Let \(A\) be a regular local ring, and let \(M\) be a finitely generated \(A\)-module. In this case, strengthen the result (6.10A) as follows.
\(M\) is projective if and only if \(\Ext^i(M, A)=0\) for all \(i>0\).
Use (a) to show that for every \(n\ge0\), \(\operatorname{pd} M \leq n\) if and only if \(\Ext^i(M, A)=0\) for all \(i>n\).
Hint
For (a), use (6.11A) and descending induction on \(i\) to show that \(\Ext^i(M, N)=0\) for all \(i>0\) and all finitely generated \(A\)-modules \(N\). Then show \(M\) is a direct summand of a free \(A\)-module.
Solution
All Ext groups and projective dimensions are over \(A\). The regular local ring \(A\) is noetherian and has finite dimension \(d\). The regular-local global-dimension theorem gives \(\operatorname{pd}_A L\le d\) for every \(A\)-module \(L\) [@Har10a, Proposition III.6.11A].
1. If \(\Ext_A^i(M,A)=0\) for every \(i>0\), then \(\Ext_A^i(M,N)=0\) for every finite \(A\)-module \(N\) and every \(i>0\).
The bound \(\operatorname{pd}_A M\le d\) gives the assertion for all \(i>d\), even for arbitrary \(N\). Proceed downwards from \(i=d\) to \(i=1\). Suppose the assertion has been proved in degree \(i+1\) for all finite modules, and let \(N\) be finite. Choose an exact sequence \[ 0\longrightarrow K\longrightarrow A^{\oplus r}\longrightarrow N\longrightarrow0. \] The kernel \(K\) is finite because \(A\) is noetherian. The long exact Ext sequence in the second variable contains \[ \Ext_A^i(M,A^{\oplus r})\longrightarrow\Ext_A^i(M,N) \longrightarrow\Ext_A^{i+1}(M,K). \] The first term is \(\Ext_A^i(M,A)^{\oplus r}=0\); the last is zero by the induction hypothesis. Thus the middle group is zero. This completes the descending induction; when \(d=0\), the initial bound already gives every required vanishing.
2. The equivalence in (a) holds.
If \(M\) is projective, its length-zero projective resolution gives \(\Ext_A^i(M,A)=0\) for all \(i>0\).
Conversely, assume these groups vanish and choose a finite free presentation \[ 0\longrightarrow K\longrightarrow A^{\oplus r}\xrightarrow{p}M\longrightarrow0. \] Again \(K\) is finite. Step 1 gives \(\Ext_A^1(M,K)=0\). The corresponding long exact sequence therefore makes \[ \Hom_A(M,A^{\oplus r})\longrightarrow\Hom_A(M,M) \] surjective. A preimage of \(\id_M\) is a section of \(p\), so \(M\) is a direct summand of a finite free module. It is therefore projective, proving (a).
3. The equivalence in (b) holds for every \(n\ge0\).
A projective resolution of length at most \(n\) makes \(\Ext_A^i(M,A)=0\) for \(i>n\).
For the converse, construct a finite-free resolution of \(M\) successively and denote its kernels by \[ K_0=M,\qquad 0\longrightarrow K_{j+1}\longrightarrow P_j\longrightarrow K_j\longrightarrow0, \] where each \(P_j\) is finite free and each \(K_j\) is finite. The long exact Ext sequences in the first variable give \[ \Ext_A^i(K_n,A)\cong\Ext_A^{i+n}(M,A) \qquad(i>0). \] If the groups on the right vanish for all \(i>0\), part (a), proved in step 2, implies that \(K_n\) is projective. For \(n\ge1\), truncation then gives a projective resolution \[ 0\longrightarrow K_n\longrightarrow P_{n-1}\longrightarrow\cdots \longrightarrow P_0\longrightarrow M\longrightarrow0 \] of length \(n\). For \(n=0\), the same argument says directly that \(M=K_0\) is projective. Hence \(\operatorname{pd}_A M\le n\) in every case.
4. Q.E.D.
Steps <1>1–<1>2 prove (a), and step 3 proves (b).