Is $M\otimes_{S} S(n)$ Isomorphic to $M(n)$? $S$ is a graded ring and $M$ is a graded $S$ module, let $S(n)$ denote the graded $S$ module by shifting the grading of $S$, i.e. $S(n)_i = S_{n+i}$.
Then do we have $M\otimes_{S} S(n)$ Isomorphic to $M(n)$?
If this is not true, is it true with the additional condition that $S$ is generated by $S_1$ as a $S_0$ algebra?

EDIT:
The reason that I am interested in this question is that I am trying to understand the proof of Hartshorne Chapter II Proposition 5.12 (b) which states If $S$ is a graded ring, assume $S$ is generated by $S_1$ as a $S_0$ algebra. For any graded $S$ module $M$, $\tilde{M}(n)\cong \widetilde{M(n)}$, Where in the proof Hartshorne said use the fact that $\widetilde{M\otimes_{S} N}\cong \tilde{M}\otimes_{O_X} \tilde{N}$.
I am able to show this fact, but I think Hartshorne was trying to let $N$ be $S(n)$, then by the isomorphism, we have $\tilde{M}(n) \cong \widetilde{M\otimes_{S} S(n)}$. Then if I can show $M\otimes_{S} S(n)\cong M(n)$, then it is done.
 A: It is true with no conditions on $M$ and $S$: the map $M\otimes_S S(n)\to M(n)$ given by sending $m\otimes s \mapsto ms$ where $m\in M_d$, $s\in S(n)_e$, and $ms\in M(n)_{d+e}$ is an isomorphims. The proof is the same as in the ungraded case: using the rules for manipulating tensors, rewrite $m\otimes s$ as $ms\otimes 1$. Keeping track of the grading, we see that if $m\in M_d$ and $s\in S(n)_e=S_{n+e}$, then $ms\in M_{d+n+e}= M(n)_{d+e}$ and all is exactly as it should be.

It is understandable to be curious about what's going on with the sheafy version, though - there can be some interesting things that happen there when the condition "$S$ is generated by $S_1$ as an $S_0$-algebra" is removed. For instance, Hartshorne's proof that $\widetilde{M}\otimes_{\mathcal{O}_X} \widetilde{N} \cong \widetilde{M\otimes_S N}$ requires that $S$ is generated in degree one. There is always a canonical map from the LHS to the RHS but it need not be an isomorphism, and there are cases where there is no isomorphism.
One source I like which explains this situation is the Stacks Project, specifically their sections on quasi-coherent sheaves on Proj and invertible sheaves on Proj.
For a more explicit description of what "generated in degree one" buys you for sheaves of the form $\mathcal{O}(n)$, the key lemma here is the following:

Lemma (Stacks 01MS): Let $S$ be a graded ring, and set $X=\operatorname{Proj} S$. Let $f\in S$ be homogeneous of degree $d>0$. The sheaves $\mathcal{O}(nd)|_{D(f)}$ are invertible, and in fact trivial, for all $n\in \Bbb Z$.

When $S$ is generated by $S_1$, this means that $X$ can be covered by open sets of the form $D(f)$ for $f\in S_1$ and thus all $\mathcal{O}(n)$ are invertible and everything works like it ought to.
