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*}

10.11 Tensor products

Definition 10.11.1. Let $R$ be a ring, $M, N, P$ be three $R$-modules. A mapping $f : M \times N \to P$ (where $M \times N$ is viewed only as Cartesian product of two $R$-modules) is said to be $R$-bilinear if for each $x \in M$ the mapping $y\mapsto f(x, y)$ of $N$ into $P$ is $R$-linear, and for each $y\in N$ the mapping $x\mapsto f(x, y)$ is also $R$-linear.

Lemma 10.11.2. Let $M, N$ be $R$-modules. Then there exists a pair $(T, g)$ where $T$ is an $R$-module, and $g : M \times N \to T$ an $R$-bilinear mapping, with the following universal property: For any $R$-module $P$ and any $R$-bilinear mapping $f : M \times N \to P$, there exists a unique $R$-linear mapping $\tilde{f} : T \to P$ such that $f = \tilde{f} \circ g$. In other words, the following diagram commutes:

\[ \xymatrix{ M \times N \ar[rr]^ f \ar[dr]_ g & & P\\ & T \ar[ur]_{\tilde f} } \]

Moreover, if $(T, g)$ and $(T', g')$ are two pairs with this property, then there exists a unique isomorphism $j : T \to T'$ such that $j\circ g = g'$.

The $R$-module $T$ which satisfies the above universal property is called the tensor product of $R$-modules $M$ and $N$, denoted as $M \otimes _ R N$.

Proof. We first prove the existence of such $R$-module $T$. Let $M, N$ be $R$-modules. Let $T$ be the quotient module $P/Q$, where $P$ is the free $R$-module $R^{(M \times N)}$ and $Q$ is the $R$-module generated by all elements of the following types: ($x\in M, y\in N$)

\begin{align*} (x + x', y) - (x, y) - (x', y), \\ (x, y + y') - (x, y) - (x, y'), \\ (ax, y) - a(x, y), \\ (x, ay) - a(x, y) \end{align*}

Let $\pi : M \times N \to T$ denote the natural map. This map is $R$-bilinear, as implied by the above relations when we check the bilinearity conditions. Denote the image $\pi (x, y) = x \otimes y$, then these elements generate $T$. Now let $f : M \times N \to P$ be an $R$-bilinear map, then we can define $f' : T \to P$ by extending the mapping $f'(x \otimes y) = f(x, y)$. Clearly $f = f'\circ \pi $. Moreover, $f'$ is uniquely determined by the value on the generating sets $\{ x \otimes y : x\in M, y\in N\} $. Suppose there is another pair $(T', g')$ satisfying the same properties. Then there is a unique $j : T \to T'$ and also $j' : T' \to T$ such that $g' = j\circ g$, $g = j'\circ g'$. But then both the maps $(j\circ j') \circ g$ and $g$ satisfies the universal properties, so by uniqueness they are equal, and hence $j'\circ j$ is identity on $T$. Similarly $(j'\circ j) \circ g' = g'$ and $j\circ j'$ is identity on $T'$. So $j$ is an isomorphism. $\square$

Lemma 10.11.3. Let $M, N, P$ be $R$-modules, then the bilinear maps

\begin{align*} (x, y) & \mapsto y \otimes x\\ (x + y, z) & \mapsto x \otimes z + y \otimes z\\ (r, x) & \mapsto rx \end{align*}

induce unique isomorphisms

\begin{align*} M \otimes _ R N & \to N \otimes _ R M, \\ (M\oplus N)\otimes _ R P & \to (M \otimes _ R P)\oplus (N \otimes _ R P), \\ R \otimes _ R M & \to M \end{align*}

Proof. Omitted. $\square$

We may generalize the tensor product of two $R$-modules to finitely many $R$-modules, and set up a correspondence between the multi-tensor product with multilinear mappings. Using almost the same construction one can prove that:

Lemma 10.11.4. Let $M_1, \ldots , M_ r$ be $R$-modules. Then there exists a pair $(T, g)$ consisting of an $R$-module T and an $R$-multilinear mapping $g : M_1\times \ldots \times M_ r \to T$ with the universal property: For any $R$-multilinear mapping $f : M_1\times \ldots \times M_ r \to P$ there exists a unique $R$-module homomorphism $f' : T \to P$ such that $f'\circ g = f$. Such a module $T$ is unique up to unique isomorphism. We denote it $M_1\otimes _ R \ldots \otimes _ R M_ r$ and we denote the universal multilinear map $(m_1, \ldots , m_ r) \mapsto m_1 \otimes \ldots \otimes m_ r$.

