cyclic rings and zero rings
Lemma 1.
Let be a positive integer and be a cyclic ring of order (http://planetmath.org/OrderRing) . Then the following are equivalent:
Proof.
To show that 1 implies 2, let have behavior . Then there exists a generator (http://planetmath.org/Generator) of the additive group of such that . Since is a zero ring, . Since , it must be the case that . By definition of behavior, divides (http://planetmath.org/Divides) . Hence, .
The fact that 2 implies 3 follows immediately from the theorem that is stated and proven at cyclic rings that are isomorphic to (http://planetmath.org/CyclicRingsThatAreIsomorphicToKmathbbZ_kn).
The fact that 3 implies 1 follows immediately since is a zero ring. ∎
Lemma 2.
Let be an infinite . Then the following are equivalent:
-
1.
is a zero ring;
-
2.
has behavior ;
-
3.
is isomorphic (http://planetmath.org/Isomorphism7) to the subring of :
Proof.
To show that 1 implies 2, the contrapositive of the theorem that is stated and proven at cyclic rings that are isomorphic to (http://planetmath.org/CyclicRingsThatAreIsomorphicToKmathbbZ) can be used. If does not have behavior , then its behavior must be positive by definition, in which case . It is clear that is not a zero ring.
To show that 2 implies 3, let be a generator of the additive group of . It can be easily verified that defined by is a ring isomorphism.
The fact that 3 implies 1 follows immediately since is a zero ring. ∎
Title | cyclic rings and zero rings |
---|---|
Canonical name | CyclicRingsAndZeroRings |
Date of creation | 2013-03-22 17:14:43 |
Last modified on | 2013-03-22 17:14:43 |
Owner | Wkbj79 (1863) |
Last modified by | Wkbj79 (1863) |
Numerical id | 9 |
Author | Wkbj79 (1863) |
Entry type | Result |
Classification | msc 13M05 |
Classification | msc 13A99 |
Classification | msc 16U99 |