description/proof of that integers ring is principal integral domain
Topics
About: ring
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of integers ring.
- The reader knows a definition of principal integral domain.
Target Context
- The reader will have a description and a proof of the proposition that the integers ring is a principal integral domain.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(\mathbb{Z}\): \(= \text{ the integers ring }\)
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Statements:
\(\mathbb{Z} \in \{\text{ the principal integral domains }\}\)
//
2: Natural Language Description
The integers ring, \(\mathbb{Z}\), is a principal integral domain.
3: Proof
Whole Strategy: Step 1: prove that \(\mathbb{Z}\) is an integral domain; Step 2: prove that each ideal of \(\mathbb{Z}\) is a principal ideal.
Step 1:
\(\mathbb{Z}\) is obviously a nonzero commutative ring.
For each \(p_1, p_2 \in \mathbb{Z}\) such that \(p_1, p_2 \neq 0\), \(p_1 p_2 \neq 0\).
So, \(\mathbb{Z}\) is an integral domain.
Step 2:
Let us prove that each ideal of \(\mathbb{Z}\) is a principal ideal.
Step 2 Strategy: Step 2-1: take any ideal; Step 2-2: take the smallest positive element of the ideal; Step 2-3: show that the principal ideal by the element is contained in the ideal; Step 2-4: show that there is no other element in the ideal.
Step 2-1:
Let \(I \subseteq \mathbb{Z}\) be any ideal.
Step 2-2:
There is the smallest positive element, \(p \in I\), which is because the positive integers set is a subset of the natural numbers set while the natural numbers set is well-ordered.
Step 2-3:
While \(p \mathbb{Z}\) is a principal ideal, \(p \mathbb{Z} \subseteq I\), because \(p \mathbb{Z} \subseteq I \mathbb{Z} = I\).
Step 2-4:
Let us see that \(p \mathbb{Z} = I\).
Let us suppose otherwise.
Let us take any \(p' \in I \setminus p \mathbb{Z}\).
There would be \(k \in \mathbb{Z}\) and \(0 \lt j \lt p\) such that \(p' = p k + j\). But \(j = p' - p k \in I\), a contradiction against \(p\)'s being the smallest positive element of \(I\).
So, \(p \mathbb{Z} = I\).
So, \(I\) is a principal integral domain.
