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Let $R$ and $S$ be unitary, commutative rings. Let $f:R\to S$ be a ring homomorphism. I want to show that there exists a bijection between the ideals of $\text{im}(f)$ and between the set of ideals of $R$ that contain $\ker(f)$.

First of, define $$A:=\{J \subseteq R | J \text{ is an ideal and } \ker(f) \subseteq J \}$$ $$B:=\{J \subseteq \text{im}(f) | J \text{ is an ideal} \}$$

Now I know I want to use the ring homomorphism theorem and I know it is closely related to the problem of solving Bijection between ideals of $R/I$ and ideals containing $I$ and Prove a bijection of ideals exists.. But I cannot really make sense of it.

We know $R/\ker(f) \cong \text{im}(f)$ and we know that for a surjective ring homomorphism $f$ and for all ideals $J$ both $f(J)$ and $f^{-1}(J)$ are also ideals. But I don't know how to extract the bijection from $R/\ker(f) \cong \text{im}(f)$.

How can I construct such a bijection?

Sera Gunn
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Jonathan
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  • You have all the key facts. Replace $S$ with $im(f)$. Given $J$ an ideal of $im(f)$, you get an ideal $f^{-1}(J)$ of $S$ as you've described. Now show that this correspondence is injective and surjective. What part of this is giving you trouble? – vociferous_rutabaga Jun 21 '17 at 19:07

2 Answers2

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The canonical projection $\pi:R\to R/\ker(f)$ induces a map (which I'll also call $\pi$) from $A$ to $B$ since, given an ideal $I\lhd R$ with $\ker(f)\subset I$, $\pi(I)\lhd R/\ker(f)$.

For a set $X$, let $P(X)$ denote the set of subsets of $X$. Note that the inverse image map $\pi^{-1}:P(R/\ker(f))\to P(R)$ restricts to a map $$\pi^{-1}:B\to A$$ since, given an ideal $J\lhd R/\ker(f)$, $\pi^{-1}(J)\lhd R$ and $I\subset \pi^{-1}(J)$.

You need to check that the compositions $\pi\circ\pi^{-1}=1_B$ and $\pi^{-1}\circ\pi=1_A$. This is spelled out in the linked posts.

David Hill
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  • What does $I\lhd R$ this symbol mean? I guess I is any Ideal of R? – Jonathan Jun 21 '17 at 19:55
  • @Jonathan It is common to use $I \lhd R$ to mean an ideal. This is borrowed from the common notation for a normal subgroup. It is not however standardized whether $I \lhd R$ allows $I = R$ or not. From context, we can infer that David is using this notation to allow $I = R$. Other people would write $I \unlhd R$. – Sera Gunn Jun 21 '17 at 20:23
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There is a natural choice for the maps. Prove the following steps.

  1. If $I$ is an ideal of $R$, then $f^{\to}(I)$ is an ideal of $S'=\operatorname{im}f$.

  2. If $J$ is an ideal of $S'$, then $f^{\gets}(J)$ is an ideal of $R$ and $\ker f\subseteq f^{\gets}(J)$.

  3. If $I$ is an ideal of $R$ and $\ker f\subseteq I$, then $ f^{\gets}(f^{\to}(I))=I$.

  4. If $J$ is an ideal of $S'$, then $f^{\to}(f^{\gets}(J))=J$.

This will end the required proof.

Note: I use the notation $f^{\to}(X)=\{f(x):x\in X\}$ and $f^{\gets}(Y)=\{x:f(x)\in Y\}$; others use $f(X)$ and $f^{-1}(Y)$ for the same purpose.

egreg
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