrak{b}\trianglelefteq R/I\}, ~ \mathfrak{a}\mapsto \nu(\mathfrak{a})$$ where $\nu ~:~ R \to R/I$ is the canonical epimorphism.

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