Remark 27.9.2. In general the map constructed in Lemma 27.9.1 above is not an isomorphism. Here is an example. Let $k$ be a field. Let $S = k[x, y, z]$ with $k$ in degree $0$ and $\deg (x) = 1$, $\deg (y) = 2$, $\deg (z) = 3$. Let $M = S(1)$ and $N = S(2)$, see Algebra, Section 10.56 for notation. Then $M \otimes _ S N = S(3)$. Note that
\begin{eqnarray*} S_ z & = & k[x, y, z, 1/z] \\ S_{(z)} & = & k[x^3/z, xy/z, y^3/z^2] \cong k[u, v, w]/(uw - v^3) \\ M_{(z)} & = & S_{(z)} \cdot x + S_{(z)} \cdot y^2/z \subset S_ z \\ N_{(z)} & = & S_{(z)} \cdot y + S_{(z)} \cdot x^2 \subset S_ z \\ S(3)_{(z)} & = & S_{(z)} \cdot z \subset S_ z \end{eqnarray*}
Consider the maximal ideal $\mathfrak m = (u, v, w) \subset S_{(z)}$. It is not hard to see that both $M_{(z)}/\mathfrak mM_{(z)}$ and $N_{(z)}/\mathfrak mN_{(z)}$ have dimension $2$ over $\kappa (\mathfrak m)$. But $S(3)_{(z)}/\mathfrak mS(3)_{(z)}$ has dimension $1$. Thus the map $M_{(z)} \otimes N_{(z)} \to S(3)_{(z)}$ is not an isomorphism.
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