Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
3  
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

2 Answers 2

up vote 6 down vote accepted

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 :)

share|improve this answer
    
Oh, I see! The induction argument only works if $I$ is prime! Thanks! –  Matt N. 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 $(M + (a))(M + (b)) \subset M + (ab)$? –  Matt N. Oct 6 '11 at 15:37
2  
@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
1  
Thank you, @AmiteshDatta ! –  Matt N. 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!

share|improve this answer
1  
Thank you! These exercises are useful! –  Matt N. 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 at 8:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.