isomorphism swapping zero and unity

Let  $(R,\,+,\,\cdot)$  be a ring with unity 1.  Define two new binary operations of $R$ as follows:

 $\displaystyle a\!\oplus\!b\;=:\;a\!+\!b\!-\!1,\qquad a\odot\!b\;=:\;a\!+\!b\!-% \!a\!\cdot\!b$ (1)

Then we see that

 $\displaystyle a\!\oplus\!1\;=\;a\;=\;1\!\oplus\!a,\qquad a\!\odot\!0\;=\;a\;=% \;0\!\odot\!a.$ (2)

But moreover, the algebraic system$(R,\,\oplus,\,\odot)$  is a unitary ring, too, and isomorphic with the original ring.

In fact, we may define the bijective mapping

 $\displaystyle f\!:\;x\,\mapsto\,1\!-\!x$ (3)

from $R$ to $R$ and verify that it is homomorphic:

 $f(a)\!\oplus\!f(b)\;=\;(1\!-\!a)\!\oplus\!(1\!-\!b)\;=\;(1\!-\!a)\!+\!(1\!-\!b% )\!-\!1\;=\;1\!-\!a\!-\!b\;=\;f(a\!+\!b),$
 $f(a)\!\odot\!f(b)\;=\;(1\!-\!a)\!\odot\!(1\!-\!b)\;=\;(1\!-\!a)\!+\!(1\!-\!b)% \!-(1\!-\!a)\!\cdot\!(1\!-\!b)\;=\;1\!-\!a\!\cdot\!b\;=\;f(a\!\cdot\!b)$

Thus  $(R,\,\oplus,\,\odot)$  as a homomorphic image (http://planetmath.org/HomomorphicImageOfGroup) of the ring  $(R,\,+,\,\cdot)$  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