# Is there a ring such that the nilradical of it is not nilpotent? [closed]

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## closed as off-topic by user26857, Jonas, Antonios-Alexandros Robotis, choco_addicted, Michael AlbaneseMar 15 at 3:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Jonas, Antonios-Alexandros Robotis, choco_addicted, Michael Albanese
If this question can be reworded to fit the rules in the help center, please edit the question.

What have you tried so far? To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. – Zev Chonoles Feb 6 '13 at 9:55
@Zev Chonoles I know that in a Noetherian ring the nil radical is nilpotent. Also, in a Artinian ring. Indeed, the ring is commutative with identity. – aliakbar Feb 6 '13 at 9:59
Do you know examples of non-Noetherian rings? Can you modify those examples so they have lots of nilpotents? Look at my answer for a hint. – Zev Chonoles Feb 6 '13 at 10:00

Hint:

• The nilradical of $k[x]/(x^2)$ is the ideal $N=(x)$, and $N^2=(0)$, so $N$ is a nilpotent ideal.

• The nilradical of $k[x,y]/(x^2,y^2)$ is the ideal $N=(x,y)$, and $N^3=(0)$, so $N$ is a nilpotent ideal.

• ...

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So consider $Z[x_{1},x_{2},...]$. Is it true? – aliakbar Feb 6 '13 at 10:08
The ring $\mathbb{Z}[x_1,x_2,\ldots]$ doesn't have any nilpotents; you need to take a quotient. Look again at how the rings were formed in my hint. – Zev Chonoles Feb 6 '13 at 10:09

Take the polynomial ring $R=\mathbb K[x_i \mid i \in \mathbb N]$ in infinitely many variables. Then take the ideal $I=(x_i^i \mid i \in \mathbb N)$. The ring $R/I$ has for each $i \in \mathbb N$ a nilpotent element $\bar x_i$ which has nilpotence index $i$. In this ring the nilradical contain for each $i \in \mathbb N$ the element $\bar x_i$, but for every $k \in \mathbb N$ we have that $\bar x^k_{k+1} \ne 0$ is an element of the $k$-th power of the nilradical. So the nilradical isn't nilpotent.

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