regular elements of finite ring
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 reduces to ). Hence, for every regular element , the square determines another such that . This implies , i.e. , and since is regular (http://planetmath.org/ZeroDivisor), we obtain that . So is idempotent, and because it also is , it must be the unity of the ring (http://planetmath.org/Unity): . Thus we see that has a unity which is a regular element and that has a multiplicative inverse , also regular. Consequently the regular elements form a group.
|Title||regular elements of finite ring|
|Date of creation||2013-03-22 15:11:09|
|Last modified on||2013-03-22 15:11:09|
|Last modified by||pahio (2872)|