description/proof of that 'rings - homomorphisms' isomorphism between fields is 'fields - homomorphisms' isomorphism
Topics
About: field
The table of contents of this article
Starting Context
- The reader knows a definition of field.
- The reader knows a definition of %structure kind name% homomorphism.
- The reader knows a definition of %category name% isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that any 'rings - homomorphisms' isomorphism between any fields is a 'fields - homomorphisms' isomorphism.
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:
\(F_1\): \(\in \{\text{ the fields }\}\)
\(F_2\): \(\in \{\text{ the fields }\}\)
\(f\): \(F_1 \to F_2\), \(\in \{\text{ the 'rings - homomorphisms' isomorphisms }\}\)
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Statements:
\(f \in \{\text{ the 'fields - homomorphisms' isomorphisms }\}\)
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2: Proof
Whole Strategy: Step 1: see that all what needs to be checked is that for each \(r_1 \in F_1\), \(f (r_1^{-1}) = f (r_1)^{-1}\); Step 2: see that \(f (r_1^{-1}) = f (r_1)^{-1}\).
Step 1:
As any field is a ring, calling \(f\) "'rings - homomorphisms' isomorphism" makes sense.
While only the differences of 'field' from 'ring' are being multiplicative commutative and each element's having the inverse, the commutativity is just about inside \(F_1\) or \(F_2\), not about \(f\), and the only concern about \(f\) is whether for each \(r_1 \in F_1\), \(f (r_1^{-1}) = f (r_1)^{-1}\) and whether for each \(r_2 \in F_2\), \(f^{-1} (r_2^{-1}) = f^{-1} (r_2)^{-1}\).
If \(f (r_1^{-1}) = f (r_1)^{-1}\) holds, \(f^{-1} (r_2^{-1}) = f^{-1} (r_2)^{-1}\) will hold likewise, because the situation is symmetric.
So, let us check that \(f (r_1^{-1}) = f (r_1)^{-1}\).
Step 2:
\(r_1 r_1^{-1} = r_1^{-1} r_1 = 1\). \(f (r_1 r_1^{-1}) = f (r_1^{-1} r_1) = f (1) = 1\). But \(f (r_1 r_1^{-1}) = f (r_1) f (r_1^{-1})\) and \(f (r_1^{-1} r_1) = f (r_1^{-1}) f (r_1)\). So, \(f (r_1) f (r_1^{-1}) = f (r_1^{-1}) f (r_1) = 1\), which means that \(f (r_1)^{-1} = f (r_1^{-1})\).
