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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 $A/I$.

Fly by Night
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Maxim_Koelt
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    Where can I find a "dishonest" proof of that? (to help understand what you mean) – Bill Dubuque May 29 '14 at 17:12
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    If you already know what the correspondence is, it shouldn't be terribly difficult to prove it is bijective (and even preserves lattice operations). – anon May 29 '14 at 17:15
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    Well, I mean a proof that does not refer to other results elsewhere, but rather starts from the start and ends at the end. – Maxim_Koelt May 29 '14 at 17:15
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    @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? – rschwieb May 29 '14 at 17:56

2 Answers2

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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 $f^{-1} (f(J)) = J + \ker{f} = J$.

Andrew Dudzik
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  • Can you explain the first inclusion $Bf(J) \subset f(AJ)$? Why can't we just jump straight to $Bf(J) \subset f(J)$ – Lemon Dec 28 '16 at 19:33
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    @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. – Andrew Dudzik Dec 28 '16 at 21:03
  • Why I can't just assume that ${a \mid a + I \in f(J)} = {a \mid a \in J} = J$? – Lucas Giraldi Jun 11 '20 at 02:32
  • @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 $f$! – Andrew Dudzik Jun 11 '20 at 11:58
  • The sixth paragraph should be $\mathcal{J}\in\mathscr{D}$. – 19021605 Mar 31 '25 at 08:35
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Consider rings $R, R'$ (commutative, with unity) and a ring homomorphism $R \overset{\varphi}{\to} R'$.

For any ideal $J$ of subring $\varphi(R)$, clearly $\color{green}{\varphi(\varphi ^{-1} (J)) = J}$.

Also, for any ideal $I$ of $R$, $\varphi ^{-1} (\varphi(I))$ $= \lbrace t \in R : \varphi(t) = \varphi(i) \text{ for some } i \in I \rbrace$ $ = \lbrace t \in R : t - i \in \ker(\varphi) \text{ for some } i \in I \rbrace$ $ = I + \ker(\varphi) $.
That is, for any ideal $I$ of $R$, $\color{green}{\varphi ^{-1} (\varphi (I)) = I + \ker(\varphi)}$.

But ideal $I + \ker(\varphi) = I$ if and only if $I \supseteq \ker(\varphi)$. This suggests :

Th: Let $R \overset{\varphi}{\to} R'$ be a ring homomorphism. There is a bijection

$$\lbrace \text{ideals of } R \text{ containing } \ker(\varphi)\rbrace \leftrightarrow \lbrace \text{ideals of } \varphi(R) \rbrace, $$

where $I$ in LHS is mapped to $\varphi(I)$.
Pf: Call LHS $\mathbf{A}$ and RHS $\mathbf{B}$. Define $\mathbf{A} \overset{f}{\to} \mathbf{B}$ by $I \mapsto \varphi(I)$, and $\mathbf{B} \overset{g}{\to} \mathbf{A}$ by $J \mapsto \varphi ^{-1}(J)$.

These are well-defined : $I \in \mathbf{A}$ implies $\varphi(I) \in \mathbf{B}$, and $J \in \mathbf{B}$ implies $\varphi ^{-1}(J) \in \mathbf{A}.$

For any $J \in \mathbf{B}$, $(f \circ g)(J)$ $= \varphi(\varphi ^{-1}(J))$ $=J$, so $f \circ g = \text{id} _{\mathbf{B}}$.
Similarly for any $I \in \mathbf{A}$, $(g \circ f)(I)$ $= \varphi ^{-1} (\varphi(I))$ $= I + \ker(\varphi)$ $=I$, so $g \circ f = \text{id} _{\mathbf{A}}$ too.

Venkata Karthik Bandaru
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  • Doesn't $\phi$ need to be surjective? – gen Dec 14 '21 at 21:31
  • No because it can always be viewed as a surjective ring homomorphism ${ R \overset{\varphi}{\to} \varphi(R) }$ – Venkata Karthik Bandaru Dec 15 '21 at 00:45
  • @gen Note that $f\circ g=\mbox{id}_B$, which implies surjectivity of $f$. On the other hand, $g\circ f=\mbox{id}_A$, which implies injectivity of $f$. (In general, existence of a left inverse is equivalent to injectivity and existence of a right inverse is equivalent to surjectivity). Note that $f$ is actually $\varphi$. So, you get bijectivity of $\varphi$. – Sam Wong Dec 30 '22 at 12:29