Tensor product of modules: Difference between revisions

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==Balanced product==

{{main|pairing}}

For a ring ''R'', a right ''R''-module ''M'', a left ''R''-module ''N'', and an abelian group ''G'', a map {{nowrapmath|''φ'': ''M'' × ''N'' → ''G''}} is said to be '''''R''-balanced''', '''''R''-middle-linear''' or an '''''R''-balanced product''' if for all ''m'', ''m''′ in ''M'', ''n'', ''n''′ in ''N'', and ''r'' in ''R'' the following hold:{{refn|{{citation |author=Nathan Jacobson |title=Basic Algebra II |edition=2nd |year=2009 |publisher=[[Dover Publications]] }}}}{{rp|126}}

:<math display="block">\begin{align}

\varphi (m, n+n') &= \varphi (m, n) + \varphi (m, n') && \text{Dl}_{\varphi} \\

\varphi (m +m', n) &= \varphi (m, n) + \varphi (m', n) && \text{Dr}_{\varphi} \\

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\end{align}</math>

The set of all such balanced products over ''R'' from {{nowrapmath|''M'' × ''N''}} to ''G'' is denoted by {{nowrapmath|L_''R''(''M'', ''N''; ''G'')}}.

If ''φ'', ''ψ'' are balanced products, then each of the operations {{nowrapmath|''φ'' + ''ψ''}} and −''φ'' defined [[pointwise]] is a balanced product. This turns the set {{nowrapmath|L_''R''(''M'', ''N''; ''G'')}} into an abelian group.

For ''M'' and ''N'' fixed, the map {{nowrapmath|''G'' ↦ L_''R''(''M'', ''N''; ''G'')}} is a [[functor]] from the [[category of abelian groups]] to itself. The morphism part is given by mapping a group homomorphism {{nowrapmath|''g'' : ''G'' → ''G''′}} to the function {{nowrapmath|''φ'' ↦ ''g'' ∘ ''φ''}}, which goes from {{nowrapmath|L_''R''(''M'', ''N''; ''G'')}} to {{nowrapmath|L_''R''(''M'', ''N''; ''G''′)}}.

;Remarks:

#Properties (Dl) and (Dr) express [[Additive map|biadditivity]] of ''φ'', which may be regarded as [[Distributive property|distributivity]] of ''φ'' over addition.

#Property (A) resembles some [[associative property]] of ''φ''.

#Every ring ''R'' is an ''R''-[[bimodule]]. So the ring multiplication {{nowrapmath|(''r'', ''r''′) ↦ ''r'' ⋅ ''r''′}} in ''R'' is an ''R''-balanced product {{nowrapmath|''R'' × ''R'' → ''R''}}.

==Definition==

For a ring ''R'', a right ''R''-module ''M'', a left ''R''-module ''N'', the '''tensor product''' over ''R''

:<math display="block">M \otimes_R N</math>

is an [[abelian group]] together with a balanced product (as defined above)

:<math display="block">\otimes : M \times N \to M \otimes_{R} N</math>

which is [[universal property|universal]] in the following sense:<ref>Hazewinkel, et al. (2004), [https://books.google.com/books?id=AibpdVNkFDYC&pg=PA95 p. 95], Prop. 4.5.1</ref>

[[File:Tensor product of modules2.svg|200px|right]]

:For every abelian group ''G'' and every balanced product <math display="block">f: M \times N \to G</math> there is a ''unique'' group homomorphism <math display="block"> \tilde{f}: M \otimes_R N \to G</math> such that <math display="block">\tilde{f} \circ \otimes = f.</math>

::<math>f: M \times N \to G</math>

:there is a ''unique'' group homomorphism

::<math> \tilde{f}: M \otimes_R N \to G</math>

:such that

::<math>\tilde{f} \circ \otimes = f.</math>

As with all [[Universal property#Existence and uniqueness|universal properties]], the above property defines the tensor product uniquely [[up to]] a unique isomorphism: any other abelian group and balanced product with the same properties will be isomorphic to {{nowrapmath|''M'' ⊗_''R'' ''N''}} and ⊗. Indeed, the mapping ⊗ is called ''canonical'', or more explicitly: the canonical mapping (or balanced product) of the tensor product.<ref>{{harvnb|Bourbaki|loc=ch. II §3.1}}</ref>

The definition does not prove the existence of {{nowrapmath|''M'' ⊗_''R'' ''N''}}; see below for a construction.

The tensor product can also be defined as a [[representable functor|representing object]] for the functor {{nowrapmath|''G'' → L_''R''(''M'',''N'';''G'')}}; explicitly, this means there is a [[natural isomorphism]]:

:<math display="block">\begin{cases}\operatorname{Hom}_{\Z} (M \otimes_R N, G) \simeq \operatorname{L}_R(M, N; G) \\ g \mapsto g \circ \otimes \end{cases}</math>

This is a succinct way of stating the universal mapping property given above. (if a priori one is given this is natural isomorphism, then <math>\otimes</math> can be recovered by taking <math>G = M \otimes_R N</math> and then mapping the identity map.)

