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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...

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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

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This seems like a very roundabout way of proving the pure algebraic statement in the question, really. – Mariano Suárez-Alvarez Sep 9 '12 at 9:35
@MarianoSuárez-Alvarez But the question has a geometric meaning. – Makoto Kato Sep 9 '12 at 9:46

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?

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I don't think this is true in general when the degree of $f$ is not $1$, since then we need not have $(M_f\otimes_{S_f} N_f)_{(0)}\cong M_{(f)}\otimes_{S_{(f)}}N_{(f)}$. For example, if $f = X_1^2$, then $m\otimes n / X_1^2$ is in the former but not necessarily in the latter. Perhaps I am mistaken but I have seen several sources claim that this is an isomorphism specifically for $deg(f) = 1$. – Tom Oldfield May 3 '15 at 10:42

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