Nilradical problem

Quote:

Originally Posted by tttcomrader

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?

$r + I \in \text{Nil}(R/I) \Longleftrightarrow \exists n: \ r^n + I=0_{\frac{R}{I}}=I \Longleftrightarrow r^n \in I \Longleftrightarrow r \in \sqrt{I} \Longleftrightarrow r+I \in \frac{\sqrt{I}}{I}.$