Are projective modules "graded projective"? Let $A^{\bullet}$ be a graded commutative algebra. Denote by $A^{\bullet}$-mod the category of graded modules over $A^{\bullet}$. Let $A$ be $A^{\bullet}$ considered as an algebra (we forgot grading). Finally let $A$-mod be category of modules over $A$. 
So we have an oblivion functor $$ Obl: A^{\bullet}-{\rm mod} \rightarrow  A-{\rm mod}.$$
Consider $P^{\bullet} \in A^{\bullet}$-mod such that $Obl(P^{\bullet}) $ is projective in category of $A$-mod.
Question: Is $P^{\bullet}$ projective in category $A^{\bullet}$-mod ?
 A: This is true and easy to prove.

Let $M\stackrel{g}\to N$, $P\stackrel{f}\to N$ be graded homomorphisms, and $P\stackrel{h}\to M$ be a homomorphism such that $f=gh$. Then there is a graded homomorphism $P\stackrel{h'}\to M$ such that $f=gh'$.

For $x_n\in P_n$ we have $f(x_n)\in N_n$. From $h(x_n)=\sum y_m$ with $y_m\in M_m$ we get $f(x_n)=gh(x_n)=\sum g(y_m)$, and since $g(y_m)\in N_m$ it follows $g(y_m)=0$ for $m\ne n$ and $f(x_n)=g(y_n)$. Now define $h':P\to M$ by $h'(\sum x_n)=\sum y_n$, where $y_n$ in the unique element in $M_n$ with $f(x_n)=g(y_n)$. 
A: Here is only a partial, positive answer: If $(A^{\bullet},{\mathfrak m})$ is a commutative, graded, and graded local ring, then the functor $M\mapsto M_{\mathfrak m}$ from $A^{\bullet}\text{-grmod}$ to $A_{\mathfrak m}\text{-mod}$ preserves many numerical invariants of modules, such as projective dimension, injective dimension, depth, dimension, or Betti numbers. In particular, it reflects projectivity. Hence, if $M$ is a finitely generated, graded $A^{\bullet}$-module such that $M$ is projective considered as an ungraded $A$-module, then also $M_{\mathfrak m}$ is projective as an $A_{\mathfrak m}$-module, so $M$ is projective as a graded $A^{\bullet}$-module.
One proof that the projective dimension doesn't change under $M\mapsto M_{\mathfrak m}$ is to note that it maps minimal projective resolutions over $A^{\bullet}$ to such over $A_{\mathfrak m}$. 
For non-local rings and arbitrary modules, I don't know an answer at the moment.
