A sharp dimension bound makes equations regular on an equidimensional Cohen-Macaulay fibre

If $A=0$ or $V(f_1,\dots ,f_i)=\mathrm{\varnothing }$, the assertion has no primes to test. Otherwise choose a maximal ideal $\mathfrak{n}$ containing all the $f_j$ and set $C=A_{\mathfrak{n}}$ and $J=(f_1,\dots ,f_i)C$. By [F1], $\dim C=d$, and by [F2], $C$ is Cohen–Macaulay. The nonzero local quotient $C\mathrm{/}J$ has dimension $e\le \dim A\mathrm{/}(f_1,\dots ,f_i)\le d-i$, where the first inequality is localization's dimension bound.

Choose a system of parameters ${\bar{y}}_1,\dots ,{\bar{y}}_e$ for $C\mathrm{/}J$ by [F3] and lift it to $y_1,\dots ,y_e\in C$. Then $(f_1,\dots ,f_i,y_1,\dots ,y_e)C$ is maximal-primary. It has $i+e\le d$ generators. The height theorem in [F3] forces $d\le i+e$, so $e=d-i$ and this tuple of $d$ elements is a system of parameters of $C$. By [F2] it is regular, hence its initial segment $f_1,\dots ,f_i$ is regular in $C$.

Every prime $\mathfrak{q}\supseteq (f_1,\dots ,f_i)$ lies in some maximal $\mathfrak{n}$ of the same kind. The sequence is regular in $A_{\mathfrak{n}}$ by step 2.1; localizing further to $A_{\mathfrak{q}}$ preserves injectivity of each multiplication map and the required nonzero successive quotients because the $f_j$ all lie in $\mathfrak{q}{A}_{\mathfrak{q}}$. Thus it is regular at $\mathfrak{q}$. AC covers the selections of maximal ideals, components and parameters used above.