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!
Best Answer
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 :)