regular elements of finite ring

Theorem.

If the finite ring $R$ has regular elements, then it has a unity.  All regular elements of $R$ form a group under the ring multiplication and with identity element the unity of $R$.  Thus the regular elements are exactly the units of the ring; the rest of the elements are the zero and the zero divisors.

Proof.  Obviously, the set of the regular elements is non-empty and closed under the multiplication.  Let’s think the multiplication table of this set.  It is a finite distinct elements (any equation  $ax=ay$  reduces to  $x=y$).  Hence, for every regular element $a$, the square $a^2$ determines another $a^{\prime }$ such that  $a^2{a}^{\prime }=a$.  This implies  $a^{\prime }(a^2{a}^{\prime })(a{a}^{\prime })=a^{\prime }a(a{a}^{\prime })$,  i.e.  $(a^{\prime }a){(a{a}^{\prime })}^2=(a^{\prime }a)(a{a}^{\prime })$,  and since $a^{\prime }a$ is regular (http://planetmath.org/ZeroDivisor),  we obtain that  ${(a{a}^{\prime })}^2=a{a}^{\prime }$.  So $a{a}^{\prime }$ is idempotent, and because it also is , it must be the unity of the ring (http://planetmath.org/Unity):  $a{a}^{\prime }=1$.  Thus we see that $R$ has a unity which is a regular element and that $a$ has a multiplicative inverse $a^{\prime }$, also regular.  Consequently the regular elements form a group.

Title	regular elements of finite ring
Canonical name	RegularElementsOfFiniteRing
Date of creation	2013-03-22 15:11:09
Last modified on	2013-03-22 15:11:09
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	19
Author	pahio (2872)
Entry type	Theorem
Classification	msc 13G05
Classification	msc 16U60
Related topic	GroupOfUnits
Related topic	PrimeResidueClass
Related topic	WedderburnsTheorem
Related topic	Unity