Let $x\in X$ be a closed point in a normal variety $X/\mathbb C$, also assume $X$ is Cohen-Macaulay. Let $m_x$ be the maximal ideal of $x$. Suppose $f: Y \to X$ is a resolution of $m_x$ such that the strict transform $\bar H$ of a general hyperplane section $H$ of $X$ through $x$ is $f$-base point free. Then $m_x\mathcal{O}_{Y}=\mathcal{O}_{Y}(-E)$ and $f^*H = \bar H+E$. It is claimed that by Cohen-Macaulay property, $${\rm mult}_xX = (E \cdot \bar H^{d-1}),$$ where $\dim X=d$.
Could anyone advise me how this formula is obtained?
Here ${\rm mult}_xX$ is defined by the following. The function $P(t)=\dim_{\mathbb C}(\sum_{i=1}^t m^{i}_x/m_x^{i+1})$ is a polynomial of degree $d$ when $t \gg 1$. Then $\frac{{\rm mult}_x}{d!} t^d$ is its leading term.
But I got confused at this point, because according to Fulton's book " Introduction to intersection theory in algebraic geometry" (page 12, Lemma (b)), if $A(=\mathcal{O}_{X,x})$ is CM, and $f_1, \ldots, f_d$ is a regular system of the ideal $m(=m_x)$, then there exists an isomorphism $$A/m[X_1, \ldots, X_d] \to \oplus_{i=0}^{\infty} m^i/m^{i+1}$$ sending $X_k$ to $f_k \in m/m^2$. So in the above case, $$\oplus_{i=0}^{\infty} m^i/m^{i+1} \simeq \mathbb C[X_1, \ldots, X_d].$$ Hence the ${\rm mult}_xX = {\rm mult}_0 \mathbb A^d=1$ which is absurd. Where did I make wrong??