Problems on Characteristic of a Ring

Prove that the characteristic of a finite ring R \mathcal{R} is finite and it is a divisor of R |\mathcal{R}| where R |\mathcal{R}| denotes the number of elements of R \mathcal{R}.
Solution:
Let (R,+,)(\mathcal{R}, +, \cdot) is finite ring.
First Part\colorbox{yellow}{First Part}
To prove, Char R\mathcal{R} is finite.\\
Since, by definition, the characteristic of a ring is the least positive integer nn (if exists) such that nα=0  αRn\cdot\alpha=0 ~\forall~ \alpha\in \mathcal{R} where 00 is the zero element of R.\mathcal{R}.\\ If no such positive integer nn exists then Char R=0.\mathcal{R}=0.\\
Therefore, in both case, the characteristic of a ring is finite whether the ring is finite or not.
Second Part\colorbox{yellow}{Second Part}
To prove, Char R\mathcal{R} is a divisor of R.|\mathcal{R}|.\\
Let R=m|\mathcal{R}|=m and Char R=n\mathcal{R}=n\\ Then we need to prove that nmn|m\\ Since (R,+,)(\mathcal{R}, +, \cdot) is ring then (R,+)(\mathcal{R}, +) is a abelian group.\\ Therefore, the Lagrange’s theorem mα=0  αR.m\cdot\alpha=0 ~\forall~ \alpha\in \mathcal{R}.\\ By division algorithm, \exists two integers qq and rr such that m=nq+r where 0r<n    mα=(nq+r)α  αR    0=n(qα)+rα  αR.    n0+rα=0  αR.    rα=0  αR.m=nq+r where 0r<n    mα=(nq+r)α  αR    0=n(qα)+rα  αR.    n0+rα=0  αR.    rα=0  αR.m=nq+r where 0r<n    mα=(nq+r)α  αR    0=n(qα)+rα  αR.    n0+rα=0  αR.    rα=0  αR.\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<nr \text{\textless} n and nn is the least positive integer such that nα=0  αR n\cdot\alpha=0 ~\forall~ \alpha\in \mathcal{R} . Therefore the only possibility is r=0r =0.
Hence m=nq    nm m=nq \implies n|m.