isomorphism swapping zero and unity
Let (R,+,⋅) be a ring with unity 1. Define two new binary operations of R as follows:
a⊕b=:a+b-1,a⊙b=:a+b-a⋅b | (1) |
Then we see that
a⊕1=a= 1⊕a,a⊙0=a= 0⊙a. | (2) |
But moreover, the algebraic system (R,⊕,⊙) is a unitary ring, too, and isomorphic with the original ring.
In fact, we may define the bijective mapping
f:x↦ 1-x | (3) |
from R to R and verify that it is homomorphic:
f(a)⊕f(b)=(1-a)⊕(1-b)=(1-a)+(1-b)-1= 1-a-b=f(a+b), |
f(a)⊙f(b)=(1-a)⊙(1-b)=(1-a)+(1-b)-(1-a)⋅(1-b)= 1-a⋅b=f(a⋅b) |
Thus (R,⊕,⊙) as a homomorphic image (http://planetmath.org/HomomorphicImageOfGroup) of the ring (R,+,⋅) is a ring, it’s a question of two isomorphic rings.
Title | isomorphism swapping zero and unity |
Canonical name | IsomorphismSwappingZeroAndUnity |
Date of creation | 2013-03-22 19:17:16 |
Last modified on | 2013-03-22 19:17:16 |
Owner | pahio (2872) |
Last modified by | pahio (2872) |
Numerical id | 7 |
Author | pahio (2872) |
Entry type | Example |
Classification | msc 16B99 |
Classification | msc 20A05 |
Classification | msc 16S50 |
Related topic | RingHomomorphism |
Related topic | EpimorphismBetweenUnitaryRings |
Related topic | Null |
Related topic | TranslationAutomorphismOfAPolynomialRing |