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Homesubdirect product of rings

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# subdirect product of rings

A ring $R$ is said to be (represented as) a *subdirect product* of a family of rings $\{R_{i}:i\in I\}$ if:

1. there is a monomorphism $\varepsilon:R\longrightarrow\prod R_{i}$, and

2. given 1., $\pi_{i}\circ\varepsilon:R\longrightarrow R_{i}$ is surjective for each $i\in I$, where $\pi_{i}:\prod R_{i}\longrightarrow R_{i}$ is the canonical projection map.

A *subdirect product* (representation) of $R$ is said to be *trivial* if one of the $\pi_{i}\circ\varepsilon:R\longrightarrow R_{i}$ is an isomorphism.

Direct products and direct sums of rings are all examples of subdirect products of rings. $\mathbb{Z}$ does not have non-trivial direct product nor non-trivial direct sum representations of rings. However, $\mathbb{Z}$ can be represented as a non-trivial subdirect product of $\mathbb{Z}/({p_{i}}^{{n_{i}}})$.

As an application of subdirect products, it can be shown that any ring can be represented as a subdirect product of subdirectly irreducible rings. Since a subdirectly irreducible commutative reduced ring is a field, a Boolean ring $B$ can be represented as a subdirect product of $\mathbb{Z}_{2}$. Furthermore, if this Boolean ring $B$ is finite, the subdirect product representation becomes a direct product representation. Consequently, $B$ has $2^{n}$ elements, where $n$ is the number of copies of $\mathbb{Z}_{2}$.

## Mathematics Subject Classification

16D70*no label found*16S60

*no label found*

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