$R$ is a ring with $\operatorname{char}(R)=mn$ where $(m,n)=1$, then there exists an ideal $A$ such that $\operatorname{char}(A)=m$?

Question

whole question :

Let $R$ be a ring with characteristic $mn$ for some positive integer $m,n$. If $(m,n)=1$, show that there exists an ideal $A$ (resp. $B$) of $R$ with characteristic $m$ (resp. $n$). In fact, $R=A+B$

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My opinion

There exists some element "$c$" satisfying order of "$c$" that is $mn$. Then consider $A=(n*c)+<n*c>$.

I think $A$ is an ideal and characteristic $m$. Likewise $B=(m*c)+<m*c>$.

But how to show that $R=A+B$?

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Please tell the proof! :)

Answer 1

Hint: For $R=A+B$, take $u,v \in \mathbb Z$ such that $1=um+vn$. Then $r=umr+vnr$ for all $r \in R$.