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

So I need to prove that elements in $\displaystyle \frac { \sqrt {I} }{I} $, say $\displaystyle ij, i \in \sqrt {I},j \in I $, and we know that $\displaystyle i^n \in I \ , \ n \geq 1 $, that we would have $\displaystyle (ij)^m = 0 \ \ \ m \geq 1 $. But what is the zero in $\displaystyle \frac {R}{I} $? Isn't it just 0?