2025-01-19

954: For 2 Rings Between Which There Is Bijection That Preserves Multiplications, if Domain Is Field, Codomain Is Field

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description/proof of that for 2 rings between which there is bijection that preserves multiplications, if domain is field, codomain is field

Topics


About: ring
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 for any 2 rings between which there is any bijection that preserves multiplications, if the domain is a field, the codomain is a field.

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:
R1: ∈{ the rings }
R2: ∈{ the rings }
f: :R1→R2, ∈{ the bijections }
//

Statements:
(
f(1)=1∧∀r1,r1′∈R1(f(r1r1′)=f(r1)f(r1′))
∧
R1∈{ the fields }
)
⟹
R2∈{ the fields }
//

2: Note


f does not need to be (or to be confirmed to be) a 'rings - homomorphisms' isomorphism or a ring homomorphism, because Proof uses only the bijective-ness and the preservation of multiplications in the forward direction.

3: Proof


Whole Strategy: Step 1: take the inverse, f−1:R2→R1; Step 2: see that R2 is commutative; Step 3: see that each element of R2 has an inverse; Step 4: conclude the proposition.

Step 1:

As f is bijective, there is the inverse, f−1:R2→R1.

Step 2:

Let us see that R2 is commutative.

Let r2,r2′∈R2 be any.

As R1 is a field, R1 is commutative.

So, f−1(r2)f−1(r2′)=f−1(r2′)f−1(r2).

f(f−1(r2)f−1(r2′))=f(f−1(r2′)f−1(r2)).

But f(f−1(r2)f−1(r2′))=f(f−1(r2))f(f−1(r2′))=r2r2′ and f(f−1(r2′)f−1(r2))=f(f−1(r2′))f(f−1(r2))=r2′r2.

So, r2r2′=r2′r2.

Step 3:

Let us see that each element of R2 has an inverse.

Let r2∈R2 be any.

As R1 is a field, f−1(r2) has the inverse, r1.

f−1(r2)r1=r1f−1(r2)=1.

f(f−1(r2)r1)=f(r1f−1(r2))=f(1)=1.

But f(f−1(r2)r1)=f(f−1(r2))f(r1)=r2f(r1) and f(r1f−1(r2))=f(r1)f(f−1(r2))=f(r1)r2.

So, r2f(r1)=f(r1)r2=1.

That means that f(r1) is an inverse of r2.

Step 4:

So, R2 is a commutative ring whose each element has an inverse, which means that R2 is a field.

References
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