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Math Help - Nilradical problem

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    Nilradical problem

    Suppose that R is a commutative ring, and let I be an ideal of R. Let  \sqrt I = \{ x \in R : x^n \in I \ , \ n \geq 1 \} , prove that  \frac { \sqrt {I} }{I}} is the nilradical of  \frac {R}{I} .

    So I need to prove that elements in  \frac { \sqrt {I} }{I} , say  ij, i \in \sqrt {I},j \in I , and we know that i^n \in I \ , \ n \geq 1 , that we would have  (ij)^m = 0 \ \ \ m \geq 1 . But what is the zero in  \frac {R}{I} ? Isn't it just 0?
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Suppose that R is a commutative ring, and let I be an ideal of R. Let  \sqrt I = \{ x \in R : x^n \in I \ , \ n \geq 1 \} , prove that  \frac{\sqrt {I}}{I} is the nilradical of  \frac {R}{I} .

    So I need to prove that elements in  \frac { \sqrt {I} }{I} , say  ij, i \in \sqrt {I},j \in I , and we know that i^n \in I \ , \ n \geq 1 , that we would have  (ij)^m = 0 \ \ \ m \geq 1 . But what is the zero in  \frac {R}{I} ? Isn't it just 0?
    r + I \in \text{Nil}(R/I) \Longleftrightarrow \exists n: \ r^n + I=0_{\frac{R}{I}}=I \Longleftrightarrow r^n \in I \Longleftrightarrow r \in \sqrt{I} \Longleftrightarrow r+I \in \frac{\sqrt{I}}{I}.
    Last edited by NonCommAlg; November 10th 2008 at 02:05 AM.
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