Assume $R$ is a local ring, $M\subseteq R$ is the maximal ideal, $I\subseteq R$ is an $M$-primary ideal and $I=\bigcap_{i=1}^n Q_i$ is a minimal irreducible decomposition of $I$ (i.e. $Q_i\subseteq R$ are irreducible ideals and the intersection of a proper subset of $\{Q_i\}$ can not equal $I$ (minimal property)).
Using the fact that for any $i$ the (vector space) dimension of $(Q_i:M)/Q_i$ over the field $R/M$ equals $1$ (because $Q_i$ are irreducible) I would like to prove that the dimension of $(I:M)/I$ over $R/M$ equals $n$. I ask (for a hint) for a proof of this fact.
My initial idea was to build a composition series of length $n$ from $I$ to $(I:M)$ but I failed to find the suitable ideals between $I$ and $(I:M)$.
This is proved in Section 3 of Computational methods in commutative algebra and algebraic geometry by W. V. Vasconcelos. From the 1998 edition of the book:
Proposition 3.1.7. Let $(R,\mathfrak{m})$ be a local commutative Noetherian ring. Let $\mathfrak{p}$ be an associated prime of an $R$-module $M$ and denote $\Delta_\mathfrak{p}(M)$ the submodule of $M$ whose elements are annihilated by $\mathfrak{p}$. The number of irreducible $\mathfrak{p}$-primary components in an irredundant irreducible decomposition of $0\subset M$ is $\dim_{k(\mathfrak{p})}(\Delta_\mathfrak{p}(M))_\mathfrak{p}$.
Your question is the case where $M=R/I$.
The main idea of the proof is to analyze what happens to the socle $\Delta_\mathfrak{p}(M)$ under the map $M \to \bigoplus_i M/M_i$ which is at least injective if $0 = \bigcap M_i$ is any decomposition of zero.
N.b. In the second edition of the book the numbering changed. I think it is Prop. 3.15 there, but I'm not sure.
A related question is here.