Similarly, given the natural identification <math>\operatorname{L}_R(M, N; G) = \operatorname{Hom}_R(M, \operatorname{Hom}_{\Z}(N, G))</math>,<ref>First, if <math>R=\Z,</math> then the claimed identification is given by <math>f \mapsto f'</math> with <math>f'(x)(y) = f(x, y)</math>. In general, <math>\operatorname{Hom}_{\Z }(N, G)</math> has the structure of a right ''R''-module by <math>(g \cdot r)(y) = g(r y)</math>. Thus, for any <math>\Z</math>-bilinear map ''f'', ''f''′ is ''R''-linear <math>\Leftrightarrow f'(xr) = f'(x) \cdot r \Leftrightarrow f(xr, y) = f(x, ry).</math></ref> one can also define {{nowrapmath|''M'' ⊗_''R'' ''N''}} by the formula

:<math display="block">\operatorname{Hom}_{\Z} (M \otimes_R N, G) \simeq \operatorname{Hom}_R(M, \operatorname{Hom}_{\Z}(N, G)).</math>

This is known as the [[tensor-hom adjunction]]; see also {{section link||Properties}}.

For each ''x'' in ''M'', ''y'' in ''N'', one writes

:{{block indent|em=1.5|text=''x'' ⊗ ''y''}}

for the image of (''x'', ''y'') under the canonical map <math>\otimes: M \times N \to M \otimes_R N</math>. It is often called a [[pure tensor]]. Strictly speaking, the correct notation would be ''x'' ⊗_''R'' ''y'' but it is conventional to drop ''R'' here. Then, immediately from the definition, there are relations:

:{|

|-

| style="width:24em;" | ''x'' ⊗ (''y'' + ''y''′) = ''x'' ⊗ ''y'' + ''x'' ⊗ ''y''′ || (Dl_⊗)

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{{math_theorem|name=Proposition|Every element of <math>M \otimes_R N</math> can be written, non-uniquely, as

:<math display="block">\sum_i x_i \otimes y_i, \, x_i \in M, y_i \in N.</math>

In other words, the image of <math>\otimes</math> generates <math>M \otimes_R N</math>. Furthermore, if ''f'' is a function defined on elements <math>x \otimes y</math> with values in an abelian group ''G'', then ''f'' extends uniquely to the homomorphism defined on the whole <math>M \otimes_R N</math> if and only if <math>f(x \otimes y)</math> is <math>\Z</math>-bilinear in ''x'' and ''y''.}}

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The proposition says that one can work with explicit elements of the tensor products instead of invoking the universal property directly each time. This is very convenient in practice. For example, if ''R'' is commutative and the left and right actions by ''R'' on modules are considered to be equivalent, then <math>M \otimes_R N </math> can naturally be furnished with the ''R''-scalar multiplication by extending

:<math display="block">r \cdot (x \otimes y) := (r \cdot x) \otimes y = x \otimes (r \cdot y)</math>

to the whole <math>M \otimes_R N</math> by the previous proposition (strictly speaking, what is needed is a bimodule structure not commutativity; see a paragraph below). Equipped with this ''R''-module structure, <math>M \otimes_R N</math> satisfies a universal property similar to the above: for any ''R''-module ''G'', there is a natural isomorphism:

:<math display="block">\begin{cases} \operatorname{Hom}_R(M \otimes_R N, G) \simeq \{R\text{-bilinear maps } M \times N \to G \}, \\ g \mapsto g \circ \otimes \end{cases}</math>

If ''R'' is not necessarily commutative but if ''M'' has a left action by a ring ''S'' (for example, ''R''), then <math>M \otimes_R N</math> can be given the left ''S''-module structure, like above, by the formula

:<math display="block">s \cdot (x \otimes y) := (s \cdot x) \otimes y.</math>

Analogously, if ''N'' has a right action by a ring ''S'', then <math>M \otimes_R N</math> becomes a right ''S''-module.<!-- Doesn't seem correct; see the example below. Strictly speaking, the ring used to form the tensor should be indicated: most modules can be considered as modules over several different rings or over the same ring with a different actions of the ring on the module elements. For example, it can be shown that {{nowrapmath|'''R''' ⊗_'''R''' '''R'''}} and {{nowrapmath|'''R''' ⊗_'''Z''' '''R'''}} are completely different from each other. However, in practice, whenever the ring is clear from context, the subscript denoting the ring may be dropped.-->

===Tensor product of linear maps and a change of base ring===

Given linear maps <math>f: M \to M'</math> of right modules over a ring ''R'' and <math>g: N \to N'</math> of left modules, there is a unique group homomorphism