Proof. Omitted. $\square$

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. $\square$

Doing induction we see that this extends to multi-tensor products. Combined with Lemma 10.11.3 we see that the tensor product operation on the category of $R$-modules is associative, commutative and distributive.

Definition 10.11.6. An abelian group $N$ is called an $(A, B)$-bimodule if it is both an $A$-module and a $B$-module, and the actions $A \to End(M)$ and $B \to End(M)$ are compatible in the sense that $(ax)b = a(xb)$ for all $a\in A, b\in B, x\in N$. Usually we denote it as $_ AN_ B$.

Lemma 10.11.7. For $A$-module $M$, $B$-module $P$ and $(A, B)$-bimodule $N$, the modules $(M \otimes _ A N)\otimes _ B P$ and $M \otimes _ A(N \otimes _ B P)$ can both be given $(A, B)$-bimodule structure, and moreover

\[ (M \otimes _ A N)\otimes _ B P \cong M \otimes _ A(N \otimes _ B P). \]

Proof. A priori $M \otimes _ A N$ is an $A$-module, but we can give it a $B$-module structure by letting

\[ (x \otimes y)b = x \otimes yb, \quad x\in M, y\in N, b\in B \]

Thus $M \otimes _ A N$ becomes an $(A, B)$-bimodule. Similarly for $N \otimes _ B P$, and thus for $(M \otimes _ A N)\otimes _ B P$ and $M \otimes _ A(N \otimes _ B P)$. By Lemma 10.11.5, these two modules are isomorphic as both as $A$-module and $B$-module via the same mapping. $\square$

Lemma 10.11.8. For any three $R$-modules $M, N, P$,

\[ \mathop{\mathrm{Hom}}\nolimits _ R(M \otimes _ R N, P) \cong \mathop{\mathrm{Hom}}\nolimits _ R(M, \mathop{\mathrm{Hom}}\nolimits _ R(N, P)) \]

Proof. An $R$-linear map $\hat{f}\in \mathop{\mathrm{Hom}}\nolimits _ R(M \otimes _ R N, P)$ corresponds to an $R$-bilinear map $f : M \times N \to P$. For each $x\in M$ the mapping $y\mapsto f(x, y)$ is $R$-linear by the universal property. Thus $f$ corresponds to a map $\phi _ f : M \to \mathop{\mathrm{Hom}}\nolimits _ R(N, P)$. This map is $R$-linear since

\[ \phi _ f(ax + y)(z) = f(ax + y, z) = af(x, z)+f(y, z) = (a\phi _ f(x)+\phi _ f(y))(z), \]

for all $a \in R$, $x \in M$, $y \in M$ and $z \in N$. Conversely, any $f \in \mathop{\mathrm{Hom}}\nolimits _ R(M, \mathop{\mathrm{Hom}}\nolimits _ R(N, P))$ defines an $R$-bilinear map $M \times N \to P$, namely $(x, y)\mapsto f(x)(y)$. So this is a natural one-to-one correspondence between the two modules $\mathop{\mathrm{Hom}}\nolimits _ R(M \otimes _ R N, P)$ and $\mathop{\mathrm{Hom}}\nolimits _ R(M, \mathop{\mathrm{Hom}}\nolimits _ R(N, P))$. $\square$

Lemma 10.11.9 (Tensor products commute with colimits). Let $(M_ i, \mu _{ij})$ be a system over the preordered set $I$. Let $N$ be an $R$-module. Then

\[ \mathop{\mathrm{colim}}\nolimits (M_ i \otimes N) \cong (\mathop{\mathrm{colim}}\nolimits M_ i)\otimes N. \]

Moreover, the isomorphism is induced by the homomorphisms $\mu _ i \otimes 1: M_ i \otimes N \to M \otimes N$ where $M = \mathop{\mathrm{colim}}\nolimits _ i M_ i$ with natural maps $\mu _ i : M_ i \to M$.

Proof. First proof. The functor $M' \mapsto M' \otimes _ R N$ is left adjoint to the functor $N' \mapsto \mathop{\mathrm{Hom}}\nolimits _ R(N, N')$ by Lemma 10.11.8. Thus $M' \mapsto M' \otimes _ R N$ commutes with all colimits, see Categories, Lemma 4.24.5.

