description/proof of that bijective field homomorphism 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 bijection.
- The reader knows a definition of %structure kind name% homomorphism.
- The reader knows a definition of %category name% isomorphism.
- The reader admits the proposition that any bijective ring homomorphism is a 'rings - homomorphisms' isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that any bijective field homomorphism 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:
\(F_1\): \(\in \{\text{ the fields }\}\)
\(F_2\): \(\in \{\text{ the fields }\}\)
\(f\): \(F_1 \to F_2\), \(\in \{\text{ the bijections }\} \cap \{\text{ the field homomorphisms }\}\)
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Statements:
\(f \in \{\text{ the 'fields - homomorphisms' isomorphisms }\}\)
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2: Note
Sometimes, the definition that dictates any bijective field homomorphism to be a 'fields - homomorphisms' isomorphism is seen, but that does not seem a good practice: 'isomorphism' is a general concept defined in the category theory and requires the inverse to be a homomorphism in the category.
In general, a bijective morphism of a category is not necessarily any %category name% isomorphism. For example, a bijective continuous map, which is a morphism of the 'topological spaces - continuous maps' category, is not necessarily a homeomorphism, which is a 'topological spaces - continuous maps' isomorphism.
As this proposition holds, some people think that that definition that requires only bijective-ness is valid, but just because this proposition holds does not mean that the general definition made in the category theory should be deformed for the field case.
3: Proof
Whole Strategy: Step 1: take the inverse, \(f^{-1}\); Step 2: see that \(f^{-1}\) is a field homomorphism.
Step 1:
As \(f\) is bijective, there is the inverse, \(f^{-1}: F_2 \to F_1\).
Step 2:
Let us see that \(f^{-1}\) is necessarily field homomorphic.
\(F_1\) and \(F_2\) are some rings and \(f\) is a bijective ring homomorphism between the rings.
By the proposition that any bijective ring homomorphism is a 'rings - homomorphisms' isomorphism, \(f\) is a 'rings - homomorphisms' isomorphism between the rings. So, \(f^{-1}\) is a ring homomorphism between the rings.
The remaining issue is only whether for each element of \(F_2\), \(f^{-1}\) maps its inverse to the inverse of its image under \(f^{-1}\).
Let \(f (r_1) \in F_2\) be any. \(f (r_1)^{-1} = f (r_1^{-1})\), because \(f\) is field homomorphic. \(f^{-1} (f (r_1)^{-1}) = f^{-1} (f (r_1^{-1})) = r_1^{-1} = {f^{-1} (f (r_1))}^{-1}\).
So, \(f^{-1}\) is a field homomorphism.
