subdirectly irreducible ring

A ring R is said to be subdirectly irreducible if every subdirect product of R is trivial.

Equivalently, a ring R is subdirectly irreducible iff the intersectionMathworldPlanetmath of all non-zero ideals of R is non-zero.


Let {Ii} be the set of all non-zero ideals of R.

(). Suppose first that R is subdirectly irreducible. If Ii=0, then R is a subdirect product of Ri:=R/Ii, for ϵ:RRi given by ϵ(r)(i)=r+Ii is injectivePlanetmathPlanetmath. If ϵ(r)=0, then rIi for all i, or rIi=0, or r=0. But then RRiRi given by rr+Ii is not an isomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath for any i, contradicting the fact that R is subdirectly irreducible. Therefore, Ii0.

(). Suppose next that Ii0. Let R be a subdirect product of some Ri, and let Ji:=ker(RRiRi). Each Ji is an ideal of R. Let J=Ji. If RRiRi is not an isomorphism (therefore not injective), Ji is non-zero. This means that if R is not subdirectly irreducible, J0. But Jker(RRi), contradicting the subdirect irreducibility of R. As a result, some Ji=0, or RRiRi is an isomorphism. ∎

As an application of the above equivalence, we have that a simple ringMathworldPlanetmath is subdirectly irreducible. In addition, a commutativePlanetmathPlanetmathPlanetmathPlanetmath subdirectly irreducible reduced ring is a field. To see this, let {Ii} be the set of all non-zero ideals of a commutative subdirectly irreducible reduced ring R, and let I=Ii. So I0 by subdirect irreducibility. Pick 0sI. Then s2RsRI. So s2R=sR since I is minimalPlanetmathPlanetmath. This means s=s2t, or 1=stsR=I, which means I=R. Now, let any 0rR, then R=IrR, so 1=pr for some pR, which means R is a field.

Title subdirectly irreducible ring
Canonical name SubdirectlyIrreducibleRing
Date of creation 2013-03-22 14:19:13
Last modified on 2013-03-22 14:19:13
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 10
Author CWoo (3771)
Entry type Definition
Classification msc 16D70