Commutative ring | Exam question

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In summary, if $I$ is an ideal of the commutative ring $A$, then there 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 $\overline J.$
  • #1
Krizalid1
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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 $\overline J.$
 
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  • #2
Too easy? Too boring? :D
 
  • #3
Krizalid said:
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 said:
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?
 
  • #4
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 \(\displaystyle A/I\):

\(\displaystyle J \leftrightarrow \overline{J} = J/I\)
 
  • #5
Deveno, I admire your divine ( ;) ) patience! I'm very lazy with latex (and in general...).
 

FAQ: Commutative ring | Exam question

What is a commutative ring?

A commutative ring is a mathematical structure that consists of a set of elements, along with two operations, addition and multiplication. These operations must satisfy certain properties, such as commutativity (a*b = b*a) and associativity (a*(b*c) = (a*b)*c). Examples of commutative rings include the set of integers and the set of polynomials.

What is the difference between a commutative ring and a non-commutative ring?

The main difference between a commutative ring and a non-commutative ring is that the multiplication operation in a commutative ring is commutative, meaning that the order in which elements are multiplied does not affect the result. In a non-commutative ring, the multiplication operation is not commutative, and the order in which elements are multiplied does matter.

What is the role of commutative rings in abstract algebra?

Commutative rings play a fundamental role in abstract algebra, as they are a basic structure that is studied in this field. They are used to study properties of more complex algebraic structures, such as fields and modules. They also have applications in various areas of mathematics, such as number theory and algebraic geometry.

What are some common properties of commutative rings?

In addition to commutativity and associativity, commutative rings also have other important properties, such as the existence of an identity element for both addition and multiplication, and the distributive property (a*(b+c) = a*b + a*c). Commutative rings can also have additional properties, such as being integral domains or having a multiplicative inverse for every non-zero element.

How are commutative rings used in real-world applications?

Commutative rings have various applications in real-world problems, such as in coding theory, cryptography, and computer science. They are also used in physics and engineering, particularly in the study of symmetry and geometry. Additionally, commutative rings are used in economics and finance, where they can be used to model and analyze systems with multiple interacting variables.

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