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Commutative ring | Exam question

Krizalid

Active member
Feb 9, 2012
118
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 $\overline J.$
 

Krizalid

Active member
Feb 9, 2012
118
Too easy? Too boring? :D
 

ModusPonens

Well-known member
Jun 26, 2012
45
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

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Almost...you want the inverse map:

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

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

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

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

\(\displaystyle \pi(x),\pi(y) \in \overline{J}\), hence \(\displaystyle \pi(x) - \pi(y) = \pi(x - y) \in \overline{J}\)

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

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

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

We want to show that \(\displaystyle ax \in \pi^{-1}(\overline{J})\).

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

\(\displaystyle \pi(ax) \in \overline{J}\), thus \(\displaystyle ax \in \pi^{-1}(\overline{J})\).

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

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

such that \(\displaystyle J = \pi^{-1}(\overline{J}) = \pi^{-1}(\overline{K})\).

Then \(\displaystyle \overline{J} = \pi(\pi^{-1}(\overline{J})) = \pi(\pi^{-1}(\overline{K})) = \overline{K}\).

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

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

\(\displaystyle \pi\) is a surjective ring homomorphsm).

This establishes a bijection between ideals of \(\displaystyle A\) containing \(\displaystyle I\) and ideals of \(\displaystyle A/I\):

\(\displaystyle J \leftrightarrow \overline{J} = J/I\)
 

ModusPonens

Well-known member
Jun 26, 2012
45
Deveno, I admire your divine ( ;) ) patience! I'm very lazy with latex (and in general...).