Let $\mathbb{Z}$ be the ring of integers and $a,b,c \in \mathbb{Z}$. Consider $R= \Bigg\{\begin{pmatrix} a & c\\ 0 & b \\ \end{pmatrix} \big| a-b\equiv c\equiv 0\mod2 \Bigg\}$. I have to find the prime radical of this ring $R$. I know that prime radical of a ring is the intersection of all prime ideals. Also, I know that in a ring with unity, prime radical is the collection of all strongly nilpotent elements. But I am having hard time to find its prime radical. In their paper titled "Armendariz rings and reduced ings" N.K Kim and Y. Lee in example 13, write the prime radical of this ring as $P(R)= \Bigg\{\begin{pmatrix} 0 & c\\ 0 & 0 \\ \end{pmatrix} \big| c\equiv 0\mod2 \Bigg\}$. But I don't know how to find it. For example, I took that suppose $I$ is a prime ideal of $R$. Now, I have to find $\cap I $.How to proceed?


1 Answer 1


It shouldn't read "Suppose $I$ is a prime ideal, then find their intersection." Rather it should read "Find all prime ideals of $R$, then find their intersection."

So, you can see you have not done that step yet: what are the prime ideals?

Part of the problem is made easier because the collection $\begin{bmatrix}0&c\\0&0\end{bmatrix}$ is a nilpotent ideal, and all prime ideals contain it. So you can reduce the task to finding prime ideals of $\{(a,b)\in\mathbb Z\times\mathbb Z\mid a-b\equiv 0 \pmod 2\}$.

What's nice is that this ring is commutative, and so the prime radical is equal to the nilradical. But it is a subring of $\mathbb Z\times\mathbb Z$ which is reduced already. Without computing the prime ideals directly, you can still reason that their intersection will be zero in the quotient ring (and this tells you something about the intersection of prime ideals in the original ring.)

P.S.: Another way to view this ring is as the subring of $\begin{bmatrix}\mathbb Z&\mathbb Z\\0&\mathbb Z\end{bmatrix}$ generated by the identity matrix and the ideal $\begin{bmatrix}2\mathbb Z&2\mathbb Z\\0&2\mathbb Z\end{bmatrix}$.

  • $\begingroup$ Thanks. It helps a lot. $\endgroup$
    – Chaudhary
    May 19 at 9:23

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