# Prove that $\text{nil}(\overline{R})=\sqrt{A}/A$

Let $R$ be a commutative ring and $A \lt R$ an ideal. Define the radical of an ideal $A$ to be $\sqrt{A}:=\lbrace x \in R \mid x^n \in A \text{ for some } x \in \mathbb{Z}^+ \rbrace$. Let ${}^{-}:R \rightarrow R/A$ be the canonical ring homomorphism. Show that $\text{nil}(\overline{R})=\sqrt{A}/A$.

I am a little confused by the notation $\text{nil}(\overline{R})$. Does it mean the set $\lbrace r+A\mid r\in R, (r+A)^n=A \rbrace$? If so, how do I show the claim? I have proved that $\sqrt{A}$ is the intersection of all prime ideals containing $A$ but I am not sure how to use it in this proof.

## 1 Answer

Let $\mathfrak{N}_A$ denote the nilradical of a ring $A$.

We can show this directly by verifying the equality.

Take $x \in \mathfrak{N}_\overline{R}$. Then by definition, $x^n = 0$ in $R/A$ so $x^n \in A$.

Conversely, take $x \in \sqrt{A}/A$. By definition of radical, this implies $x^n \in A$ which as we just saw, implies $x \in \mathfrak{N}$.

You should note that there is a very nice characterization of both radical and nilradical that gives us this result, and other results, much easier.

Proposition: Let $\mathfrak{a}$ be an ideal in a ring $A$. Then $$\sqrt{\mathfrak{a}} = \bigcap_{\mathfrak{a} \subseteq p} p$$

Namely, that the radical is the intersection of all prime ideals containing the ideal. Then in particular, the nilradical is the radical of the $0$ ideal so we are asking about the intersection of all prime ideals in $R/A$ which are precisely the intersection of all prime ideals in $R$ containing $A$. Now apply the Lattice Isomorphism Theorem.