description/proof of that ring homomorphism between fields is field homomorphism
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 ring.
- The reader knows a definition of %structure kind name% homomorphism.
Target Context
- The reader will have a description and a proof of the proposition that any ring homomorphism between any fields is a field homomorphism.
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:
\(F_1\): \(\in \{\text{ the fields }\}\)
\(F_2\): \(\in \{\text{ the fields }\}\)
\(f\): \(F_1 \to F_2\), \(\in \{\text{ the ring homomorphisms }\}\)
//
Statements:
\(f \in \{\text{ the field homomorphisms }\}\)
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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 \(f\) "ring homomorphism" makes sense.
While only the differences of 'field' from 'ring' are being multiplicative commutative and each nonzero 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 \(r_1 \in F_1\), \(f (r_1^{-1}) = f (r_1)^{-1}\).
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 \(f (r_1)^{-1} = f (r_1^{-1})\).
