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