:<math display="block">\begin{cases}f \otimes g: M \otimes _R N \to M' \otimes_R N' \\ x \otimes y \mapsto f(x) \otimes g(y) \end{cases}</math>

The construction has a consequence that tensoring is a functor: each right ''R''-module ''M'' determines the functor

:<math display="block">M \otimes_R -: R\text{-Mod} \longrightarrow \text{Ab}</math>

from the [[category of modules|category of left modules]] to the category of abelian groups that sends ''N'' to {{nowrapmath|''M'' ⊗ ''N''}} and a module homomorphism ''f'' to the group homomorphism {{nowrapmath|1 ⊗ ''f''}}.<!-- unfortunately, this doesn't quite work as written unless ''S'' is an ''R''-algebra (say ''S'' = ''R''). '''Example''': Let ''M'' be a right ''R''-module. A left action on ''M'' by a ring ''S'' is the same thing as a group homomorphism:

:<math display="block">S \otimes_R M \to M</math>.

Tensoring this with a left ''R''-module ''N'' results in

:<math display="block">(S \otimes_R M) \otimes_R N \to M \otimes_R N</math>.

Here, the group on the left is really <math>S \otimes_R (M \otimes_R N)</math> by associativity (see below) and so this shows <math>M \otimes_R N</math> is a left ''S''-module.-->

If <math>f: R \to S</math> is a ring homomorphism and if ''M'' is a right ''S''-module and ''N'' a left ''S''-module, then there is the canonical ''surjective'' homomorphism:

:<math display="block">M \otimes_R N \to M \otimes_S N</math>

induced by

:<math display="block">M \times N \overset{\otimes_S} \longrightarrow M \otimes_S N.</math><ref>{{harvnb|Bourbaki|loc=ch. II §3.2.}}</ref>

The resulting map is surjective since pure tensors {{nowrapmath|''x'' ⊗ ''y''}} generate the whole module. In particular, taking ''R'' to be <math>\Z</math> this shows every tensor product of modules is a quotient of a tensor product of abelian groups.

{{see also|Tensor product#Tensor product of linear maps}}

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It is possible to extend the definition to a tensor product of any number of modules over the same commutative ring. For example, the universal property of

{{block indent|em=1.5|text=''M''₁ ⊗ ''M''₂ ⊗ ''M''₃}}

:''M''₁ ⊗ ''M''₂ ⊗ ''M''₃

is that each trilinear map on

{{block indent|em=1.5|text=''M''₁ × ''M''₂ × ''M''₃ → ''Z''}}

:''M''₁ × ''M''₂ × ''M''₃ → ''Z''

corresponds to a unique linear map

{{block indent|em=1.5|text=''M''₁ ⊗ ''M''₂ ⊗ ''M''₃ → ''Z''.}}

:''M''₁ ⊗ ''M''₂ ⊗ ''M''₃ → ''Z''.

The binary tensor product is associative: (''M''₁ ⊗ ''M''₂) ⊗ ''M''₃ is naturally isomorphic to ''M''₁ ⊗ (''M''₂ ⊗ ''M''₃). The tensor product of three modules defined by the universal property of trilinear maps is isomorphic to both of these iterated tensor products.

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*For an ''R''₁-''R''₂-[[bimodule]] ''M''₁₂ and a left ''R''₂-module ''M''₂₀, <math>M_{12}\otimes_{R_2} M_{20}</math> is a left ''R''₁-module.

*For a right ''R''₂-module ''M''₀₂ and an ''R''₂-''R''₃-[[bimodule]] ''M''₂₃, <math>M_{02}\otimes_{R_2} M_{23}</math> is a right ''R''₃-module.

*(associativity) For a right ''R''₁-module ''M''₀₁, an ''R''₁-''R''₂-bimodule ''M''₁₂, and a left ''R''₂-module ''M''₂₀ we have:<ref>{{harvnb|Bourbaki|loc=ch. II §3.8}}</ref> <math display="block">\left (M_{01} \otimes_{R_1} M_{12} \right ) \otimes_{R_2} M_{20} = M_{01} \otimes_{R_1} \left (M_{12} \otimes_{R_2} M_{20} \right ).</math>

*Since ''R'' is an ''R''-''R''-bimodule, we have <math>R\otimes_R R = R</math> with the ring multiplication <math>mn =: m \otimes_R n</math> as its canonical balanced product.

*(associativity) For a right ''R''₁-module ''M''₀₁, an ''R''₁-''R''₂-bimodule ''M''₁₂, and a left ''R''₂-module ''M''₂₀ we have:<ref>{{harvnb|Bourbaki|loc=ch. II §3.8}}</ref>

::<math>\left (M_{01} \otimes_{R_1} M_{12} \right ) \otimes_{R_2} M_{20} = M_{01} \otimes_{R_1} \left (M_{12}\otimes_{R_2} M_{20} \right ).</math>

*Since ''R'' is an ''R''-''R''-bimodule, we have <math>R\otimes_R R = R</math> with the ring multiplication <math>mn =: m\otimes_R n</math> as its canonical balanced product.

