878: Quotient Ring of Integers Ring by Prime Principal Ideal Is Field

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description/proof of that quotient ring of integers ring by prime principal ideal is field

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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 ideal of ring.
• The reader knows a definition of quotient ring of ring by ideal.
• The reader knows a definition of field.
• The reader admits the proposition that the quotient ring of any commutative ring by any ideal is a commutative ring.
• The reader admits the proposition that the integers ring is a principal integral domain.
• The reader admits the proposition that any principal integral domain is a greatest common divisors domain, and for each 2 elements on the principal integral domain, each of the greatest common divisors of the 2 elements is a one by which the sum of the principal ideals by the 2 elements is the principal ideal.

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Target Context

• The reader will have a description and a proof of the proposition that the quotient ring of the integers ring by any prime principal ideal is a field.

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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:
$\mathbb{Z}$: $=\text{ the integers ring }$
$p$: $\in \{\text{ the prime numbers }\}$
$\mathbb{Z}/(p\mathbb{Z})$: $=\text{ the quotient ring of }\mathbb{Z}\text{ by }p\mathbb{Z}$
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Statements:
$\mathbb{Z}/(p\mathbb{Z})\in \{\text{ the fields }\}$
//

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2: Natural Language Description

For the integers ring, $\mathbb{Z}$, and any prime number, $p$, the quotient ring, $\mathbb{Z}/(p\mathbb{Z})$, is a field.

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3: Proof

Whole Strategy: Step 1: prove that $\mathbb{Z}/(p\mathbb{Z})$ is a commutative ring; Step 2: prove that each nonzero element of $\mathbb{Z}/(p\mathbb{Z})$ has an inverse.

Step 1:

$\mathbb{Z}/(p\mathbb{Z})$ is a commutative ring, because $\mathbb{Z}$ is a commutative ring, by the proposition that the quotient ring of any commutative ring by any ideal is a commutative ring.

Step 2:

Let us prove that each nonzero element of $\mathbb{Z}/(p\mathbb{Z})$ has an inverse.

Let $[p^{\prime }]\in \mathbb{Z}/(p\mathbb{Z})$ be any such that $[p^{\prime }]\ne 0$.

We are searching for a $[p^{″}]\in \mathbb{Z}/(p\mathbb{Z})$ such that $[p^{\prime }][p^{″}]=[1]$ (then, $[p^{″}][p^{\prime }]=[1]$ will follow from the commutativity). $[p^{\prime }][p^{″}]=[p^{\prime }{p}^{″}]$. It is about whether there are some $p^{″},p^{‴}\in \mathbb{Z}$ such that $p^{\prime }{p}^{″}+p{p}^{‴}=1$.

As $p$ is a prime number, $gcd(p^{\prime },p)=1$. As $\mathbb{Z}$ is a principal integral domain, by the proposition that the integers ring is a principal integral domain, by the proposition that any principal integral domain is a greatest common divisors domain, and for each 2 elements on the principal integral domain, each of the greatest common divisors of the 2 elements is a one by which the sum of the principal ideals by the 2 elements is the principal ideal, there are indeed such $p^{″}$ and $p^{‴}$: $1\in 1\mathbb{Z}=p\mathbb{Z}+p^{\prime }\mathbb{Z}$.

So, $[p^{\prime }]$ has an inverse.

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References

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