The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.11.5. The homomorphisms

\[ (M \otimes _ R N)\otimes _ R P \to M \otimes _ R N \otimes _ R P \to M \otimes _ R (N \otimes _ R P) \]

such that $f((x \otimes y)\otimes z) = x \otimes y \otimes z$ and $g(x \otimes y \otimes z) = x \otimes (y \otimes z)$, $x\in M, y\in N, z\in P$ are well-defined and are isomorphisms.

Proof. We shall prove $f$ is well-defined and is an isomorphism, and this proof carries analogously to $g$. Fix any $z\in P$, then the mapping $(x, y)\mapsto x \otimes y \otimes z$, $x\in M, y\in N$, is $R$-bilinear in $x$ and $y$, and hence induces homomorphism $f_ z : M \otimes N \to M \otimes N \otimes P$ which sends $f_ z(x \otimes y) = x \otimes y \otimes z$. Then consider $(M \otimes N)\times P \to M \otimes N \otimes P$ given by $(w, z)\mapsto f_ z(w)$. The map is $R$-bilinear and thus induces $f : (M \otimes _ R N)\otimes _ R P \to M \otimes _ R N \otimes _ R P$ and $f((x \otimes y)\otimes z) = x \otimes y \otimes z$. To construct the inverse, we note that the map $\pi : M \times N \times P \to (M \otimes N)\otimes P$ is $R$-trilinear. Therefore, it induces an $R$-linear map $h : M \otimes N \otimes P \to (M \otimes N)\otimes P$ which agrees with the universal property. Here we see that $h(x \otimes y \otimes z) = (x \otimes y)\otimes z$. From the explicit expression of $f$ and $h$, $f\circ h$ and $h\circ f$ are identity maps of $M \otimes N \otimes P$ and $(M \otimes N)\otimes P$ respectively, hence $f$ is our desired isomorphism. $\square$


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