Characteristic of endomorphism ring of $\mathbb{Z}_n$.
- Tolaso J Kos
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Characteristic of endomorphism ring of $\mathbb{Z}_n$.
Evaluate the characteristic of endomorphism ring of $\mathbb{Z}_n, \; n \in \mathbb{N}$, that is:
$${\rm char}\left\{ {\rm End} \left(\mathbb{Z}_n\right) \right\}$$
$${\rm char}\left\{ {\rm End} \left(\mathbb{Z}_n\right) \right\}$$
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- Grigorios Kostakos
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Re: Characteristic of endomorphism ring of $\mathbb{Z}_n$.
Every endomorphism $f_r:\mathbb{Z}_n\longrightarrow\mathbb{Z}_n$ is of the form $$\mathbb{Z}_n\ni[k]_n\longmapsto f_r([k]_n)=\frac{n}{r}\,[k]_n\,,$$ where $r$ is a divisor of $n$. It is obvious that $$(\forall\, f_r\in {\rm{End}}(\mathbb{Z}_n))(\forall\, [k]_n\in \mathbb{Z}_n)\quad n\,f_r([k]_n)=n\,\frac{n}{r}\,[k]_n=[0]_n\,.$$ We claim that $n$ it is the smallest positive integer such that $n\,f_r=0$. Indeed, let's take the identity automorphism $f_n$ and suppose that there is a positive integer $q<n$ such that $$(\forall\, [k]_n\in \mathbb{Z}_n) \quad q\,f_n([k]_n)=[0]_n\,.$$ Then must $$q\,f_n([1]_n)=q\,\frac{n}{n}\,[1]_n=q\,[1]_n=[0]_n\quad\stackrel{(*)}{\Longrightarrow}\quad n\,\big|\, q\,.$$ A contradiction. So, $n$ is the smallest positive integer such that $(\forall\, f_r\in {\rm{End}}(\mathbb{Z}_n))\quad n\,f_r=0$ and ${\rm{char}}({\rm{End}}(\mathbb{Z}_n))=n$.
$(*)\quad \circ\!(a^k)=1\quad \Rightarrow \quad \circ(a)\,\big|\,k\,.$
$(*)\quad \circ\!(a^k)=1\quad \Rightarrow \quad \circ(a)\,\big|\,k\,.$
Grigorios Kostakos
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Re: Characteristic of endomorphism ring of $\mathbb{Z}_n$.
Second solution (almost similar).
Let \(\displaystyle{f\in\rm{End}_{\mathbb{Z}}(\mathbb{Z}_{n})}\). Then,
\(\displaystyle{(n\,f)(x)=n\,f(x)=0\,,\forall\,x\in\mathbb{Z}_{n}}\). So,
\(\displaystyle{n\,f=\mathbb{O}\,,\forall\,f\in\rm{End}_{\mathbb{Z}}(\mathbb{Z}_{n})}\).
Let \(\displaystyle{m\in\mathbb{N}}\) such that \(\displaystyle{m\,f=\mathbb{O}\,,\forall\,f\in\rm{End}_{\mathbb{Z}}(\mathbb{Z}_{n})}\).
Especially, for \(\displaystyle{f=Id_{\mathbb{Z}_{n}}}\) we get
\(\displaystyle{(m\,Id)(x)=0\,,\forall\,x\in\mathbb{Z}_{n}\iff m\,x=0\,,\forall\,x\in\mathbb{Z}_{n}}\).
Since \(\displaystyle{\rm{char}(\mathbb{Z}_{n})=n}\), we have that \(\displaystyle{m\leq n}\).
Therefore, \(\displaystyle{\rm{char}(\rm{End}_{\mathbb{Z}}(\mathbb{Z}_{n}))=n}\).
Let \(\displaystyle{f\in\rm{End}_{\mathbb{Z}}(\mathbb{Z}_{n})}\). Then,
\(\displaystyle{(n\,f)(x)=n\,f(x)=0\,,\forall\,x\in\mathbb{Z}_{n}}\). So,
\(\displaystyle{n\,f=\mathbb{O}\,,\forall\,f\in\rm{End}_{\mathbb{Z}}(\mathbb{Z}_{n})}\).
Let \(\displaystyle{m\in\mathbb{N}}\) such that \(\displaystyle{m\,f=\mathbb{O}\,,\forall\,f\in\rm{End}_{\mathbb{Z}}(\mathbb{Z}_{n})}\).
Especially, for \(\displaystyle{f=Id_{\mathbb{Z}_{n}}}\) we get
\(\displaystyle{(m\,Id)(x)=0\,,\forall\,x\in\mathbb{Z}_{n}\iff m\,x=0\,,\forall\,x\in\mathbb{Z}_{n}}\).
Since \(\displaystyle{\rm{char}(\mathbb{Z}_{n})=n}\), we have that \(\displaystyle{m\leq n}\).
Therefore, \(\displaystyle{\rm{char}(\rm{End}_{\mathbb{Z}}(\mathbb{Z}_{n}))=n}\).
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