description/proof of that bijective ring homomorphism is 'rings - homomorphisms' isomorphism
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of ring.
- 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.
Target Context
- The reader will have a description and a proof of the proposition that any bijective ring homomorphism is a 'rings - 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:
\(R_1\): \(\in \{\text{ the rings }\}\)
\(R_2\): \(\in \{\text{ the rings }\}\)
\(f\): \(R_1 \to R_2\), \(\in \{\text{ the bijections }\} \cap \{\text{ the ring homomorphisms }\}\)
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Statements:
\(f \in \{\text{ the 'rings - homomorphisms' isomorphisms }\}\)
//
2: Note
Sometimes, the definition that dictates any bijective ring homomorphism to be a 'rings - 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 ring case.
3: Proof
Whole Strategy: Step 1: take the inverse, \(f^{-1}\); Step 2: see that \(f^{-1}\) is a ring homomorphism.
Step 1:
As \(f\) is bijective, there is the inverse, \(f^{-1}: R_2 \to R_1\).
Step 2:
Let us see that \(f^{-1}\) is necessarily ring homomorphic.
\(R_1\) and \(R_2\) are some additive groups and \(f\) is a bijective group homomorphism between the additive groups.
By the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism, \(f\) is a 'groups - homomorphisms' isomorphism between the additive groups. So, \(f^{-1}\) is a group homomorphism between the additive groups.
\(f^{-1} (1) = 1\), because \(f (1) = 1\) and \(1 = f^{-1} (f (1)) = f^{-1} (1)\).
For each \(f (r_1), f (r'_1) \in R_2\), \(f^{-1} (f (r_1) f (r'_1)) = f^{-1} (f (r_1)) f^{-1} (f (r'_1))\), because \(f (r_1 r'_1) = f (r_1) f (r'_1)\) and \(f^{-1} (f (r_1)) f^{-1} (f (r'_1)) = r_1 r'_1 = f^{-1} (f (r_1 r'_1)) = f^{-1} (f (r_1) f (r'_1))\).
So, \(f^{-1}\) is a ring homomorphism.
