Let $R=\mathbb Z_{36}$. Find $N(\langle 0 \rangle), N(\langle 4\rangle), N(\langle 6 \rangle)$.
Let $R$ be a commutative ring and let $A$ be any ideal of $R$. Then the nilradical of A, $N(A)=\{r\in R: r^n \in A \text{ for some positive integer } n\}$ is an ideal of $R$.
I understand the definition, but I'm not entirely sure how to compute in general the nilradicals of these principal ideals.
I recognize that $N(\langle 0 \rangle)$ is really the same as the nilpotent elements in $R$, which in this case would be the nilpotent elements in $\mathbb Z_{36}$. I found here that in $\mathbb Z_n$ an element $a$ is nilpotent iff every prime divisor of $n$ also divides $a$. Therefore, since $36=2^23^2$ the nilpotent elements in $\mathbb Z_{36}$ are $6$ and $12$. (Edit: I just realized that these would not be the only nilradical elements, by the theorem I just stated. Whoops! It would also include $18,24,30,36=0$)
Extending this logic to the other ones, we see that $4=2^2$ and $6=2\cdot 3$, so elements of $N(\langle 4 \rangle)$ must be divisible by $2$ and elements of $N(\langle 6 \rangle)$ must be divisible by $2$ and $3$.
Therefore, $N(\langle 4 \rangle) = \mathbb Z_{36} \cap 2\mathbb Z$, and $N(\langle 6 \rangle) = \mathbb Z_{36}\cap 6\mathbb Z$.
Is this correct?
Best Answer
Yes, it is correct, up to a minor problem of notation: $\mathbb Z_{36}\cap k\mathbb Z$ makes no sense ($\mathbb Z_{36}$ is a set of $\bmod{36}$-classes, whereas $k\mathbb Z$ is a set of integers). The correct notation is $k\mathbb Z_{36}$, or simply $\langle k\rangle$. So yes: $N(\langle4\rangle)=\langle2\rangle$ and $N(\langle0\rangle)=N(\langle6\rangle)=\langle6\rangle$.
More generaly, in $\mathbb Z_n$, $N(\langle m\rangle)=\langle$rad$(\gcd(m,n))\rangle$: see How can I calculate the radical of an ideal in ring ${\Bbb Z}_n$?