description/proof of that characteristic of ring with inverses is 0 or prime number
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of characteristic of ring.
Target Context
- The reader will have a description and a proof of the proposition that the characteristic of any ring with inverses is 0 or a prime number.
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:
\(R\): \(\in \{\text{ the rings }\}\), such that \(\forall r \in R \setminus \{0\} (\exists r^{-1} \in R (r r^{-1} = r^{-1} r = 1))\)
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Statements:
\(Ch (R) = 0 \lor Ch (R) \in \{\text{ the prime numbers }\}\)
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2: Note
\(R\) does not need to be commutative, so, \(R\) does not need to be any field.
Any field is a ring with inverses, and so, the characteristic of any field is \(0\) or a prime number.
3: Proof
Whole Strategy: Step 1: suppose that \(Ch (R) \neq 0\); Step 2: suppose that \(Ch (R) = m n\) where \(m, n \in \mathbb{N} \setminus \{0, 1\}\), and find a contradiction.
Step 1:
Let us suppose that \(Ch (R) \neq 0\).
Step 2:
Let us suppose that \(Ch (R)\) was not any prime number.
That would mean that \(Ch (R) = m n\) where \(m, n \in \mathbb{N} \setminus \{0, 1\}\).
\((m n) \cdot 1 = m \cdot (n \cdot 1) = 0\) where \(n \cdot 1 \neq 0\).
\(n \cdot 1\) has an inverse, \((n \cdot 1)^{-1} \in R\).
\((n \cdot 1)^{-1} (m \cdot (n \cdot 1)) = (n \cdot 1)^{-1} 0 = 0\), but the left hand side is \((n \cdot 1)^{-1} ((n \cdot 1) + ... + (n \cdot 1)) = (n \cdot 1)^{-1} (n \cdot 1) + ... + (n \cdot 1)^{-1} (n \cdot 1) = 1 + ... + 1 = m \cdot 1\), so, \(m \cdot 1 = 0\), a contradiction against the supposition that \(m n\) such that \(m \lt m n\) was the smallest such.
So, \(Ch (R)\) is a prime number.
