# Projective dimension over restriction of scalars

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.

• 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.) – user26857 Apr 29 at 15:08

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 $$\mathrm{p.dim}_AM\leq d+\mathrm{p.dim}_BM$$.
• 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 $\mathrm{p.dim}_AP\leq d$. – Andrew Hubery Apr 29 at 18:45
• 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 $A$. – Andrew Hubery Apr 30 at 5:30