$pd_{\Lambda /I} B < \infty$ induces $pd_{\Lambda} B < \infty$ Let $\Lambda$ be an artin algebra. $I$ is an ideal in $\Lambda$ with $pd _{\Lambda} \Lambda/I < \infty$.($pd _{\Lambda}M$ means the projective dimension of $\Lambda$-module $M$). Suppose $B$ is a $\Lambda$-module that is also a $\Lambda /I$-module, and $pd_{\Lambda /I} B < \infty$. How to get that $pd_{\Lambda} B < \infty$?
 A: This follows from the following fact: if $L,M$ and $N$ are $\Lambda$-modules sitting in a short exact sequence 
$$
0\to L \to M \to N \to 0,
$$
and if two of the modules $L,M$ and $N$ have finite projective dimension, then so does the third one.  A proof of this is given here.
Applying induction, one can prove the following: assume that $L_n, L_{n-1}, \ldots, L_0$ and $M$ are $\Lambda$-modules sitting in an exact sequence
$$
 0\to L_n\to L_{n-1} \to \ldots \to L_0 \to M \to 0.
$$
If $L_n, L_{n-1}, \ldots, L_0$ all have finite projective dimension, then so does $M$.
We can now prove the proposition in the statement of the question.  Assume that $\Lambda/I$ has finite projective dimension as a $\Lambda$-module, and let $B$ be a $\Lambda$-module which is also a $\Lambda/I$-module, and such that its projective dimension over $\Lambda/I$ is finite.  Let 
$$
0\to P_n\to P_{n-1}\to \ldots \to P_0 \to B \to 0
$$
be a projective resolution of $B$ as a $\Lambda/I$-module.  Then this exact sequence is still an exact sequence of $\Lambda$-modules, and by our assumptions, $P_n, P_{n-1}, \ldots, P_0$ have finite projective dimension as $\Lambda$-modules.  Therefore, by the above, $B$ also have finite projective dimension as a $\Lambda$-module.
