# 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 intersection^{} of *all* non-zero ideals of $R$ is non-zero.

###### Proof.

Let $\{{I}_{i}\}$ be the set of all non-zero ideals of $R$.

$(\Rightarrow )$. Suppose first that $R$ is subdirectly irreducible. If $\bigcap {I}_{i}=0$, then $R$ is a subdirect product of ${R}_{i}:=R/{I}_{i}$, for $\u03f5:R\to \prod {R}_{i}$ given by $\u03f5(r)(i)=r+{I}_{i}$ is injective^{}. If $\u03f5(r)=0$, then $r\in {I}_{i}$ for all $i$, or $r\in \bigcap {I}_{i}=0$, or $r=0$. But then $R\to \prod {R}_{i}\to {R}_{i}$ given by $r\mapsto r+{I}_{i}$ is not an isomorphism^{} for any $i$, contradicting the fact that $R$ is subdirectly irreducible. Therefore, $\bigcap {I}_{i}\ne 0$.

$(\Leftarrow )$. Suppose next that $\bigcap {I}_{i}\ne 0$. Let $R$ be a subdirect product of some ${R}_{i}$, and let ${J}_{i}:=\mathrm{ker}(R\to \prod {R}_{i}\to {R}_{i})$. Each ${J}_{i}$ is an ideal of $R$. Let $J=\bigcap {J}_{i}$. If $R\to \prod {R}_{i}\to {R}_{i}$ is not an isomorphism (therefore not injective), ${J}_{i}$ is non-zero. This means that if $R$ is not subdirectly irreducible, $J\ne 0$. But $J\subseteq \mathrm{ker}(R\to \prod {R}_{i})$, contradicting the subdirect irreducibility of $R$. As a result, some ${J}_{i}=0$, or $R\to \prod {R}_{i}\to {R}_{i}$ is an isomorphism. ∎

As an application of the above equivalence, we have that a simple ring^{} is subdirectly irreducible. In addition, a commutative^{} subdirectly irreducible reduced ring is a field. To see this, let $\{{I}_{i}\}$ be the set of all non-zero ideals of a commutative subdirectly irreducible reduced ring $R$, and let $I=\bigcap {I}_{i}$. So $I\ne 0$ by subdirect irreducibility. Pick $0\ne s\in I$. Then ${s}^{2}R\subseteq sR\subseteq I$. So ${s}^{2}R=sR$ since $I$ is minimal^{}. This means $s={s}^{2}t$, or $1=st\in sR=I$, which means $I=R$. Now, let any $0\ne r\in R$, then $R=I\subseteq rR$, so $1=pr$ for some $p\in R$, 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 |