0

If $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is an exact sequence, show that

  1. $p d(B) \leq \max \{p d(A), p d(C)\}$ with equality except when $p d(C)= p d(A)+1$.

  2. $id (B) \leq \max \{i d(A), i d(C)\}$ with equality except when $i d(A)= i d(C)+1$.

  3. $f d(B) \leq \max \{f d(A), f d(C)\}$ with equality except when $f d(C)= f d(A)+1$.

Here $A$ is a right $R$-module, $pd(A)$ is the projective dimension of $A$, $id(A)$ is the injective dimension of $A$, $fd(A)$ is the flat dimension of $A$.

Suppose that $0\to P'_{3}\to P'_{2}\to P'_{1}\to P'_{0}\to A$ is a projective resolution of $A$, and $0\to 0\to 0\to P''_{1} \to P''_{0}\to C$ is a projective resolution of $C$, then by Horseshoe lemma, there is a projective resolution of $B$: $$ 0\to P'_{3}\to P'_{2}\to P'_{1}\oplus P''_{1}\to P'_{0}\oplus P''_{0}\to B, $$ so $p d(B) = 4 = \max \{p d(A), p d(C)\}$.

How to prove that $p d(B) \leq \max \{p d(A), p d(C)\}$ with equality except when $p d(C)= p d(A)+1$? And could you please tell me how to prove 3? Thanks!

Ryze
  • 1,136
  • 1
    I think you'd be better off using the long exact sequences of Ext and Tor instead of the horseshoe lemma, especially for the equality parts. – Zeek Jun 21 '20 at 15:24
  • @Zeek Excuse me, Could you please give me more details? – Ryze Jun 22 '20 at 02:55
  • If $pd(A)=n$, then $\text{Ext}^{n+k}(A,M)=0$ for all $M\in\text{Mod}(R)$ and $k>0$. By considering the long exact sequence obtained from $\text{Hom}(-,M)$ for particular choices of $M$, you can then look at where terms vanish. For example, you cannot have $pd(B)>pd(A),pd(C)$ because you would have an exact sequence with a non-zero terms bound by zero terms, which is impossible. – Zeek Jun 22 '20 at 11:43
  • @Zeek Thank you for your help! – Ryze Jun 23 '20 at 06:42

0 Answers0