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Let $A$ be a ring and $\mathfrak{b}$ be an ideal of $A$. Prove that the assignment $$\mathfrak{c} \mapsto \mathfrak{c}/\mathfrak{b}$$ induces a one-to-one correspondence between the ideals of $A$ that contain $\mathfrak{b}$ and the ideals of $A/\mathfrak{b}$.

This is a problem I've been given, typed exactly as I received it. I find it a bit unclear, but I think $\mathfrak{c}$ is meant to be an ideal of $A$ which contains $\mathfrak{b}$. Then I must prove that the map given is bijective.

I've seen quotient rings, but not between ideals before. Should I treat it as I would any other quotient ring?

Assuming what I've said is correct, if the map above is $f$, then I must show that $f(\mathfrak{c})=\mathfrak{c}/\mathfrak{b}=\lbrace x+y : x \in \mathfrak{c}, y \in \mathfrak{b} \rbrace$ is a bijection?

Ducky
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2 Answers2

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In order to better distinguish between elements and ideals I'll use Fraktur letters for the latter: $\mathfrak{b}$ and $\mathfrak{c}$.

With $\mathfrak{c}/\mathfrak{b}$ the ideal $$ \{x+\mathfrak{b}:x\in \mathfrak{c}\} $$ is meant.

You have to prove that

  1. $\mathfrak{c}/\mathfrak{b}$ is an ideal of $A/\mathfrak{b}$
  2. Every ideal of $A/\mathfrak{b}$ is of this form for a unique ideal $\mathfrak{c}$ of $A$ such that $\mathfrak{c}\supseteq\mathfrak{b}$.

The proof of 1 is easy, just a verification. For 2, the hint is

given an ideal $\mathfrak{d}$ of $A/\mathfrak{b}$, consider $\mathfrak{c}=\{x\in A:x+\mathfrak{b}\in\mathfrak{d}\}$

egreg
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Hint: $b$ induces a congruence on $A$, and $c/b$ is a set of certain congruence classes modulo $b$. How do you reconstruct $c$ from $c/b$?

yo'
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