Nilpotents of \(\mathbb{Z}_n\)

Groups, Rings, Domains, Modules, etc, Galois theory
Post Reply
User avatar
Grigorios Kostakos
Founder
Founder
Posts: 461
Joined: Mon Nov 09, 2015 1:36 am
Location: Ioannina, Greece

Nilpotents of \(\mathbb{Z}_n\)

#1

Post by Grigorios Kostakos »

Let $n=p_1^{b_1}p_2^{b_2}\cdots p_s^{b_s}$ be the prime factorization of the positive integer $n$. Find conditions such that the ring $\mathbb{Z}_n$ has no-zero nilpotent elements.
Grigorios Kostakos
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Nilpotents of \(\mathbb{Z}_n\)

#2

Post by Papapetros Vaggelis »

If \(\displaystyle{b_i=0\,,\forall\,i\in\left\{1,...,k\right\}}\), then \(\displaystyle{\mathbb{Z}_{n}=\mathbb{Z}_{1}}\)

and this ring has not no-zero nilpotent elements.

Let \(\displaystyle{b_i\geq 1}\) for some \(\displaystyle{i\in\left\{1,...,k\right\}}\). Then,

\(\displaystyle{n>1}\).

Answer

If there exists \(\displaystyle{j\in\left\{1,...,s\right\}}\) such that

\(\displaystyle{b_{j}\geq 2}\), then \(\displaystyle{p_{j}^2\mid n}\) and the ring \(\displaystyle{\left(\mathbb{Z}_{n},+,\cdot\right)}\)

is not semisimple with \(\displaystyle{J(\mathbb{Z}_{n})=\langle{p_1...p_s\rangle}}\).

Then, \(\displaystyle{a=[p_1...p_s]\neq [0]}\) and \(\displaystyle{a^m=0}\) where

\(\displaystyle{m=\max\,\left\{b_1,...,b_s\right\}}\).

Suppose now that \(\displaystyle{b_{j}<2\,,\forall\,j\in\left\{1,...,s\right\}}\), then the positive

integer \(\displaystyle{n>1}\) is not divided by a square of integer, so the ring \(\displaystyle{\mathbb{Z}_{n}}\)

is semisiple and \(\displaystyle{J(\mathbb{Z}_{n})=\left\{0\right\}}\) and this ring has not

no-zero nilpotents elements.

So, the ring \(\displaystyle{\left(Z_{n},+,\cdot\right)}\) has no-zero nilpotents elements if,

and only if, it is not semisimple.

Some comments

If \(\displaystyle{\left(R,+,\cdot\right)}\) is a commutative Artinian ring, then

\(\displaystyle{J(R)=\left\{r\in R: \exists\,n\in\mathbb{N}\,,r^n=0\right\}}\).

In our case, the ring \(\displaystyle{\mathbb{Z}_{n}}\) is Noehterian and Artinian.
User avatar
Grigorios Kostakos
Founder
Founder
Posts: 461
Joined: Mon Nov 09, 2015 1:36 am
Location: Ioannina, Greece

Re: Nilpotents of \(\mathbb{Z}_n\)

#3

Post by Grigorios Kostakos »

We will need the lemma: Let $m,\,n$ be positive integers such that $n\,\big|\,k^r$, for some positive integer $r$. The prime number $p$ is a factor of $n$ iff is a factor of $k$.
Proof: $\color{gray}{``\Rightarrow``}$
$$p\,\big|\,n\quad\Longrightarrow\quad p\,\big|\,k^{r}\quad\stackrel{p\,{\text{prime}}}{=\!=\!=\!\Longrightarrow}\quad p\,\big|\,k\,.$$
$\color{gray}{``\Leftarrow``}$ Let $k=p_1^{b_1}\,p_2^{b_2}\cdots{p}_{s}^{b_s}$ be the prime factorization of $k$. Then $k=p_1^{rb_1}\,p_2^{rb_2}\cdots{p}_{s}^{rb_s}$ and because $n\,\big|\,k^r$, we have that $n=p_1^{a_1}\,p_2^{a_2}\cdots{p}_{s}^{a_s}$, with $a_i\leqslant{r\,b}_i\,, \;i\in\{1,2,\ldots, s\}$. So, every prime divisor of $k$ is a divisor of $n$. $\small\blacksquare$

Let $n=p_1^{a_1}\,p_2^{a_2}\cdots{p}_{s}^{a_s}$ be the prime factorization of positive integer $n$. Then, by the lemma, we have that $k=p_1^{b_1}\,p_2^{b_2}\cdots{p}_{s}^{b_s}$ with $b_i\leqslant a_i\,, \;i\in\{1,2,\ldots, s\}$. Additively, if $b_i=a_i$ every $i\in\{1,2,\ldots, s\}$, then $k=n$.
Let $[k]_{n}$ is a (non-trivial) zero divisor of the ring $\mathbb{Z}_n$. Then exists $r\in\mathbb{N}\,,\; r\geqslant2$, such that
$$[k]_{n}^{r}=[0]_{n}\quad \Rightarrow\quad n\,\big|\,k^r\,.$$
Then, it must be $a_i\leqslant{r\,b}_i$ for every $i\in\{1,2,\ldots, s\}$. Finally, we have that there exists $r\in\mathbb{N}\,,\; r\geqslant2$, such that
\begin{align}
b_i\leqslant a_i\leqslant{r\,b}_i
\end{align} where $k=p_1^{b_1}\,p_2^{b_2}\cdots{p}_{s}^{b_s}$ and $n=p_1^{a_1}\,p_2^{a_2}\cdots{p}_{s}^{a_s}$ are the prime factorizations of $k$ and $n$.
In the special case that we have $a_i=1$ for every $i\in\{1,2,\ldots, s\}$, by $(1)$, we have that $b_i=1$ for every $i\in\{1,2,\ldots, s\}$. Therefor $[k]_{n}=[n]_{n}=[0]_{n}$ and there no zero-divisors in $\mathbb{Z}_n$.
Grigorios Kostakos
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 8 guests