993: Characteristic of Ring with Inverses Is 0 or Prime Number

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description/proof of that characteristic of ring with inverses is 0 or prime number

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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: Note
• 3: Proof

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Starting Context

• The reader knows a definition of characteristic of ring.

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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.

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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:
$R$: $\in \{\text{ the rings }\}$, such that $\mathrm{\forall }r\in R\setminus \{0\}(\mathrm{\exists }{r}^{-1}\in R(r{r}^{-1}=r^{-1}r=1))$
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Statements:
$Ch(R)=0\vee Ch(R)\in \{\text{ the prime numbers }\}$
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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.

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3: Proof

Whole Strategy: Step 1: suppose that $Ch(R)\ne 0$; Step 2: suppose that $Ch(R)=mn$ where $m,n\in \mathbb{N}\setminus \{0,1\}$, and find a contradiction.

Step 1:

Let us suppose that $Ch(R)\ne 0$.

Step 2:

Let us suppose that $Ch(R)$ was not any prime number.

That would mean that $Ch(R)=mn$ where $m,n\in \mathbb{N}\setminus \{0,1\}$.

$(mn)\cdot 1=m\cdot (n\cdot 1)=0$ where $n\cdot 1\ne 0$.

$n\cdot 1$ 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 $mn$ such that $m<mn$ was the smallest such.

So, $Ch(R)$ is a prime number.

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References

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