# Commutative ring | Exam question

#### Krizalid

##### Active member
This is a question I gave in one of my exams. 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
Too easy? Too boring? #### ModusPonens

##### Well-known member
This is a question I gave in one of my exams. 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? 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
Almost...you want the inverse map:

$$\displaystyle \pi^{-1} (A/I) \to P(A)$$, 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
Deveno, I admire your divine ( ) patience! I'm very lazy with latex (and in general...).