Depth as the first nonzero Ext degree

If $IM=M$, [F1] makes $M=0$. Its depth is infinite by definition and every displayed Ext group vanishes. Hence assume $M\mathrm{\ne }0$ and $IM\mathrm{\ne }M$. Write $E^i(T)={\mathrm{Ext}}_R^i(R\mathrm{/}I,T)$ and $d=\inf \{i:E^i(M)\mathrm{\ne }0\}$. If $d=0$, a nonzero element of $E^0(M)=(0{:}_{M}I)$ is killed by every $x\in I$, so no $M$-regular element exists in $I$ and depth is zero.

Suppose $d>0$, allowing $d=\mathrm{\infty }$. Then $E^0(M)=0$. If an associated prime $\mathfrak{p}={\mathrm{Ann}}_R(m)$ contained $I$, its nonzero associated element $m$ would belong to $(0{:}_{M}I)$, a contradiction. By [L1], choose an $M$-regular $x\in I$ and put $Q=M\mathrm{/}xM\mathrm{\ne }0$. Applying $E^{\ast }(-)$ to $0\to M\stackrel{x}{\to }M\to Q\to 0$, multiplication by $x$ on each $E^i(M)$ is zero because $x$ annihilates $R\mathrm{/}I$. Thus $0\to E^i(M)\to E^i(Q)\to E^{i+1}(M)\to 0$ is exact for every $i\ge 0$. If $d<\mathrm{\infty }$, the first nonzero degree for $Q$ is $d-1$; if $d=\mathrm{\infty }$, every $E^i(Q)$ vanishes.

Induct on finite $d$. The case $d=0$ is step 1.1. For $d>0$, step 1.2 and induction give an $M\mathrm{/}xM$-regular sequence of length $d-1$, so prepending $x$ gives an $M$-regular sequence of length $d$. Conversely, for any $M$-regular first element $y\in I$, the same argument gives first nonzero Ext degree $d-1$ on $M\mathrm{/}yM$; induction bounds every sequence beginning with $y$ by $d$. Therefore ${\mathrm{depth}}_I(M)=d$. If $d=\mathrm{\infty }$, step 1.2 applies successively to every nonzero quotient and keeps all its Ext groups zero; finite induction constructs regular sequences of every length. Their supremum, hence depth, is $\mathrm{\infty }$.