Could someone provide a reference that includes a full and honest proof of the Correspondence Theorem for rings?

Let $A$ be a multiplicative ring with identity and $I$ an ideal of $A$. There is a one-to-one correspondence between the ideals of $A$ that contain $I$ and the ideals of the quotient ring $A/I$.

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    $\begingroup$ Where can I find a "dishonest" proof of that? (to help understand what you mean) $\endgroup$ – Bill Dubuque May 29 '14 at 17:12
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    $\begingroup$ If you already know what the correspondence is, it shouldn't be terribly difficult to prove it is bijective (and even preserves lattice operations). $\endgroup$ – blue May 29 '14 at 17:15
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    $\begingroup$ Well, I mean a proof that does not refer to other results elsewhere, but rather starts from the start and ends at the end. $\endgroup$ – Maxim_Koelt May 29 '14 at 17:15
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    $\begingroup$ @Maxim_Koelt Usually this theorem appears almost immediately after the definition of rings and ideals, and so it does not refer to anything more exotic. Possibly the correspondence theorem for abelian groups is invoked to speed things up, but that hardly seems like a problem. What algebra books have you already checked and rejected? $\endgroup$ – rschwieb May 29 '14 at 17:56

My reference is this post. Note that it is not necessary for $A$ to have an identity.

Let $\mathscr{C}$ and $\mathscr{D}$ denote, respectively, the collection of ideals of $A$ containing $I$, and the collection of ideals of $A/I$.

Define: $f:\mathscr{C}\to\mathscr{D}$ by $f(J) = \{a + I \mid a\in J\}\subset A/I$.

Define $g:\mathscr{D}\to\mathscr{C}$ by $g(\mathcal{J}) = \{a\mid a+I \in \mathcal{J}\} \subset A$.

(I omit the proofs that these give ideals, but these follow trivially from the definitions.)

If $\mathcal{J}\in\mathscr{C}$, then $(f\circ g)(\mathcal{J}) = \{a+I \mid a\in g(\mathcal{J})\} = \{a+I \mid a+I\in\mathcal{J}\} = \mathcal{J}$ .

If $J\in\mathscr{C}$, then $(g\circ f)(J) = \{a\mid a+I \in f(J)\} = \{a \mid a+I=b+I\text{ for some } b\in J\}$ $= \{a \mid a\in b+I\text{ for some } b\in J\}$.

This last set clearly contains $J$. But $a\in b+I \implies (a-b)\in I\subset J \implies a=b+J\implies a\in J$. So $(g\circ f)(J) = J$, and we have shown that $f$ and $g$ establish a bijection.

It probably doesn't get proved like this very often, because it's a lot shorter if we're allowed to use a few definitions. It is, however, the same proof.

If $f:A\to B$ is a surjective ring homomorphism, and $J$ is an ideal of $A$ containing $\ker f$, then $f(J)$ is an ideal of $B$, because $Bf(J) \subset f(AJ) \subset f(J)$.

If $J'$ is an ideal of $S$, then $f^{-1} (J')$ is an ideal of $R$.

We have $f(f^{-1}(J')) = J'$ trivially, and $f^{-1} (f(J)) = J + \ker{f} = J$.

  • $\begingroup$ Can you explain the first inclusion $Bf(J) \subset f(AJ)$? Why can't we just jump straight to $Bf(J) \subset f(J)$ $\endgroup$ – Hawk Dec 28 '16 at 19:33
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    $\begingroup$ @Hawk $B = f(A)$ by surjectivity, so $Bf(J) = f(A)f(J) \subset f(AJ)$, since $f$ is a homomorphism. (actually, we have equality here for unital rings, but we don't really need this) Then we use $AJ\subset J$, since $J$ is an ideal. I don't see how we can "jump" the first step. $\endgroup$ – Slade Dec 28 '16 at 21:03
  • $\begingroup$ Why I can't just assume that $\{a \mid a + I \in f(J)\} = \{a \mid a \in J\} = J$? $\endgroup$ – Lucas Giraldi A. Coimbra Jun 11 '20 at 2:32
  • $\begingroup$ @LucasGiraldiA.Coimbra The question isn't why you can't, but why you think you can. Certainly the second equality is clear (it's almost an axiom of set theory), but how can you "assume" the first without proving it? Surely you need to know something about the relationship between $I$ and $J$, not to mention $f$! $\endgroup$ – Slade Jun 11 '20 at 11:58

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