I've been thinking about the nilradical and I am wondering if the nilradical is the smallest, non-zero ideal of the ring.

The reason why I'm asking is the following:

Every ideal contains $0$. If $x \in R$ is nilpotent this implies that $0 = x^n$ is in every ideal and therefore, by an induction argument ($x \cdot x^{n-1}$), so is $x$. Therefore the nilradical is a subset of the intersection of all non-zero ideals. Therefore it is the smallest ideal in the ring.

What I'm not so sure about though is what to do about $0$. I could take the intersection over all ideals, including $(0)$ and the nilradical would still be a subset. Which would be a (wrong) proof that the nilradical is the zero ideal... Many thanks for your help!

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Let $A$ be a commutative ring. Your argument proves that the nilradical of $A$ is contained in the intersection of all prime ideals of $A$. In fact, the converse holds (but is less trivial): the nilradical of $A$ is equal to the intersection of all prime ideals of $A$. However, the nilradical of the ring $A=\mathbb{Z}/(8\mathbb{Z})$ is not the smallest non-zero ideal of $A$. Indeed, the nilradical of $A$ is $(2\mathbb{Z})/(8\mathbb{Z})$ and the non-zero ideal $(4\mathbb{Z})/(8\mathbb{Z})$ is strictly contained in $(2\mathbb{Z})/(8\mathbb{Z})$. – Amitesh Datta Oct 6 '11 at 11:43

It is false that any nilpotent element of a ring is contained in every ideal. For example, in the ring $R[x,y]/(x^n)$ where $R$ is any ring and $n>1$, the element $x^n=0$ is every ideal but $x^r$ is not in any ideal of the form $(g)$ for $g\in R[y]$ when $r<n$, for example. To be even more concrete, consider the example of $\mathbb{Z}[x]/(x^2)$, in which $x^2=0$ but $x$ is not in the ideal generated by $2$, for example.

Also, "smallest non-zero ideal of the ring" doesn't make sense for most rings. For example, in $\mathbb{Z}$, there is no minimal non-zero ideal; given any non-zero ideal $(n)$, the ideal $(2n)$ is smaller, and still non-zero. So there are no minimal non-zero ideals in $\mathbb{Z}$. In general, the poset of non-zero ideals may have many minimal elements: if $n\in\mathbb{Z}$ is a product of $k$ distinct prime numbers $q_1,\ldots,q_k$, then $\mathbb{Z}/n\mathbb{Z}$ has $k$ distinct minimal non-zero ideals, the $a_i\mathbb{Z}/n\mathbb{Z}$ where $a_i=q_1q_2\cdots q_{i-1}q_{i+1}\cdots q_k$.

What is true is that the nilradical of $R$ is the intersection of all the prime ideals of the ring $R$ (and this includes the zero ideal if $R$ is an integral domain).

This is because, if $P\subset R$ is a prime ideal, then $x^n=0\in P$ does imply that $x\in P$ (by the definition of prime ideal), so $$x\in\text{nil}(R)\implies x\in\bigcap_{P\subset R \text{ prime} }P,$$ and if $x\notin\text{nil}(R)$, then the collection $\Sigma$ of ideals of $R$ not containing $1,x,x^2,\ldots$ is a partially ordered set under inclusion, and it has some maximal element $M$ (using Zorn's lemma). This maximal element $M$ must be a prime ideal, because if $a\notin M$ and $b\notin M$, then $M+(a)$ and $M+(b)$ are both ideals of $R$ strictly containing $M$, hence containing powers of $x$ (because $M$ is maximal among ideals not containing powers of $x$). Thus the ideal $M+(ab)\supseteq (M+(a))(M+(b))$ contains a power of $x$, hence $M+(ab)$ strictly contains $M$, hence $ab\notin M$. Thus we have shown that $a,b\notin M\implies ab\notin M$, so $M$ is a prime ideal not containing $x$, and therefore we have shown $$x\notin\text{nil}(R)\implies x\notin\bigcap_{P\subset R \text{ prime} }P.$$ Therefore $$\text{nil}(R)=\bigcap_{P\subset R \text{ prime} }P.$$

Amitesh's exercises are also excellent, as usual :)

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Oh, I see! The induction argument only works if $I$ is prime! Thanks! – Rudy the Reindeer Oct 6 '11 at 15:32
Can I ask you one more question? Is $M + (ab) = (M + (a))(M + (b))$? i.e. is it also true that $(M + (a))(M + (b)) \subset M + (ab)$? – Rudy the Reindeer Oct 6 '11 at 15:37
@Matt Yes, it is true since $(M+(a))(M+(b))\subseteq MM+(a)M+M(b)+(a)(b)\subseteq M+(ab)$ (where we have used the assumption that $M$ is an ideal of $R$). – Amitesh Datta Oct 6 '11 at 22:26
Thank you, @AmiteshDatta ! – Rudy the Reindeer Oct 7 '11 at 15:03

Zev's answer is excellent as usual. The following exercises serve to supplement Zev's answer and my comment above.

The following theorem is very important; it is proven in Zev's answer above:

Theorem Let $A$ be a commutative ring. The nilradical of $A$ is equal to the intersection of all prime ideals of $A$.

You should make sure that you understand this theorem and its proof (given in Zev's answer above) before attempting the following exercises.

Exercise 1: Let $p,q$ be prime ideals of a commutative ring $A$. If $p\cap q$ is a prime ideal of $A$, then prove that either $p\subseteq q$ or $q\subseteq p$. More generally, prove that if $p_1,\dots,p_n$ are prime ideals of $A$ and if $p_1\cap\cdots\cap p_n\subseteq P$, then $p_i\subseteq P$ for some $1\leq i\leq n$.

Exercise 2: Let $A$ be a commutative ring. Let $\{p_i\}_{i\in I}$ be a family of prime ideals of $A$ totally ordered by inclusion. Prove that $\bigcap_{i\in I} p_i$ is a prime ideal of $A$.

Exercise 3: Let $A$ be a commutative ring. A prime ideal $p$ of $A$ is said to be a minimal prime ideal of $A$ if there exists no prime ideal $q$ of $A$ strictly contained in $p$. Prove that if $P$ is a prime ideal of $A$, then there exists a minimal prime ideal $Q$ of $A$ contained in $P$. (Hint: use Zorn's lemma and Exercise 2.)

Exercise 4: If $A$ is each of the following commutative rings, then determine the minimal prime ideal(s) of $A$:

(a) $A$ is an integral domain;

(b) $A=\mathbb{Z}/(n\mathbb{Z})$ for some positive integer $n$;

(c) $A=k[x]/(f(x))$ for some non-constant polynomial $f(x)$ where $k$ is a field.

Exercise 5 (Challenge): Let $A$ be a Noetherian ring. Prove that $A$ has finitely many minimal prime ideals. (Hint: it suffices to prove that the nilradical of $A$ is the intersection of finitely many prime ideals of $A$ by Exercise 1. Assume, for a contradiction, that this is not the case and note that the set of ideals $I$ of $A$ that are not the intersection of finitely many prime ideals of $A$ is non-empty. Since $A$ is a Noetherian ring, it follows that there is an ideal $I$ of $A$ maximal with respect to the property of not being an intersection of finitely many prime ideals of $A$. Derive a contradiction.)

I hope this helps!

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Thank you! These exercises are useful! – Rudy the Reindeer Oct 6 '11 at 15:26
The first exercise is not what I think you want. You want $P$ to be prime and the ideals in the intersection to be just ideals. – Pedro Tamaroff Aug 29 '14 at 8:32