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# zero ring

Zero rings are commutative under multiplication. For if $Z$ is a zero ring, $0_{Z}$ is its additive identity, and $x,y\in Z$, then $xy=0_{Z}=yx.$

Every zero ring is a nilpotent ring. For if $Z$ is a zero ring, then $Z^{2}=\{0_{Z}\}$.

Since every subring of a ring must contain its zero element, every subring of a ring is an ideal, and a zero ring has no prime ideals.

The simplest zero ring is ${\mathbb{Z}}_{1}=\{0\}$. Up to isomorphism, this is the only zero ring that has a multiplicative identity.

Zero rings exist in abundance. They can be constructed from any ring. If $R$ is a ring, then

$\left\{\left.\left(\begin{array}[]{cc}r&-r\\ r&-r\end{array}\right)\right|r\in R\right\}$ |

considered as a subring of ${\mathbf{M}}_{{2\operatorname{x}2}}(R)$ (with standard matrix addition and multiplication) is a zero ring. Moreover, the cardinality of this subset of ${\mathbf{M}}_{{2\operatorname{x}2}}(R)$ is the same as that of $R$.

Moreover, zero rings can be constructed from any abelian group. If $G$ is a group with identity $e_{G}$, it can be made into a zero ring by declaring its addition to be its group operation and defining its multiplication by $a\cdot b=e_{G}$ for any $a,b\in G$.

Every finite zero ring can be written as a direct product of cyclic rings, which must also be zero rings themselves. This follows from the fundamental theorem of finite abelian groups. Thus, if $p_{1},\ldots,p_{m}$ are distinct primes, $a_{1},\ldots,a_{m}$ are positive integers, and $\displaystyle n=\prod_{{j=1}}^{m}{p_{j}}^{{a_{j}}}$, then the number of zero rings of order $n$ is $\displaystyle\prod_{{j=1}}^{m}p(a_{j})$, where $p$ denotes the partition function.

## Mathematics Subject Classification

16U99*no label found*13M05

*no label found*13A99

*no label found*

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better link by Mathprof ✓

broken by Mathprof ✘

mention by CWoo ✓

even more abundant by rspuzio ✓