2025-01-19

961: 'Rings - Homomorphisms' Isomorphism Between Fields Is 'Fields - Homomorphisms' Isomorphism

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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



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:
F1: { the fields }
F2: { the fields }
f: F1F2, { the 'rings - homomorphisms' isomorphisms }
//

Statements:
f{ the 'fields - homomorphisms' isomorphisms }
//


2: Proof


Whole Strategy: Step 1: see that all what needs to be checked is that for each r1F1, f(r11)=f(r1)1; Step 2: see that f(r11)=f(r1)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 F1 or F2, not about f, and the only concern about f is whether for each r1F1, f(r11)=f(r1)1 and whether for each r2F2, f1(r21)=f1(r2)1.

If f(r11)=f(r1)1 holds, f1(r21)=f1(r2)1 will hold likewise, because the situation is symmetric.

So, let us check that f(r11)=f(r1)1.

Step 2:

r1r11=r11r1=1. f(r1r11)=f(r11r1)=f(1)=1. But f(r1r11)=f(r1)f(r11) and f(r11r1)=f(r11)f(r1). So, f(r1)f(r11)=f(r11)f(r1)=1, which means that f(r1)1=f(r11).


References


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