# Thread: characteristic of a ring

1. ## characteristic of a ring

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

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

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

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

the second part doesn't look correct to me! are you sure $\displaystyle char R_i$ 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 $\displaystyle char R = n > 0$ 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 $\displaystyle char R_i=p_i^{r_i}.$

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

6. 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 $\displaystyle char R_i=p_i^{r_i}.$
would you please show that $\displaystyle I_i + I_j = R$ for $\displaystyle i \neq j$ ? Also why is $\displaystyle p_i^{r_i}$ the least positive integer satisfying $\displaystyle p_i^{r_i}.1_{R_i}=0$ ?