Problems on Characteristic of a Ring

Solution:

Let $(\mathcal{R}, +, \cdot)$ is finite ring.

$\colorbox{yellow}{First Part}$

To prove, Char $\mathcal{R}$ is finite.$\\$

Since, by definition, the characteristic of a ring is the least positive integer $n$ (if exists) such that $$n\cdot\alpha=0 ~\forall~ \alpha\in \mathcal{R}$$ where $0$ is the zero element of $\mathcal{R}.\\$ If no such positive integer $n$ exists then Char $\mathcal{R}=0.\\$

Therefore, in both case, the characteristic of a ring is finite whether the ring is finite or not.

$\colorbox{yellow}{Second Part}$

To prove, Char $\mathcal{R}$ is a divisor of $|\mathcal{R}|.\\$

Let $|\mathcal{R}|=m$ and Char $\mathcal{R}=n\\$ Then we need to prove that $n|m\\$ Since $(\mathcal{R}, +, \cdot)$ is ring then $(\mathcal{R}, +)$ is a abelian group.$\\$ Therefore, the Lagrange’s theorem $m\cdot\alpha=0 ~\forall~ \alpha\in \mathcal{R}.\\$ By division algorithm, $\exists$ two integers $q$ and $r$ such that $$\begin{array}{rl}{\displaystyle }& {\displaystyle m=nq+rwhere0\le r\text{<}n}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle m\cdot \alpha =(nq+r)\cdot \alpha \mathrm{\forall }\alpha \in \mathcal{R}}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle 0=n\cdot (q\cdot \alpha )+r\cdot \alpha \mathrm{\forall }\alpha \in \mathcal{R}\mathrm{.}}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle n\cdot 0+r\cdot \alpha =0\mathrm{\forall }\alpha \in \mathcal{R}\mathrm{.}}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle r\cdot \alpha =0\mathrm{\forall }\alpha \in \mathcal{R}\mathrm{.}}\end{array}$$$$\begin{array}{rl}{\displaystyle }& {\displaystyle m=nq+rwhere0\le r\text{<}n}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle m\cdot \alpha =(nq+r)\cdot \alpha \mathrm{\forall }\alpha \in \mathcal{R}}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle 0=n\cdot (q\cdot \alpha )+r\cdot \alpha \mathrm{\forall }\alpha \in \mathcal{R}\mathrm{.}}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle n\cdot 0+r\cdot \alpha =0\mathrm{\forall }\alpha \in \mathcal{R}\mathrm{.}}\\ {\displaystyle \hspace{0.25em} ⟹\hspace{0.25em} }& {\displaystyle r\cdot \alpha =0\mathrm{\forall }\alpha \in \mathcal{R}\mathrm{.}}\end{array}$$$$\begin{align*} & m=nq+r~where~ 0 \le r \text{\textless} n \\ \implies & m\cdot\alpha=(nq+r)\cdot\alpha~\forall~ \alpha\in \mathcal{R} \\ \implies & 0=n\cdot(q\cdot\alpha)+r\cdot\alpha~\forall~ \alpha\in \mathcal{R}. \\ \implies & n\cdot 0+r\cdot\alpha=0~\forall~ \alpha\in \mathcal{R}. \\ \implies & r\cdot\alpha=0 ~\forall~ \alpha\in \mathcal{R}. \end{align*}$$ But $r \text{\textless} n$ and $n$ is the least positive integer such that $n\cdot\alpha=0 ~\forall~ \alpha\in \mathcal{R}$. Therefore the only possibility is $r =0$.

Hence $$m=nq \implies n|m$$.