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

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?

• Why is $A$ contained in every prime ideal of $R$?? This is certainly not true in general... – SEWillB Jun 2 '17 at 17:33
• I don't understand why do you think that $A$ is contained in every prime ideal of $R$. – Xam Jun 2 '17 at 17:37
• Maybe I am wrong, I don't know. But this is how I understood the theorem "The nil radical of an ideal A is the intersection of the prime ideals of R containing A". I think of it as an intersection of all the primes of R such that A is included in each of them. – User1999 Jun 2 '17 at 17:50

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).

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

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

• Thank you very much for your example, but I have difficulties in understanding it, since I am not very familiar with this ring. – User1999 Jun 2 '17 at 17:53

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 $\{a\}$?

• Thank you very much! I got really confused while reading it. Of couse not, but {a} is included in the intersection. – User1999 Jun 2 '17 at 17:38