Characteristic of a simple ring is either prime or $0$ 
If $R$ is a simple ring, prove that characteristic of $R$ is either prime or $0$.

Note: in the question the presence of unity is not mentioned, can we do without it?
 A: If the characteristic is $mn$, where $(m,n)=1$, then by the Chinese Remainder Theorem,
$$R\simeq R\otimes_{\mathbf Z}\mathbf Z/mn\mathbf Z \simeq (R\otimes_{\mathbf Z}\mathbf Z/m\mathbf Z) \oplus (R\otimes_{\mathbf Z}\mathbf Z/n\mathbf Z).$$
A: To start, I'm going to assume $R$ has a unit element $1$ such that $1r = r1 = r$ for all $r \in R$.
This being the case, let $n \in \Bbb Z_+$, the positive integers, be the smallest such that
$1 + 1 + . . . + 1 \, (n \, \text{times}) = 0.\tag{1}$
To avoid trivialities, we assume $n > 1$; otherwise we have $1 = 0$ hence, for all $r \in R$, $r = r1 = 0$, and the ring $R$ collapses to $\{0\}$.  If $n$ is composite, say
$n = ab$ where $1 < a, b < n$, then consider the principal ideal $aR = \{ ar, \, r \in R \}$, where by $a$ we mean $1 + 1 + . . . + 1 \, (a \, \text{times})$.  Since $ar = ra$ for all $r \in R$, this ideal is two-sided.  Furthermore, $a = a1 \in aR$; we have $a \ne 0$ by the choice of $n$ as the smallest integer number of ones which, when added together, yield $0$; since $0 \ne a \in aR$, the ideal $aR \ne 0$.  Also, $1 \notin aR$, since otherwise we would have $1 = as$ for some $s \in R$; then $b =bas = ns = 0$; but this possibility too contradicts the choice of $n$ as smallest such.  Thus
$aR$ would be a two-sided proper ideal of $R$, contradicting the simplicity of $R$.  This shows that the characteristic of $R$, if not zero, is prime.
In the event that $R$ does not have a unit element, I think we still may be able to define its characteristic as $n$ by requiring that
$r + r + . . . + r \, (n \text{times}) = 0 \tag{2}$
for all $r \in R$.  If we adopt the notation $mr = r + r + . . . + r \, (m \text{times})$ for all $m \in \Bbb Z_+, r \in R$, then we can still consider sets of the form $mR$ consisting of all elements of the form $mr$.  It is easy to see this set is an ideal, two-sided, in $R$, since $mr - ms = m(r - s)$ and $s(mr) = m(sr) \in R$ as well as $(mr)s = m(rs) \in mR$.  (I leave the easy details of the verifications of these assertions to the reader.)  If, as above, $n = ab$ is composite, then the ideal $aR$ is not the zero ideal, since there must be some $r \in R$ such that $ar \ne 0$.  Likewise, $aR \ne R$, for if this were the case, then for every $s \in R$ we would have $s = ar$ for some $r \in R$; then $bs = b(ar) = (ba)r = 0 \,$ for every $s \in R$, contradicting the choice of $n$ as the smallest positive integer having this property.  So $aR$ is once again a proper, two-sided ideal of $R$, contradicting its simplicity as a ring.
I guess the proposition can be made to fly either way, whether $R$ is unital or not! QED!
It is important to remember, in reading the above, to distinguish between $mr$, where $m \in \Bbb Z_+, r \in R$, and $rs$ with $r, s \in R$.  The former always refers to $m$ copies of $r$ added together; the latter is the product $rs$ for $r, s \in R$.  Hopefully the context makes these two usages clear when they occur.
Hope this helps!  Cheerio,
and as always,
Fiat Lux!!!
A: The characteristic of $R$ is the least positive integer $k$ such that $kr=0$, for all $r\in R$, or the characteristic is $0$ if such an integer doesn't exist.
For an integer $m>0$, the set $mR=\{mr:r\in R\}$ is clearly an ideal of $R$. Since $R$ is simple, either $mR=\{0\}$ or $mR=R$. If $k>0$ is the characteristic of $R$, then, by definition, $kR=\{0\}$ and $mR\ne\{0\}$, for $0<m<k$.
Let $k=mn$ be composite, with $m>1$ and $n>1$. Then $mR\ne\{0\}$ and $nR\ne\{0\}$, so $nR=R$; therefore $kR=mnR=m(nR)=mR\ne\{0\}$, a contradiction.
