Is the proof of Auslander-Buchsbaum formula in Rotman's Advanced Modern Algebra wrong? 
Suppose that $(R,m,k)$ is a commutative, local, Noetherian ring. If $F$ is a free $R$-module then $\operatorname{Ext}^i_R(k,F)=0$ for $i\geq0$.

This statement is part of a proof for the Auslander-Buchsbaum formula given in Rotman's Advanced Modern Algebra (2002), page 1000, proposition 11.181.
 A: The proof can be corrected like this:
We have the long exact sequence
$$\operatorname{Ext}^i(k,F) \to \operatorname{Ext}^i(k,B)\to \operatorname{Ext}^{i+1}(k,\Omega) \to \operatorname{Ext}^{i+1}(k,F).$$
Let $d$ be the depth of $R$, which coincides with the depth of $F$. We have $\operatorname{Ext}^i(k,F)=0$ for all $i<d$. Thus $\operatorname{Ext}^i(k,B)\to \operatorname{Ext}^{i+1}(k,\Omega)$ is an isomorphism, whenever $i+1<d$.
From this you can deduce the desired $\operatorname{depth} \Omega = 1+\operatorname{depth} B$ whenever $\operatorname{depth} B < d$.
In the case $\operatorname{depth} B=d$, we can deduce $\operatorname{depth} \Omega =d$. By induction on the projective dimension, we can use Auslander-Buchsbaum for $\Omega$ and get that $\Omega$ is free. Hence $\operatorname{pd} B =1$. And actually you have to do this case by hand to finally prove the whole theorem. You find the proof of this case in Matsumura's Commutative Ring Theory and also in Twenty-Four Hours of Local Cohomology by Iyengar S. et al. 
Summarizing, all authors (except Rotman :D) prove the theorem like this:


*

*Show it for $\operatorname{pd}B=0$ (easy)

*Show it for $\operatorname{pd}B=1$ (not so easy, since you really have to look at the map in the long exact sequence)

*Show it for $\operatorname{pd}B > 1$ (Then the depth of $B$ is automatically smaller than the depth of $R$) with the induction machinery, as explained above. (Completely formal, hence fairly easy)

