characteristic of a ring

Jan 2010
69
3
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

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May 2008
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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"?
 
Jan 2010
69
3
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
May 2008
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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 \(\displaystyle char R_i=p_i^{r_i}.\)
 
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Jan 2010
69
3
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 \(\displaystyle p(>0)\)?
 
Jan 2010
69
3
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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