Maximal regular sequences have a common Ext length

Put $M_j=M\mathrm{/}(x_1,\dots ,x_j)M$ for $0\le j\le r$, so $M_0=M$ and $Q=M_r$. By [F1], $Q$ is nonzero and finite, and $IQ\mathrm{\ne }Q$. Maximality and [L1] give a prime $\mathfrak{p}\in {\mathrm{Ass}}_R(Q)$ containing $I$. By the definition in [L1], some nonzero $q\in Q$ has ${\mathrm{Ann}}_R(q)=\mathfrak{p}$, hence $Iq=0$. Therefore $E^0(Q):={\mathrm{Ext}}_R^0(R\mathrm{/}I,Q)\cong (0{:}_{Q}I)$ is nonzero.

For $1\le j\le r$, regularity gives $0\to M_{j-1}\stackrel{x_j}{\to }{M}_{j-1}\to M_j\to 0$. Write $E^i(T)={\mathrm{Ext}}_R^i(R\mathrm{/}I,T)$. Since $x_j\in I$ annihilates $R\mathrm{/}I$, multiplication by $x_j$ on every $E^i(M_{j-1})$ is zero. The long exact sequence of [L2] therefore yields, for each $i\ge 0$, a short exact sequence $0\to E^i(M_{j-1})\to E^i(M_j)\to E^{i+1}(M_{j-1})\to 0$. Moreover $E^0(M_{j-1})=(0{:}_{M_{j-1}}I)=0$: an $I$-annihilated element is killed by the injective multiplication map $x_j$.

Suppose the first nonzero $E^i(M_j)$ occurs in degree $d_j$. Starting with $E^0(M_{j-1})=0$, the sequences in step 1.2 show inductively that $E^i(M_{j-1})=0$ for $0\le i\le d_j$, and at $i=d_j$ give $E^{d_j+1}(M_{j-1})\cong E^{d_j}(M_j)\mathrm{\ne }0$. Thus the first nonzero degree rises by one when moving backwards across $x_j$. Step 1.1 gives $d_r=0$; repeating this finite argument for $j=r,r-1,\dots ,1$ gives $d_0=r$. Hence the displayed Ext minimum equals the length of every maximal sequence, so all have the same length.