Localizing and taking degree zero commutes with tensor product Let $S$ be a graded ring ($S_n=0$ for $n<0$), $f\in S$ a homogeneous element, and $M, N$ two graded $S$-modules. I'm trying to prove that $$(M\otimes_S N)_{(f)}\simeq M_{(f)}\otimes_{S_{(f)}}N_{(f)},$$ where $M_{(f)}$ is the degree zero component of the localized module $M_f$. 
I got the map $M_{(f)}\otimes_{S_{(f)}}N_{(f)} \to (M\otimes_S N)_{(f)}$ but don't know how to define its inverse. I'd appreciate suggestions...
 A: We assume the basic knowledge for projective schemes (e.g. Hartshorne).
Let $S$ be a graded commutative ring ($S_i=0$ for $i<0$).
Let $S_+ = \bigoplus_{n>0} S_n$. $S_+$ is an ideal of S.
Let Proj($S$) be the set of homogeneous prime ideals $\mathfrak{p}$ such that $\mathfrak{p}$ does not contain $S_+$.
Let $f \in S_d$ ($d > 0$).
Let $D_+(f)$ = $\{\mathfrak{p} \in$ Proj($S$); $f$ is not contained in $\mathfrak{p}\}$.
$S_f$ is a graded ring with grading $(S_f)_n$ = $\{\frac{x}{f^k}; x \in S_{n + kd}, k \geq 0\}$.
We denote $(S_f)_0$ by $S_{(f)}$.
Let $M$ be a graded $S$-module.
$M_f$ is a graded $S_f$-module with grading $(M_f)_n$ = $\{\frac{x}{f^k}; x \in M_{n + kd}, k \geq 0\}$.
We denote $(M_f)_0$ by $M_{(f)}$.
$M_{(f)}$ is an $S_{(f)}$-module.
Lemma 1
Let $M$ be a graded $S$-module. Let $X$ = Proj($S$).
There exists a quasi-coherent $\mathcal{O}_X$-module $\tilde M$ such that 
$\tilde M(D_+(f)) = M_{(f)}$ for $f \in S_d$ $(d>0)$.
The restriction homomorphism $M(D_+(f)) \rightarrow M(D_+(fg))$ is the canonical $M_{(f)} \rightarrow M_{(fg)}$.
The proof is well known.
Let $M$ and $N$ be graded $S$-modules.
$M$ and $N$ can be regarded as graded $\mathbb{Z}$-modules.
$M\otimes_\mathbb{Z} N$ is a graded $\mathbb{Z}$-modules by defining
$(M\otimes_\mathbb{Z} N)_p = \bigoplus_{m+n=p} M_m\otimes_\mathbb{Z} N_n$.
Let $P$ be the $\mathbb{Z}$-submodule of $M\otimes_\mathbb{Z} N$ generated by $sx\otimes y - x\otimes sy$ for $x \in M$, $y \in N$, $s \in S$.
Since $M\otimes_S N = (M\otimes_\mathbb{Z} N)/P$ and $P$ is a graded $\mathbb{Z}$-submodule, $M\otimes_S N$ is a graded $S$-module.
Let $f \in S_d$ ($d > 0$). 
The canonical injections
$M_{(f)} \rightarrow M_f$, $N_{(f)} \rightarrow N_f$, $S_{(f)} \rightarrow S_f$
induce a homomorphism $M_{(f)} \otimes_{S_{(f)}} N_{(f)} \rightarrow M_f \otimes_{S_f} N_f$.
Since the canonical isomorphism $M_f \otimes_{S_f} N_f \rightarrow (M\otimes_S N)_f$ preserves the degrees, we get a homomorphism:
$$\lambda_f:M_{(f)} \otimes_{S_{(f)}} N_{(f)} \rightarrow (M\otimes_S N)_{(f)}.$$ 
Let $x \in M_{md}, y \in N_{nd}$ $(m >0, n> 0)$.
Then $\lambda_f(x/f^m\otimes y/f^n) = (x\otimes y)/f^{m+n}$.
Hence we get a morphism
$\lambda:\tilde M \otimes_{\mathcal{O}_X} \tilde N \rightarrow (M\otimes_S N)^-$. 
We define a $\mathbb{Z}$-bilinear map $M_m \times N_n \rightarrow M_{(f)} \otimes_{S_{(f)}} N_{(f)}$ by asigning $(x,y)$ to $x/f^m\otimes y/f^n$ (if $m < 0$, we write $x/f^m$ for $f^{-m}x/1$).
These maps induce a $\mathbb{Z}$-linear map $M \otimes_{\mathbb{Z}} N \rightarrow M_{(f)} \otimes_{S_{(f)}} N_{(f)}$.
If $s \in S_q$, $x \in M_m, y \in N_n$, this map sends $sx\otimes y$ and $x\otimes sy$ to the same $s/f^q(x/f^m\otimes y/f^n)$.
Hence we get a homomorphism $\gamma_f:M \otimes_S N \rightarrow M_{(f)} \otimes_{S_{(f)}} N_{(f)}$. If we regard $S_{(f)}$ as an $S$-algebra by the canonical homomorphism $S \rightarrow S_{(f)}$ sending $s \in S_q$ to $s/f^q$, $\gamma_f$ is an $S$-linear.
Lemma 2
$\lambda_f$ is an isomorphism for $f \in S_1$.
Proof:
Suppose
$\lambda_f(x_i/f^{m_i}\otimes y_i/f^{n_i}) = 0$, where $x_i \in M_{m_i}, y_i \in N_{n_i}$.
Then $\Sigma_i (x_i\otimes y_i)/f^{m_i+n_i} = 0$ in $(M\otimes_S N)_{(f)}$.
Hence there exists an integer $r > 0$ such that $f^r(\Sigma_i x_i\otimes y_i) = 0$ in $M\otimes_S N$. Hence $\Sigma_i f^rx_i\otimes y_i = 0$ in $M\otimes_S N$.
Since $\gamma_f(\Sigma_i f^rx_i\otimes y_i) = 0$, $\Sigma_i f^rx_i/f^{r + m_i}\otimes y_i/f^{n_i} = 0$ in $M_{(f)} \otimes_{S_{(f)}} N_{(f)}$.
Hence $\lambda_f$ is injective.
Clearly $\lambda_f$ is surjective.
QED
Proposition
Suppose the ideal $S_+$ is generated by $S_1$.
Let $g \in S_d$ $(d>0)$.
Then $\lambda_g$ is an isomorphism.
Proof:
Proj($S$) is a union of $D_+(f)$, $f \in S_1$.
Hence $\lambda:\tilde M \otimes_{\mathcal{O}_X} \tilde N \rightarrow (M\otimes_S N)^-$ is an isomorphism by Lemma 1 and Lemma 2.
Hence $\lambda_g$ is an isomorphism.
QED
A: By the ungraded localization theory, we know that the obvious map $\phi:M_f\otimes_{S_f}N_f\to(M\otimes_SN)_f$ is an isomorphism.
Now the domain and codomain of this maps are $\mathbb Z$-graded modules, and the map is homogeneous: it induces isomorphisms in each homogeneous piece —in particular, those of degree zero. We need only check that these degree-zero components of the domain and codomain of $\phi$ are the modules we want. Can you do that?