Second direct proof. Let $P = \mathop{\mathrm{colim}}\nolimits (M_ i \otimes N)$, $M = \mathop{\mathrm{colim}}\nolimits M_ i$. Then for all $i\leq j$, the following diagram commutes:

\[ \xymatrix{ M_ i \otimes N \ar[r]_{\mu _ i \otimes 1} \ar[d]_{\mu _{ij} \otimes 1} & M \otimes N \ar[d]^{\text{id}} \\ M_ j \otimes N \ar[r]^{\mu _ j \otimes 1} & M \otimes N } \]

By Lemma 10.8.7, these maps induce a unique homomorphism $\psi : P \to M \otimes N$, with $\lambda _ i : M_ i \otimes N \to P$ given by $\lambda _ i = \pi \circ (\iota _ i \otimes 1)$.

To construct the inverse map, for each $i\in I$, there is the canonical $R$-bilinear mapping $g_ i : M_ i \times N \to M_ i \otimes N$. This induces a unique mapping $\widehat{\phi } : M \times N \to P$ such that $\widehat{\phi } \circ (\mu _ i \times 1) = \lambda _ i \circ g_ i$. It is $R$-bilinear. Thus it induces an $R$-linear mapping $\phi : M \otimes N \to P$. From the commutative diagram below:

\[ \xymatrix{ M_ i \times N \ar[r]^{g_ i} \ar[d]^{\mu _ i \times \text{id}} & M_ i \otimes N\ar[r]_{\text{id}} \ar[d]_{\lambda _ i} & M_ i \otimes N \ar[d]_{\mu _ i \otimes \text{id}} \ar[rd]^{\lambda _ i} \\ M \times N \ar[r]^{\widehat{\phi }} & P \ar[r]^{\psi } & M \otimes N \ar[r]^{\phi } & P } \]

we see that $\psi \circ \widehat{\phi } = g$, the canonical $R$-bilinear mapping $g : M \times N \to M \otimes N$. So $\psi \circ \phi $ is identity on $M \otimes N$. From the right-hand square and triangle, $\phi \circ \psi $ is also identity on $P$. $\square$

Lemma 10.11.10. Let

\begin{align*} M_1\xrightarrow {f} M_2\xrightarrow {g} M_3 \to 0 \end{align*}

be an exact sequence of $R$-modules and homomorphisms, and let $N$ be any $R$-module. Then the sequence
\begin{equation} \label{algebra-equation-2ndex} M_1\otimes N\xrightarrow {f \otimes 1} M_2\otimes N \xrightarrow {g \otimes 1} M_3\otimes N \to 0 \end{equation}

is exact. In other words, the functor $- \otimes _ R N$ is right exact, in the sense that tensoring each term in the original right exact sequence preserves the exactness.

Proof. We apply the functor $\mathop{\mathrm{Hom}}\nolimits (-, \mathop{\mathrm{Hom}}\nolimits (N, P))$ to the first exact sequence. We obtain

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits (M_3, \mathop{\mathrm{Hom}}\nolimits (N, P)) \to \mathop{\mathrm{Hom}}\nolimits (M_2, \mathop{\mathrm{Hom}}\nolimits (N, P)) \to \mathop{\mathrm{Hom}}\nolimits (M_1, \mathop{\mathrm{Hom}}\nolimits (N, P)) \]

By Lemma 10.11.8, we have

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits (M_3 \otimes N, P) \to \mathop{\mathrm{Hom}}\nolimits (M_2 \otimes N, P) \to \mathop{\mathrm{Hom}}\nolimits (M_1 \otimes N, P) \]

Using the pullback property again, we arrive at the desired exact sequence. $\square$

Remark 10.11.11. However, tensor product does NOT preserve exact sequences in general. In other words, if $M_1 \to M_2 \to M_3$ is exact, then it is not necessarily true that $M_1 \otimes N \to M_2 \otimes N \to M_3 \otimes N$ is exact for arbitrary $R$-module $N$.

Example 10.11.12. Consider the injective map $2 : \mathbf{Z}\to \mathbf{Z}$ viewed as a map of $\mathbf{Z}$-modules. Let $N = \mathbf{Z}/2$. Then the induced map $\mathbf{Z} \otimes \mathbf{Z}/2 \to \mathbf{Z} \otimes \mathbf{Z}/2$ is NOT injective. This is because for $x \otimes y\in \mathbf{Z} \otimes \mathbf{Z}/2$,

\[ (2 \otimes 1)(x \otimes y) = 2x \otimes y = x \otimes 2y = x \otimes 0 = 0 \]

Therefore the induced map is the zero map while $\mathbf{Z} \otimes N\neq 0$.

