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

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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