Recently I encountered a problem (the first exercise from chapter four of Atiyah & McDonald's Introduction to Commutative Algebra) stating that if $\mathfrak{a}$ is a decomposable ideal of $A$ (a commutative ring with unity), then the prime spectrum of $A/ \mathfrak{a}$ has finitely many irreducible components. This follows easily from the recognition that the maximal irreducible subspaces of $\textrm{ Spec } (A / \mathfrak{a})$ are precisely the "zero loci" of the minimal prime ideals of $A /\mathfrak{a}$.
I'm curious about the converse - the proof isn't easily reversed since the notion of minimal ideals of $\mathfrak{a} $ doesn't make sense before we know what we are trying to prove. My intuition says that it is false based on the general premise that images are badly-behaved (and the fact that it isn't part of the exercise.) However, I've had some difficulty constructing a counterexample, so that the main purpose of this post is to ask for a reasonable procedure or heuristic for doing so (or, of course, proof that my intuition is false.)
If it helps, if $\textrm{Spec }(A / \mathfrak{a})$ is irreducible then the nilradical $\mathcal{R}_{A /\mathfrak{a}}$ is prime so that $r(\mathfrak{a}) = \rho^{-1} ( \mathcal{R}_{A / \mathfrak{a}} ) = \displaystyle\cap_{i=1}^n \rho^{-1} (p_i),$ where $p_i$ are the minimal prime ideals of $A /\mathfrak{a}$ and $\rho $ is the associated projection, is also prime. Thanks!