The intersection of prime ideals of $R$ containing $A$

Question

Let $R$ be a ring and let $A$ be an ideal of $R$.

The nilradical of $A$ is the intersection of the prime ideals of $R$ containing $A$.

Since, $A$ is contained in every prime ideal of $R$, does it imply that the intersection of the prime ideals of $R$ containing $A$ is equal to $A$ itself? If this is true, then it means that $\operatorname{nilrad}(A) = A$.

May you please give me a counterexample to show why this cannot happen?

Thank you in advance.

Answer 1

It is the radical of $A$ that is the intersection of the prime ideals containing $A$ not the nil-radical. The radical is denoted by $\sqrt A$. Always $A\subseteq\sqrt A$ but in general equality doesn't hold.

As an example, in $\Bbb Z$, $\sqrt{n\Bbb Z}=m\Bbb Z$ where $m$ is the product of the prime factors of $n$ (but taken once only).

Answer 2

Consider $A=(0)$ in the ring $R=\mathbb{C}[x]/(x^2)$.

Its nilradical is the prime ideal $(x)$ wich is different from $A$.

Answer 3

The set $\{a\}$ is contained in each of the sets $\{a,b,c\}$, $\{a,b,d\}$, $\{a,b,e\}$. Does that imply that the intersection of $\{a,b,c\}$, $\{a,b,d\}$, $\{a,b,e\}$ is equal to $\{a\}$?