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Let $A$ and $B$ be two rings with a ring homomorphism $f: A\to B$ such that $B$ has finite projective dimension over $A$. Is it true that any module which has finite projective dimension as $B$-module must also have finite projective dimension over $A$ (as $A$-module via restriction of scalars).

I was trying using “Ext” criterion but somewhere I was using that “ Ext” well behave with restriction of scalars. I don’t know any such result. Can anyone help me? Thanks in advance.

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  • $\begingroup$ Yes. The projective dimension over $A$ is less than or equal to the sum of the other two projective dimensions that you mentioned. (The only proof I know uses spectral sequences.) $\endgroup$ – user26857 Apr 29 at 15:08
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Given a short exact sequence of $A$-modules $0\to L\to M\to N\to 0$, we have $\mathrm{p.dim}_AN\leq\max\{\mathrm{p.dim}_AM,1+\mathrm{p.dim}_AL\}$. Using this, we see that if $d:=\mathrm{p.dim}_AB$ and $M$ is a $B$-module, then $\mathrm{p.dim}_AM\leq d+\mathrm{p.dim}_BM$.

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  • $\begingroup$ Can you explain why exact sequence result implies required result? $\endgroup$ – Sunny Rathore Apr 29 at 15:57
  • $\begingroup$ Take $M$ a $B$-module with $\mathrm{p.dim}_BM=n$ and $0\to L\to P\to M\to 0$ with $P$ a projective $B$-module. Then $\mathrm{p.dim}_BL=n-1$, so by induction $\mathrm{p.dim}_AL\leq d+n-1$, and also $\mathrm{p.dim}_AP\leq d$. $\endgroup$ – Andrew Hubery Apr 29 at 18:45
  • $\begingroup$ Just last thing to ask why projective module over B has finite projective dimension over A? $\endgroup$ – Sunny Rathore Apr 30 at 4:43
  • $\begingroup$ In your setup you say that $B$ has finite projective dimension over $A$. Since any projective $B$-module is a summand of copies of $B$, it too will have finite projective dimension over $A$. $\endgroup$ – Andrew Hubery Apr 30 at 5:30
  • $\begingroup$ Could you please elaborate this argument? $\endgroup$ – Sunny Rathore Apr 30 at 5:41

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