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The

One problem in the proof is that $\operatorname{pd}_R(R/I)$ might be infinite, so you can't apply Auslander-Buchsbaum the final time.

I can't seem to come up with an example where $M$ has finite projective dimension while $\operatorname{ann} M$ has infinite projective dimension, but they must exist.

Later: The famous Dutta-Hochster-McLaughlin example does the trick. It is a module $M$ of finite projective dimension over $R = k[x,y,z,w]/(xy-zw)$, with annihilator $(x,y,z,w)^3 + (x,z)$.

2 added 281 characters in body

The problem in the proof is that $\operatorname{pd}_R(R/I)$ might be infinite, so you can't apply Auslander-Buchsbaum the final time.

I can't seem to come up with an example where $M$ has finite projective dimension while $\operatorname{ann} M$ has infinite projective dimension, but they must exist.

Later: The famous Dutta-Hochster-McLaughlin example does the trick. It is a module $M$ of finite projective dimension over $R = k[x,y,z,w]/(xy-zw)$, with annihilator $(x,y,z,w)^3 + (x,z)$.

1

The problem in the proof is that $\operatorname{pd}_R(R/I)$ might be infinite, so you can't apply Auslander-Buchsbaum the final time.

I can't seem to come up with an example where $M$ has finite projective dimension while $\operatorname{ann} M$ has infinite projective dimension, but they must exist.