999: Field with 0 Removed Is Multiplicative Group

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description/proof of that field with 0 removed is multiplicative group

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Topics

About: field
About: group

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The table of contents of this article

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

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Starting Context

• The reader knows a definition of field.
• The reader knows a definition of group.

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Target Context

• The reader will have a description and a proof of the proposition that any field with 0 removed is a multiplicative group.

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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:
$F$: $\in \{\text{ the fields }\}$
$F^{\times }$: $=F\setminus \{0\}$
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Statements:
$F^{\times }\in \{\text{ the groups }\}$ with respect to the multiplication
//

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2: Proof

Whole Strategy: confirm that the criteria for being a group are satisfied; Step 1: see that $F^{\times }$ is closed under the multiplication; Step 2: see that $1\in F^{\times }$; Step 3: see that for each $r\in F^{\times }$, there is an inverse, $r^{-1}\in F^{\times }$; Step 4: see that the multiplication is associative.

Step 1:

Let us see that $F^{\times }$ is closed under the multiplication.

Let $r_1,r_2\in F^{\times }$ be any. While, of course, $r_1{r}_2\in F$, the issue is that $r_1{r}_2\ne 0$. Let us suppose that $r_1{r}_2=0$. As $r_1\ne 0$, there would be the inverse, $r_1^{-1}$, and $r_2=r_1^{-1}{r}_1{r}_2=r_1^{-1}0=0$, a contradiction. So, $r_1{r}_2\ne 0$ (more succinctly, it is because any field is an integral domain).

Step 2:

$1\in F^{\times }$.

Step 3:

Let us see that for each $r\in F^{\times }$, there is an inverse, $r^{-1}\in F^{\times }$.

By the definition of field, there is an inverse, $r^{-1}\in F$. The issue is that $r^{-1}\ne 0$. But as $r{r}^{-1}=1\ne 0$, $r^{-1}\ne 0$.

Step 4:

The multiplication is associative, because it is so in the ambient $F$.

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References

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