689: Integers Ring Is Principal Integral Domain

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description/proof of that integers ring is principal integral domain

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Topics

About: ring

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of integers ring.
• The reader knows a definition of principal integral domain.

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Target Context

• The reader will have a description and a proof of the proposition that the integers ring is a principal integral domain.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$\mathbb{Z}$: $=\text{ the integers ring }$
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Statements:
$\mathbb{Z}\in \{\text{ the principal integral domains }\}$
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2: Natural Language Description

The integers ring, $\mathbb{Z}$, is a principal integral domain.

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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 $p_1,p_2\ne 0$, $p_1{p}_2\ne 0$.

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 $p^{\prime }\in I\setminus p\mathbb{Z}$.

There would be $k\in \mathbb{Z}$ and $0<j<p$ such that $p^{\prime }=pk+j$. But $j=p^{\prime }-pk\in I$, a contradiction against $p$'s being the smallest positive element of $I$.

So, $p\mathbb{Z}=I$.

So, $I$ is a principal integral domain.

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References

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