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I've been following MIT's old opencourseware class on commutative algebra. For one problem, I want to find the nilpotent and idempotent elements of $\mathbb{Z}/(n)$, where $n=\prod p_i^{n_i}$.

I know $\mathbb{Z}/(n)\cong\prod\mathbb{Z}/(p_i^{n_i})$. Also, $\text{nilrad}(\mathbb{Z}/(p_i^{n_i}))=\bigcap P$ where the intersection ranges over $\text{Spec}(\mathbb{Z}/(p_i^{n_i}))$. But the prime ideals of $\mathbb{Z}/(p_i^{n_i})$ are in bijection with the primes of $\mathbb{Z}$ containing $(p_i^{n_i})$, which is just $(p_i)$. So the nilpotent elements is the image $(p_i)/(p_i^{n_i})$ under the canonical projection. With that, I could theoretically find the nilpotent elements in each factor, and then find the nilpotent elements of $\mathbb{Z}/(n)$ by the Chinese remainder theorem. Is there a more concrete way to express the set of nilpotenet elements, or is this the best we can do?

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If $p^n$ (and hence $p$) divides $x^2-x=x(x -1)$, where $x$ is an integer, what can you say? –  Dylan Moreland Mar 6 '12 at 7:45
    
@DylanMoreland I think I see what you're getting at. If $x$ is idempotent in $\mathbb{Z}/(p^n)$, then $p\mid x^2-x=x(x-1)$, so $p\mid x$ or $p\mid x-1$. But $x$ and $x-1$ are coprime, so in fact $p^n\mid x$ or $p^n\mid x-1$. So $x=0$ or $x=1$ are the idepotents in $\mathbb{Z}/(p^n)$. Then the idempotents of $\mathbb{Z}/(n)$ could be found by the Chinese remainder theorem again. Is that all one can say? –  hmIII Mar 6 '12 at 7:53
    
@hmIII If your $n$ is square free the quotient ring is a reduced ring (exercise). –  user38268 Mar 9 '12 at 12:49

2 Answers 2

up vote 3 down vote accepted

This answer is about the idempotents (I gather from your comment above, @hmIII, that you want to know also about idempotents). As you write, they are got from the CRT, i.e. the isomorphism $$\mathbb{Z}/ n \cong \prod \mathbb{Z}/ p_i^{n_i}.$$ Let $e_i$ be the element of $\mathbb{Z}/ n$ that is $1$ module $p_i^{n_i}$ and $0$ modulo every other $p_j^{n_j}$. These give the primitive idempotents; it is a general principle that every idempotent $e$ is a sum $$e=\sum_{i \in S} e_i$$ (the set $S$ might be empty) of these primitive idempotents.

To write down particular representatives for the $e_i$ one might proceed as follows: let $m_i=\prod_{j \neq i } p_i^{n_i}$ and compute (using the Euclidean algorithm) a representative $l_i \in \mathbb{Z}$ for the inverse $m_i^{-1}$ of $m_i$ modulo $p_i^{n_i}$. Then $e_i=l_i m_i \ \mathrm{mod} \ n$ is the $i$th primitive idempotent. This amounts to a constructive proof of the CRT; you write "is that all you can say", and the answer is that constructing these idempotents is essentially equivalent to proving the CRT, for the isomorphism can be specified explicitly as $(a_i)_{i \in I} \mapsto \sum a_i e_i \in \mathbb{Z} / n$, where $I$ is an index set for the prime factors of $n$ and $(a_i) \in \prod_{i \in I} \mathbb{Z} / p_i^{n_i}$.

Worked examples: (1) Let $n=80=2^4 5$. We have $13=5^{-1}$ mod $16$ and $1=16^{-1}$ mod $5$ so the primitive idempotents are $e_1=13*5=65$ and $e_2=1*16=16$. Together with $0$ and $1$ these are the only idempotents.

(2) Let $n=60=2^2*3*5$. We have $(3*5)^{-1}=3$ mod $4$, $(4*5)^{-1}=2$ mod $3$, and $(4*3)^{-1}=3$ mod $5$ so the primitive idempotents are $e_1=3*15=45$, $e_2=2*20=40$, and $e_3=3*12=36$. There are $8$ idempotents total, obtained by summing over all possible subsets of $\{e_1,e_2,e_3\}$.

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Thanks for the answer and examples Steve. So really the answer is to give some algorithm for finding the nilpotent/idempotents? As a confirmation, is my thinking on finding the nilpotent elements correct? –  hmIII Mar 6 '12 at 16:51
    
Basically the answer to both your questions is yes. But it's somewhat easier to understand the nilpotents concretely, since they are just those elements of $\mathbb{Z}/n$ represented by integers divisible by $\prod p_i$ (just as in your question and lhf's answer). –  S123 Mar 6 '12 at 17:26
    
Also, it's probably worth your time to think about this: there are algorithms, and then there are algorithms. The algorithm for finding the representatives of the idempotents is very fast (assuming, of course, you are given the factorization of $n$...); it's essentially the Euclidean algorithm for computing a GCD. This should be contrasted with many important number-theoretic tasks for which no known polynomial time algorithm exists! –  S123 Mar 6 '12 at 17:29

Perhaps this is not the way you want to go but the nilpotent elements mod $n$ correspond to the numbers divisible by all the prime factors of $n$. In other words, the multiples of the radical of $n$.

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Dear lhf: $+1$, but I think you mean "divisible by the prime factors of $n$" instead of "having exactly the same prime factors as $n$". (Example: $6$ is nilpotent mod $2$.) –  Pierre-Yves Gaillard Mar 6 '12 at 12:14
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@Pierre-YvesGaillard, thanks for the correction; I've edited my answer. A better example is perhaps 30 which is nilpotent mod 48. –  lhf Mar 6 '12 at 12:26

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