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.

Proof.

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:R→∏Ri given by ϵ⁢(r)⁢(i)=r+Ii is injectivePlanetmathPlanetmath. If ϵ⁢(r)=0, then r∈Ii for all i, or r∈⋂Ii=0, or r=0. But then R→∏Ri→Ri given by r↦r+Ii is not an isomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath for any i, contradicting the fact that R is subdirectly irreducible. Therefore, ⋂Ii≠0.

(⇐). Suppose next that ⋂Ii≠0. Let R be a subdirect product of some Ri, and let Ji:=ker⁡(R→∏Ri→Ri). Each Ji is an ideal of R. Let J=⋂Ji. If R→∏Ri→Ri is not an isomorphism (therefore not injective), Ji is non-zero. This means that if R is not subdirectly irreducible, J≠0. But J⊆ker⁡(R→∏Ri), contradicting the subdirect irreducibility of R. As a result, some Ji=0, or R→∏Ri→Ri 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 I≠0 by subdirect irreducibility. Pick 0≠s∈I. Then s2⁢R⊆s⁢R⊆I. So s2⁢R=s⁢R since I is minimalPlanetmathPlanetmath. This means s=s2⁢t, or 1=s⁢t∈s⁢R=I, which means I=R. Now, let any 0≠r∈R, then R=I⊆r⁢R, so 1=p⁢r for some p∈R, 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