description/proof of that quotient ring of integers ring by prime principal ideal is field
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 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.
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.
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:
\(\mathbb{Z}\): \(= \text{ the integers ring }\)
\(p\): \(\in \{\text{ the prime numbers }\}\)
\(\mathbb{Z} / (p \mathbb{Z})\): \(= \text{ the quotient ring of } \mathbb{Z} \text{ by } p \mathbb{Z}\)
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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.
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 \([p'] \in \mathbb{Z} / (p \mathbb{Z})\) be any such that \([p'] \neq 0\).
We are searching for a \([p''] \in \mathbb{Z} / (p \mathbb{Z})\) such that \([p'] [p''] = [1]\) (then, \([p''] [p'] = [1]\) will follow from the commutativity). \([p'] [p''] = [p' p'']\). It is about whether there are some \(p'', p''' \in \mathbb{Z}\) such that \(p' p'' + p p''' = 1\).
As \(p\) is a prime number, \(gcd (p', p) = 1\). 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 \(p'''\): \(1 \in 1 \mathbb{Z} = p \mathbb{Z} + p' \mathbb{Z}\).
So, \([p']\) has an inverse.
