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We have a ring

$R=\begin{pmatrix} \mathbb{Z} & 0 \\ \mathbb{Z} & \mathbb{Z} \end{pmatrix}$

Let $I=\begin{pmatrix} 12\mathbb{Z} & 0 \\ 3\mathbb{Z} & 3\mathbb{Z} \end{pmatrix}$

How do I show that $R/I$ is artinian ring? $R/I=I$, the only ideal that isn't equal to I is $\begin{pmatrix} 0 & 0 \\ 3\mathbb{Z} & 3\mathbb{Z} \end{pmatrix}$ and $\begin{pmatrix} 12\mathbb{Z} & 0 \\ 3\mathbb{Z} & 0 \end{pmatrix}$. These are both DCC condition is satisfied with them.

So then you just conclude it's aritian. Also, the niradical elements are just the bottom left corner of the matrix.

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1 Answer

I think you are trying too hard. This ring is Artinian because it is finite: it has only $12*3*3=72$ elements.

Let $\mathbb{Z}_n$ denote the ring of integers mod $n$. The quotient is $$ R/I=\begin{pmatrix} \mathbb{Z}_{12} & 0 \\ \mathbb{Z}_3 & \mathbb{Z}_3 \end{pmatrix} $$

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