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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