Let $M'\subset M$ be a submodule and $M'=\cap_{i=1}^n M_i$ with $M_i$ a $P_i$-primary submodule.
From Theorem 3.10 in Eisenbud: The decomposition above is irredundant if no $M_i$ can be dropped. The decomposition above is minimal if there is no such intersection with fewer terms.
What's the difference between the two? The first says that for any $i=1,\dots,n$, $\cap_{j\in\{1,\dots,n\}-\{i\}}M_i$ properly contains $M'$, right? What does it mean to be minimal? I.e., what does it mean "there is no such intersection with fewer terms"? The only interpretation I can think of is: if there is such an intersection with fewer terms, then it is not equal to $M'$. But this coincides with the notion of being irreducible, doesn't it?
What minimal means is that $n$ is as small as possible. In other words, there does not exist any $m<n$ and primary submodules $N_1,\dots,N_m\subseteq M$ such that $M'=\bigcap_{i=1}^m N_i$. So, a minimal decomposition is irredundant, because if we could drop one of the $M_i$, that would give us a decomposition with $n-1$ terms instead of $n$ terms. But an irredundant decomposition might not be minimal, because we might be able to find a decomposition with less than $n$ terms which uses totally different submodules, instead of just dropping one of the $M_i$.