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
- The reader knows a definition of ring.
- The reader knows a definition of bijection.
- The reader knows a definition of field.
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:
\(R_1\): \(\in \{\text{ the rings }\}\)
\(R_2\): \(\in \{\text{ the rings }\}\)
\(f\): \(: R_1 \to R_2\), \(\in \{\text{ the bijections }\}\)
//
Statements:
(
\(f (1) = 1 \land \forall r_1, r'_1 \in R_1 (f (r_1 r'_1) = f (r_1) f (r'_1))\)
\(\land\)
\(R_1 \in \{\text{ the fields }\}\)
)
\(\implies\)
\(R_2 \in \{\text{ the fields }\}\)
//
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, \(f^{-1}: R_2 \to R_1\); Step 2: see that \(R_2\) is commutative; Step 3: see that each element of \(R_2\) has an inverse; Step 4: conclude the proposition.
Step 1:
As \(f\) is bijective, there is the inverse, \(f^{-1}: R_2 \to R_1\).
Step 2:
Let us see that \(R_2\) is commutative.
Let \(r_2, r'_2 \in R_2\) be any.
As \(R_1\) is a field, \(R_1\) is commutative.
So, \(f^{-1} (r_2) f^{-1} (r'_2) = f^{-1} (r'_2) f^{-1} (r_2)\).
\(f (f^{-1} (r_2) f^{-1} (r'_2)) = f (f^{-1} (r'_2) f^{-1} (r_2))\).
But \(f (f^{-1} (r_2) f^{-1} (r'_2)) = f (f^{-1} (r_2)) f (f^{-1} (r'_2)) = r_2 r'_2\) and \(f (f^{-1} (r'_2) f^{-1} (r_2)) = f (f^{-1} (r'_2)) f (f^{-1} (r_2)) = r'_2 r_2\).
So, \(r_2 r'_2 = r'_2 r_2\).
Step 3:
Let us see that each element of \(R_2\) has an inverse.
Let \(r_2 \in R_2\) be any.
As \(R_1\) is a field, \(f^{-1} (r_2)\) has the inverse, \(r_1\).
\(f^{-1} (r_2) r_1 = r_1 f^{-1} (r_2) = 1\).
\(f (f^{-1} (r_2) r_1) = f (r_1 f^{-1} (r_2)) = f (1) = 1\).
But \(f (f^{-1} (r_2) r_1) = f (f^{-1} (r_2)) f (r_1) = r_2 f (r_1)\) and \(f (r_1 f^{-1} (r_2)) = f (r_1) f (f^{-1} (r_2)) = f (r_1) r_2\).
So, \(r_2 f (r_1) = f (r_1) r_2 = 1\).
That means that \(f (r_1)\) is an inverse of \(r_2\).
Step 4:
So, \(R_2\) is a commutative ring whose each element has an inverse, which means that \(R_2\) is a field.
