Originally Posted by

**NonCommAlg** well, let $\displaystyle char R = n > 0$ and consider the prime factorization of n: $\displaystyle n=\prod_{i=1}^t p_i^{r_i}.$ let $\displaystyle I_i=p_i^{r_i}R, \ \ 1 \leq i \leq t.$ see that $\displaystyle I_i + I_j = R$ for $\displaystyle i \neq j$ and $\displaystyle I_1I_2 \cdots I_t = \{0\}.$

thus, by the Chinese remainder theorem for rings, we have $\displaystyle R \cong R/I_1 \times R/I_2 \times \cdots \times R/I_t.$ finally put $\displaystyle R_i = R/I_i$ and see that $\displaystyle char R_i=p_i^{r_i}.$