MHB Commutative ring | Exam question

[Krizalid1]

This is a question I gave in one of my exams. :D

If $I$ is an ideal of the commutative ring $A$ (with unity), then prove that for each ideal $\overline J$ of the quotient ring $A/I$, exists an ideal $J$ of $A$ that contains the ideal $I$ and the lateral classes of $I$ defined by the elements of $J$ are the elements of $\overline J.$

 

[Krizalid1]

Too easy? Too boring? :D

 

[ModusPonens]

This is a question I gave in one of my exams. :D

If $I$ is an ideal of the commutative ring $A$ (with unity), then prove that for each ideal $\overline J$ of the quotient ring $A/I$, exists an ideal $J$ of $A$ that contains the ideal $I$ and the lateral classes of $I$ defined by the elements of $J$ are the elements of $\overline J.$

Too easy? Too boring? :D

I didn't even see it. It seems simple. Let me try.

$\pi: A \longrightarrow A / I$?

Is it that simple?

 

[Deveno]

Almost...you want the inverse map:

[math]\pi^{-1}:P(A/I) \to P(A)[/math], where [math]P(S)[/math] is the power set of [math]S[/math].

Then given an ideal [math]\overline{J}[/math] of [math]A/I[/math], we have to show that [math]\pi^{-1}(\overline{J})[/math] is an ideal of [math]A[/math] containing [math]I[/math].

Since any ideal [math]\overline{J}[/math] of [math]A/I[/math] contains [math]I = 0 + I[/math], we have that [math]\pi^{-1}(0 + I) = \text{ker}(\pi) = I \subseteq \pi^{-1}(\overline{J})[/math].

Next, we have to show [math]\pi^{-1}(\overline{J})[/math] is an ideal. So let [math]x,y \in \pi^{-1}(\overline{J})[/math]. This means that:

[math]\pi(x),\pi(y) \in \overline{J}[/math], hence [math]\pi(x) - \pi(y) = \pi(x - y) \in \overline{J}[/math]

(since [math]\overline{J}[/math] is an ideal, and [math]\pi[/math] is a ring homomorphism) hence [math]x - y \in \pi^{-1}(\overline{J})[/math].

This shows that [math]\pi^{-1}(\overline{J})[/math] is an additive subgroup of [math]A[/math].

Now let [math]a \in A[/math] be arbitrary, and likewise choose an arbitrary [math]x \in \pi^{-1}(\overline{J})[/math].

We want to show that [math] ax \in \pi^{-1}(\overline{J})[/math].

But: [math]\pi(ax) = \pi(a)\pi(x)[/math] and [math]\pi(x) \in \overline{J}[/math], so since [math]\pi(a) \in \pi(A) = A/I[/math], and [math]\overline{J}[/math] is an ideal of [math]A/I[/math],

[math]\pi(ax) \in \overline{J}[/math], thus [math]ax \in \pi^{-1}(\overline{J})[/math].

Since [math]A[/math] is commutative, this suffices to show that [math]\pi^{-1}(\overline{J})[/math] is an ideal of [math]A[/math].

Next, we want to show that this map, restricted to the ideals of [math]A/I[/math], is injective. So let [math]\overline{J},\overline{K}[/math] be two ideals of [math]A/I[/math],

such that [math]J = \pi^{-1}(\overline{J}) = \pi^{-1}(\overline{K})[/math].

Then [math]\overline{J} = \pi(\pi^{-1}(\overline{J})) = \pi(\pi^{-1}(\overline{K})) = \overline{K}[/math].

Furthermore, if [math]J[/math] is ANY ideal of [math]A[/math] containing [math]I[/math], we have the ideal [math]\pi(J) = J/I[/math] of [math]A/I[/math] with:

[math]J = \pi^{-1}(J/I)[/math] (this is NOT true of sets in general, and only holds in this case because

[math]\pi[/math] is a surjective ring homomorphsm).

This establishes a bijection between ideals of [math]A[/math] containing [math]I[/math] and ideals of [math]A/I[/math]:

[math]J \leftrightarrow \overline{J} = J/I[/math]

 

[ModusPonens]

Deveno, I admire your divine ( ;) ) patience! I'm very lazy with latex (and in general...).