Nilpotents of \(\mathbb{Z}_n\)
- Grigorios Kostakos
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Nilpotents of \(\mathbb{Z}_n\)
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
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Re: Nilpotents of \(\mathbb{Z}_n\)
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.
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.
- Grigorios Kostakos
- Founder
- Posts: 461
- Joined: Mon Nov 09, 2015 1:36 am
- Location: Ioannina, Greece
Re: Nilpotents of \(\mathbb{Z}_n\)
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 $\mathbb{Z}_n$.
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
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