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Math Help - characteristic of a ring

  1. #1
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    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.
    Last edited by xixi; June 3rd 2010 at 05:31 AM.
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  2. #2
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    Quote Originally Posted by xixi View Post
    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"?
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    Quote Originally Posted by NonCommAlg View Post
    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 ?
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    Quote Originally Posted by xixi View Post
    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}.
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    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)?
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  6. #6
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    Quote Originally Posted by NonCommAlg View Post
    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 ?
    Last edited by xixi; June 3rd 2010 at 05:30 AM.
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