Let $R$ be a commutative ring with unity, and let $\{ \mathfrak{p}_i \}_I$ be an indexing of all of the prime ideals in $R$.
Then, it is known that $\bigcap_I \mathfrak{p}_i$ is equal to $N$, the nilradical of $R$.
From a first course in number fields, I know that the intersection of ideals can sometimes be considered as the $\text{lcm}$ of the ideals in question. Thus, this naturally seems to suggest the question: what is the product of every prime ideal in $R$?
Define:
$$ \prod_I \mathfrak{p}_i = \bigcap_{J \subseteq I \text{ finite}}\bigg(\prod_J \mathfrak{p}_j \bigg)$$
Is there any similarly appealing characterisation of this ideal, as there is for the intersection of all prime ideals?
In general, it holds that $\prod_{i=1}^n \mathfrak{a}_i \subseteq \mathfrak{a}_1 \cap \cdots \cap \mathfrak{a}_n$ with equality, for example, if the $(\mathfrak{a}_i)$ are pairwise coprime. In any case, we clearly have:
$$\prod_{I} \mathfrak{p}_i \subseteq N$$
Thus, the two notions will certainly not disagree in any integral domain.
The prime ideals of $R$ clearly cannot be coprime if $R$ is a local ring, so it seems sensible to try look at local rings for clues. However, the local rings I have met only have one prime ideal -- so we trivially have equality -- and I have only just started studying localisation, so I cannot construct other examples myself for now.
Here is an example where the two notions disagree: Let $k$ be a field. If $R=k[x,y]/(x^2)$ then its nilradical is $(x)$. However $(x,y-a)$ is a prime ideal for all $a\in k$, so the product of prime ideals is zero.