# regular elements of finite ring

###### Theorem.

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  $a^{\prime}(a^{2}a^{\prime})(aa^{\prime})=a^{\prime}a(aa^{\prime})$,  i.e.  $(a^{\prime}a)(aa^{\prime})^{2}=(a^{\prime}a)(aa^{\prime})$,  and since $a^{\prime}a$ is regular (http://planetmath.org/ZeroDivisor),  we obtain that  $(aa^{\prime})^{2}=aa^{\prime}$.  So $aa^{\prime}$ is idempotent  , and because it also is , it must be the unity of the ring (http://planetmath.org/Unity):  $aa^{\prime}=1$.  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 RegularElementsOfFiniteRing 2013-03-22 15:11:09 2013-03-22 15:11:09 pahio (2872) pahio (2872) 19 pahio (2872) Theorem msc 13G05 msc 16U60 GroupOfUnits PrimeResidueClass WedderburnsTheorem Unity