description/proof of that range of field homomorphism is subfield of codomain
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 admits the proposition that the range of any ring homomorphism is a subring of the codomain.
Target Context
- The reader will have a description and a proof of the proposition that the range of any field homomorphism is a subfield of the codomain.
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:
\(F_1\): \(\in \{\text{ the fields }\}\)
\(F_2\): \(\in \{\text{ the fields }\}\)
\(f\): \(:F_1 \to F_2\), \(\in \{\text{ the field homomorphisms }\}\)
//
Statements:
\(f (F_1) \in \{\text{ the fields }\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(f (F_1)\) is a subring of \(F_2\); Step 2: see that \(f (F_1)\) is commutative under the multiplication; Step 3: see that each element of \(f (F_1)\) has an inverse.
Step 1:
\(F_1\) and \(F_2\) are some rings and \(f\) is a ring homomorphism.
By the proposition that the range of any ring homomorphism is a subring of the codomain, \(f (F_1)\) is a subring of \(F_2\).
Step 2:
\(f (F_1)\) is commutative under the multiplication, because the multiplication is inherited from ambient \(F_2\), which is commutative under the multiplication.
Step 3:
Let us see that each element of \(f (F_1)\) has an inverse.
Let \(f (r_1) \in f (F_1)\) be any.
\(f ({r_1}^{-1}) \in f (F_1)\) is an inverse of \(f (r_1)\): \(f (r_1) f ({r_1}^{-1}) = f (r_1 {r_1}^{-1}) = f (1) = 1\) and \(f ({r_1}^{-1}) f (r_1) = f ({r_1}^{-1} r_1) = f (1) = 1\).
