A nilpotent element in $\mathbb{Z}/ 96\mathbb{Z}$

Question

I have to find a nonzero nilpotent element in the ring $\mathbb{Z}/ 96\mathbb{Z}$.

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If we take $a\in \mathbb{Z}/ 96\mathbb{Z}$, $a$ is nilpotent if there exists $n\in\mathbb{Z}$ such that $a^n = 0$.

I have tried with a lot of different elements but I can't find it... Is there any trick to find a nilpotent element in $\mathbb{Z}/ 96\mathbb{Z}$ ?

Answer 1

Note that $96=32×3=2^5×3$. Note that $6^5=2^5×3^5$. So $6$ is a nulpotent element. In fact any $a$ that is a multiple of both $2$ and $3$ will do, such as $a=6$.

Can you generalize to $\mathbb{Z}/M\mathbb{Z}$; $M$ any integer that is not square-free?

Answer 2

Question: I have tried with a lot of different elements but I can't find it... Is there any trick to find a nilpotent element in $Z/96Z$?

Answer: Since $96=2^5 \times 3$ and since $(2^5,3)=1$, by the crt there is an isomorphism of rings

$$\mathbb{Z}/96\mathbb{Z} \cong B:=\mathbb{Z}/2^5\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z},$$

and the element $0 \neq (2,0)\in B$ satisfies $(2,0)^5=(2^5,0)=(0,0)$. Similarly $(2^i,0)$ for $i=1,2,3,4$ is also nonzero and nilpotent.

Note: If $\mathfrak{m}_1:=((2),(1)), \mathfrak{m}_2:=((1),(0))$ it follows $\mathfrak{m}_i$ are the maximal ideals of $B$ and $\mathfrak{m}_1 \cap \mathfrak{m}_2=\mathfrak{m}_1\mathfrak{m}_2=((2),0)$ is the nilradical, which is nontrivial.

In general if $n:=p_1^{l_1}\cdots p_d^{l_d}$ with $p_i\neq p_j$ primes for all $i\neq j$, and $l_i\geq 2$ for some $i$, the ring $R:=\mathbb{Z}/n\mathbb{Z}$ will have non-trivial nilpotent elements. The nilradical in $R$ will be the ideal

$$ nil(R):=((p_1),..,(p_d))$$

in the crl-decomposition

$$R \cong \mathbb{Z}/(p_1^{l_1}) \mathbb{Z} \oplus \cdots \oplus \mathbb{Z}/(p_d^{l_d}) \mathbb{Z}$$

and $nil(R)$ will be nontrivial if $l_i\geq 2$ for some $i$. If $l_i=1$ for all $i$, it follows $R$ is a product of fields which is a reduced ring.