# characteristic of a ring

#### xixi

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 $$\displaystyle \forall i , 1\leq i \leq t$$.

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#### NonCommAlg

MHF Hall of Honor
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"?

#### xixi

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 ?

#### NonCommAlg

MHF Hall of Honor
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 $$\displaystyle char R_i=p_i^{r_i}.$$

• xixi

#### xixi

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 $$\displaystyle p(>0)$$?

#### xixi

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 $$\displaystyle p_i^{r_i}.1_{R_i}=0$$ ?

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