projective dimension All rings and algebra in this question are commutative and contains unity.
Suppose $M$ is an $A$ module and $A$ a $R$ algebra. If $pd_R(M) < \infty$, then will that imply $pd_A(M) < \infty$? 
  In particular if $M = \frac{R[X_1, X_2, \ldots,X_n]_N}{I}$, $A = R[X_1, X_2, \ldots,X_n]_N$ where $R[X_1, X_2, \ldots,X_n]$ is polynomial algebra , $N$ a multiplicative closed set and $I$ an ideal, is the answer affirmative?
 A: In general, the answer is no.  E.g. suppose that $R$ is a field.  Every module over a field is free, and so has projective dimension $0$.  But if we take $A = k[x]/(x^2),$ then the $A$-module $M := k$ (with $x$ acting via zero) has infinite projective dimension over $A$.
However, your particular case has $A$ being a localization of $R$.  In this case the answer is yes:
let $P^{\bullet} \to M$ be a finite length projective resolution of $M$ as an $R$-module.
Then using the facts (two basic properties of localizations) that (a) $M \cong A\otimes_R M$, and (b) $A$ is flat over $R$, we see that $A\otimes_R P^{\bullet}$ is a finite length projective resolution of $M$ as an $A$-module.
Edit: This "answer" doesn't actually answer the question, because in the question there are really three rings involved, $R$, $R[X_1,\ldots,X_n]$, and then the localization $A := R[X_1,\ldots,X_n]_A$.    So in the above answer,
the ring called $R$ should actually be $R[X_1,\ldots,X_n]$, and the question becomes: if $M$ is an $R[X_1,\ldots,X_n]$-module which is of finite projective dimension as an $R$-module, is it of finite projective dimension as an
$R[X_1,\ldots,X_n]$-module?   I believe that the answer is again yes.
If this is true, it is surely well-known (and if it's false, I apologize for the blunder!); in any case, here is an attempt at a proof.
Firstly, by induction on $n$ (i.e. adjoin variables one at a time) we reduce to
the case $n = 1$, so $M$ is an $R[x]$-module, of finite projective dimension as an $R$-module.
Now take a projective resolution of $M$ as an $R[x]$-module; since $R[x]$ is projective as an $R$-module, this is also a projective resolution of $M$ as an $R$-module, and after some point in this resolution, the objects become projective as $R$-modules.  So it suffices to show that an $R[x]$-module which is projective as an $R$-module has finite projective dimension over $R[x]$.
Now adding a complement to this projective $R$-module, we can make it free
over $R$, and we can make it an $R[x]$-module just by having $x$ act via zero on the complement that we added.
So we reduce to showing that an $R[x]$-module that is free over $R$ has finite projective dimension over $R$.  (This step is probably superfluous, but it helps me psychologically to really be working with a free $R$-module.) 
Let $F$ be our free $R$-module; the action of $x$ then corresponds to an
element $\phi \in End_R(F).$
Now write $F[x] := R[x]\otimes_R F$, and consider the complex of $R[x]$-modules
$$ 0 \to F[x] \buildrel x - \phi \over\longrightarrow F[x] \to F \to 0.$$
This is a presentation of $M$, and is short exact, so indeed we see that $F$ has finite projective dimension over $R[x]$ (in fact it has projective dimension equal to one), so we are done. (In fact the proof shows that the projective dimension of our original $M$ over $R[x]$ is at most one greater than
its projective dimension over $R$.)
