regular elements of finite ring
Theorem.
If the finite ring has regular elements, then it has a unity. All regular elements of form a group under the ring multiplication and with identity element the unity of . 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 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 |
---|---|
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 |