What does it mean if $V(I)\subset\operatorname{Spec} R$ doesn't contain any irreducible component of $\operatorname{Spec} R$?

$R$ is a commutative ring with unity, $I\subset R$ is an ideal. Denote with $V(I)$ the subset of $\operatorname{Spec}R$ corresponding to the prime ideals $p\subset R$ such that $I\subseteq p$. What does it mean that $V(I)$ doesn't contain any irreducible components of $\operatorname{Spec}R$? If the irreducible components are the sets of the form $V(p)$ for a prime ideal $p\subset R$, I thought that any $V(I)$ contained at least $V(m)$, where $m\subset R$ is a maximal ideal containing $I$, whose existence is guaranteed by Zorn's lemma.

Solution 1:

You need to distinguish between "irreducible subset" and "irreducible component". Irreducible subsets are indeed those of the form $V(p)$ for $p$ a prime ideal, while an irreducible component is a maximal irreducible subset.

As an example think about the union of the axes $\operatorname{Spec}k[x,y]/(xy)\subseteq\mathbb A^2$. This has two irreducible components $V(x)$ and $V(y)$, corresponding to the two axes, but an irreducible subset like $\{(0,1)\}=V(x,y-1)$ is not an irreducible component (nor does it contain one) because it's not maximal.