Remark 10.11.13. For $R$-modules $N$, if the functor $-\otimes _ R N$ is exact, i.e. tensoring with $N$ preserves all exact sequences, then $N$ is said to be flat $R$-module. We will discuss this later in Section 10.38.

Lemma 10.11.14. Let $R$ be a ring. Let $M$ and $N$ be $R$-modules.

  1. If $N$ and $M$ are finite, then so is $M \otimes _ R N$.

  2. If $N$ and $M$ are finitely presented, then so is $M \otimes _ R N$.

Proof. Suppose $M$ is finite. Then choose a presentation $0 \to K \to R^{\oplus n} \to M \to 0$. This gives an exact sequence $K \otimes _ R N \to N^{\oplus n} \to M \otimes _ R N \to 0$ by Lemma 10.11.10. We conclude that if $N$ is finite too then $M \otimes _ R N$ is a quotient of a finite module, hence finite, see Lemma 10.5.3. Similarly, if both $N$ and $M$ are finitely presented, then we see that $K$ is finite and that $M \otimes _ R N$ is a quotient of the finitely presented module $N^{\oplus n}$ by a finite module, namely $K \otimes _ R N$, and hence finitely presented, see Lemma 10.5.3. $\square$

Lemma 10.11.15. Let $M$ be an $R$-module. Then the $S^{-1}R$-modules $S^{-1}M$ and $S^{-1}R \otimes _ R M$ are canonically isomorphic, and the canonical isomorphism $f : S^{-1}R \otimes _ R M \to S^{-1}M$ is given by

\[ f((a/s) \otimes m) = am/s, \forall a \in R, m \in M, s \in S \]

Proof. Obviously, the map $f' : S^{-1}R \times M \to S^{-1}M$ given by $f((a/s, m)) = am/s$ is bilinear, and thus by the universal property, this map induces a unique $S^{-1}R$-module homomorphism $f : S^{-1}R \otimes _ R M \to S^{-1}M$ as in the statement of the lemma. Actually every element in $S^{-1}M$ is of the form $m/s$, $m\in M, s\in S$ and every element in $S^{-1}R \otimes _ R M$ is of the form $1/s \otimes m$. To see the latter fact, write an element in $S^{-1}R \otimes _ R M$ as

\[ \sum _ k \frac{a_ k}{s_ k} \otimes m_ k = \sum _ k \frac{a_ k t_ k}{s} \otimes m_ k = \frac{1}{s} \otimes \sum _ k {a_ k t_ k}m_ k = \frac{1}{s} \otimes m \]

Where $m = \sum _ k {a_ k t_ k}m_ k$. Then it is obvious that $f$ is surjective, and if $f(\frac{1}{s} \otimes m) = m/s = 0$ then there exists $t'\in S$ with $tm = 0$ in $M$. Then we have

\[ \frac{1}{s} \otimes m = \frac{1}{st} \otimes tm = \frac{1}{st} \otimes 0 = 0 \]

Therefore $f$ is injective. $\square$

Lemma 10.11.16. Let $M, N$ be $R$-modules, then there is a canonical $S^{-1}R$-module isomorphism $f : S^{-1}M \otimes _{S^{-1}R}S^{-1}N \to S^{-1}(M \otimes _ R N)$, given by

\[ f((m/s)\otimes (n/t)) = (m \otimes n)/st \]

Proof. We may use Lemma 10.11.7 and Lemma 10.11.15 repeatedly to see that these two $S^{-1}R$-modules are isomorphic, noting that $S^{-1}R$ is an $(R, S^{-1}R)$-bimodule:

\begin{align*} S^{-1}(M \otimes _ R N) & \cong S^{-1}R \otimes _ R (M \otimes _ R N)\\ & \cong S^{-1}M \otimes _ R N\\ & \cong (S^{-1}M \otimes _{S^{-1}R}S^{-1}R)\otimes _ R N\\ & \cong S^{-1}M \otimes _{S^{-1}R}(S^{-1}R \otimes _ R N)\\ & \cong S^{-1}M \otimes _{S^{-1}R}S^{-1}N \end{align*}

This isomorphism is easily seen to be the one stated in the lemma. $\square$

Comments (4)

Comment #967 by Tim on

There is a \textit command after the first lemma that doesn't seem to compile.

Comment #1103 by Ben on

Minor issue, but there is an unpaired parenthesis in the definition of in Lemma 10.11.16.

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