# Thread: characteristic of a ring

1. ## characteristic of a ring

Let $R$ be a commutative ring . If $\mathbb{Z}$ isn't a prime subring of $R$ then prove that $R$ has positive characteristic and therefore the ring $R$ can be written as a product $R=R_1\times...\times R_t$ where $char R_i$ is a prime integer for $\forall i , 1\leq i \leq t$.

2. Originally Posted by xixi
Let $R$ be a ring . If $\mathbb{Z}$ isn't a prime subring of $R$ then prove that $R$ has positive characteristic and therefore the ring $R$ can be written as a product $R=R_1\times...\times R_t$ where $char R_i$ is a prime integer for $\forall i , 1\leq i \leq t$.
for the first part define the map $f: \mathbb{Z} \longrightarrow R$ by $f(n)=n1_R.$ then $f$ is not injective, since $\mathbb{Z}$ is not a subring of $R.$ thus $\ker f \neq \{0\}$, i.e. there exists some $n>0$ such that $n1_R=0.$

the second part doesn't look correct to me! are you sure $char R_i$ shouldn't be a "prime power" instead of "prime"?

3. Originally Posted by NonCommAlg
for the first part define the map $f: \mathbb{Z} \longrightarrow R$ by $f(n)=n1_R.$ then $f$ is not injective, since $\mathbb{Z}$ is not a subring of $R.$ thus $\ker f \neq \{0\}$, i.e. there exists some $n>0$ such that $n1_R=0.$

the second part doesn't look correct to me! are you sure $char R_i$ shouldn't be a "prime power" instead of "prime"?
Thank you very much for the first part , about the second part ; there is written prime integer in the statement but how can you solve it if it was "prime power" instead ?

4. Originally Posted by xixi
Thank you very much for the first part , about the second part ; there is written prime integer in the statement but how can you solve it if it was "prime power" instead ?
well, let $char R = n > 0$ and consider the prime factorization of n: $n=\prod_{i=1}^t p_i^{r_i}.$ let $I_i=p_i^{r_i}R, \ \ 1 \leq i \leq t.$ see that $I_i + I_j = R$ for $i \neq j$ and $I_1I_2 \cdots I_t = \{0\}.$

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

5. Thank you very much , I have another question ; now by this assertion can we assume that without loss of generality $R$ has a prime characteristic , say $p(>0)$?

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

thus, by the Chinese remainder theorem for rings, we have $R \cong R/I_1 \times R/I_2 \times \cdots \times R/I_t.$ finally put $R_i = R/I_i$ and see that $char R_i=p_i^{r_i}.$
would you please show that $I_i + I_j = R$ for $i \neq j$ ? Also why is $p_i^{r_i}$ the least positive integer satisfying $p_i^{r_i}.1_{R_i}=0$ ?