# Idempotent and nilpotent elements of ring $\mathbb Z_{1125}$

So I have this problem:

Find the idempotent and nilpotent elements of ring $\mathbb Z_{1125}$

This is what I have tried:

I factorized $1125$ as : $3^2\cdot5^3$. Now I think that the idempotent values of $Z_{3^2}$ are $\{0,1\}$ and of $Z_{5^3}$ are $\{0,1\}$ but I don't know why, and I am not sure. I got stuck here, don't know what to do. Can you guys please explain how I should go in solving this? Also it would be nice if you explain how to find the nilpotent elements as well. Thank you.

• Try to trite up what it means for an element to be idempotent or nilpotent in terms of equations (i.e. something will need to be divisible by $1125$). Commented Jun 5, 2015 at 12:42
• well for an element to be idempotent means that $a^2=a$. But what if $n=125$ or $5^3$, surely there must be another method of finding this out rather than taking every number from $0,1,...,124$ and powering it up to the power of $2$ and dividing it by $125$ to find if it is idempotent, that takes a lot of time... Commented Jun 5, 2015 at 12:46
• As I said, write it in terms of "something divides $1125$" and think of prime divisors. Commented Jun 5, 2015 at 12:48
• Or, as you seem to want to do this by using the chinese remainder theorem, think of when something is nilpotent or idempotent in a product if rings. Commented Jun 5, 2015 at 12:51
• I just said that for an element to be a idempotent if must satisfy this rule $a^2=a$ but what if $n=125$, then what ?!? I verify $125$ numbers?! Commented Jun 5, 2015 at 12:53

Since you've factored $\mathbb{Z}/1125\mathbb{Z}\cong \mathbb{Z}/9\mathbb{Z}\times \mathbb{Z}/125 \mathbb{Z}$ by the Chinese remainder theorem.

Both $\mathbb{Z}/9\mathbb{Z}$ and $\mathbb{Z}/125 \mathbb{Z}$ are local rings so their only idempotents are $0,1$ respectively.

If $(x,y)$ was an idempotent, then $(x,y)^2=(x^2,y^2)=(x,y)$ so the only idempotents are the four combinations of $0,1$.

An element is nilpotent if it has a divisor by a prime number in the respective ring. (i.e. $(3,5)$ is nilpotent just raise it to the power $3$). You can find all of these elements.

• yeah, but how do I know that they only have $0,1$ as idempotents, what if it involves another number which as more idempotents than $0,1$ , how do I find those ? That's what I am curious about. Commented Jun 5, 2015 at 12:58
• @ciorianu the only idempotents in a local ring are 0,1.
– Eoin
Commented Jun 5, 2015 at 13:00
• So if $Z_{p^k}$ has $p$ prime that means the idempotents of that are $0,1$ ? Commented Jun 5, 2015 at 13:03
• Yes. You should try to find proof that a local ring has only 0,1 as idempotents. And that $Z/(p^k)$ is a local ring (which is on this site, I've written it up before).
– Eoin
Commented Jun 5, 2015 at 13:07
• Using that it holds for local rings and that this is a local ring seems like a bit overkill (even though it is a neat thing to know). It is really straightforward to show directly. Commented Jun 5, 2015 at 13:08

For find the nilpotent elements of ring $Z_{1125}$ :
Only maximal ideals of $z_{1125}$ are $<\overline3>$ and $<\overline5>$ and in $z_n$ every prime ideal is maximal ideal , since $nilpotent\, elements=\cap \{I;I\,\,is\,\,prime\,\,ideal\,\, of\,\, Z_{1125}\}$ so $nilpotent\, elements=<\overline3>\cap<\overline5>$ .
For idempotent elements : Let $\overline a^2=\overline a$ so $\overline a^2-\overline a=\overline 0$
$(a^2-a)|1125 ‎\Longrightarrow a(a-1) |1125 \,, since\,\, a(a-1)\, is\,\, even\,\, so\,\, a(a-1)=0$ it's means $a=0\,\, or\,\,1$