Characteristic of Quotient Ring

Question

Consider a ring $R$ with characteristic $n \gt 0$. Let $I$ be an ideal of $R$ with characteristic $m$. I have proved that $m$ divides $n$. Now I am interested about the characteristic of $R/I$. One of my friends claim that char$(R/I)=n/m$. Is this true? If so, how can this be proved?

Answer 1

No, this is false. Let $R$ be any ring. Then consider $S=R\times R$, let $I$ be $R\times\{0\}$ in $S$. Note that $S$ is an ideal since $(a,b)\cdot(r,0)=(ar,0)\in I$. Also note that if $R$ has characteristic $n$, then $S$ also has characteristic $n$. Note that $R$ is isomorphic to $I$, so $I$ also has characteristic $n$. However $S/I$ does not have characteristic 1 since it is clearly not the zero ring, and in fact is isomorphic to $R$ (since each congruence class in $S/I$ has a natural representative of the form $(0,s)$ and these are all distinct congruence classes) so has characteristic $n$.

However this is true if $I$ is the largest ideal with characteristic $m$. Let $R$ be a ring of characteristic $n$. Let $I_m=\{r\in R: mr=0\}$. Then $I_m$ is a two sided ideal of $R$ since $m (rs)= (mr)s=0s=0=s0=s(mr)=m(sr)$ by the distributive laws. Now consider $R/I_m$. Let $n=mk$, so that $n/m=k$ and I don't have to write $n/m$. Suppose $a + I_m \in R/I_m$. Then $k(a+I_m)=ka+I_m$. In order to show that this is $0$ in $R/I_m$, we just need to show that $ka\in I_m$. However $m(ka)=mka=na=0$ since $a\in R$ which has characteristic $n$. Therefore by definition of $I_m$, $ka\in I_m$ and $k(a+I_m)=0+I_m$. Since this is true for all $k$, $R/I_m$ has characteristic at most $k$. Now we just need to show that there is some element of $R/I_m$ of additive order $k$. However since $R$ has characteristic $n$, there is some element $r$ of $R$ such that $nr=0$, and $n$ is the least such integer. Then suppose $l(r+I_m)=0$ for some $l < k$, then $lr\in I_m$, which means that $mlr =0$ in $R$, but since $l<k$, $ml<mk=n$, contradicting the choice of $r$. Therefore $R/I_m$ has characteristic $k=n/m$.