description/proof of that field with 0 removed is multiplicative group
Topics
About: field
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of field.
- The reader knows a definition of group.
Target Context
- The reader will have a description and a proof of the proposition that any field with 0 removed is a multiplicative group.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(F^{\times}\): \(= F \setminus \{0\}\)
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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 \(r_1 r_2 \neq 0\). Let us suppose that \(r_1 r_2 = 0\). As \(r_1 \neq 0\), 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, \(r_1 r_2 \neq 0\) (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 \(r^{-1} \neq 0\). But as \(r r^{-1} = 1 \neq 0\), \(r^{-1} \neq 0\).
Step 4:
The multiplication is associative, because it is so in the ambient \(F\).