===Modules over commutative rings===

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*(identity) <math>R \otimes_R M = M.</math>

*(associativity) <math>(M \otimes_R N) \otimes_R P = M \otimes_R (N \otimes_R P).</math><ref>The first three properties (plus identities on morphisms) say that the category of ''R''-modules, with ''R'' commutative, forms a [[symmetric monoidal category]].</ref> Thus <math>M \otimes_R N \otimes_R P</math> is well-defined.

*(symmetry) <math>M \otimes_R N = N \otimes_R M.</math> In fact, for any permutation ''σ'' of the set {1, ..., ''n''}, there is a unique isomorphism: <math display="block">\begin{cases} M_1 \otimes_R \cdots \otimes_R M_n \longrightarrow M_{\sigma(1)} \otimes_R \cdots \otimes_R M_{\sigma(n)} \\ x_1 \otimes \cdots \otimes x_n \longmapsto x_{\sigma(1)} \otimes \cdots \otimes x_{\sigma(n)} \end{cases}</math>

*(symmetrydistributive property) <math>M \otimes_R (N \oplus P) = N(M \otimes_R N) \oplus (M \otimes_R P).</math> In fact, for<math anydisplay="block">M permutation\otimes_R ''σ''\left of(\bigoplus\nolimits_{i the\in setI} N_i \right ) = \bigoplus\nolimits_{1,i ...\in I} \left ( M \otimes_R N_i \right ),</math> for an [[index set]] ''nI''}, thereof isarbitrary a unique isomorphism:[[cardinality]].

*(commutes with finite product) for any finitely many <math>N_i</math>, <math display="block">M \otimes_R \prod_{i = 1}^n N_i = \prod_{i = 1}^nM \otimes_R N_i.</math>

::<math>\begin{cases} M_1 \otimes_R \cdots \otimes_R M_n \longrightarrow M_{\sigma(1)} \otimes_R \cdots \otimes_R M_{\sigma(n)} \\ x_1 \otimes \cdots \otimes x_n \longmapsto x_{\sigma(1)} \otimes \cdots \otimes x_{\sigma(n)} \end{cases}</math>

*(commutes with [[localization of a module|localization]]) for any multiplicatively closed subset ''S'' of ''R'', <math display="block">S^{-1}(M \otimes_R N) = S^{-1}M \otimes_{S^{-1}R} S^{-1}N</math> as <math>S^{-1} R</math>-module. Since <math>S^{-1} R</math> is an ''R''-algebra and <math>S^{-1} - = S^{-1} R \otimes_R -</math>, this is a special case of:

*(commutes with base extension) If ''S'' is an ''R''-algebra, writing <math>-_{S} = S \otimes_R -</math>, <math display="block">(M \otimes_R N)_S = M_S \otimes_S N_S;</math><ref>Proof: (using associativity in a general form) <math>(M \otimes_R N)_S = (S \otimes_R M) \otimes_R N = M_S \otimes_R N = M_S \otimes_S S \otimes_R N = M_S \otimes_S N_S</math></ref> cf. {{section link||Extension of scalars}}.

*(distributive property) <math>M \otimes_R (N \oplus P) = (M \otimes_R N) \oplus (M \otimes_R P).</math> In fact,

*(commutes with direct limit) for any direct system of ''R''-modules ''M''_''i'', <math display="block">(\varinjlim M_i) \otimes_R N = \varinjlim (M_i \otimes_R N).</math>

::<math>M \otimes_R \left (\bigoplus\nolimits_{i \in I} N_i \right ) = \bigoplus\nolimits_{i \in I} \left ( M \otimes_R N_i \right ),</math>

*(tensoring is right exact) if <math display="block">0 \to N' \overset{f}\to N \overset{g}\to N'' \to 0</math> is an exact sequence of ''R''-modules, then <math display="block">M \otimes_R N' \overset{1 \otimes f}\to M \otimes_R N \overset{1 \otimes g}\to M \otimes_R N'' \to 0</math> is an exact sequence of ''R''-modules, where <math>(1 \otimes f)(x \otimes y) = x \otimes f(y).</math> This is a consequence of:

:for an [[index set]] ''I'' of arbitrary [[cardinality]].

*(commutes with finite product) for any finitely many <math>N_i</math>,

::<math>M \otimes_R \prod_{i = 1}^n N_i = \prod_{i = 1}^nM \otimes_R N_i.</math>

*(commutes with [[localization of a module|localization]]) for any multiplicatively closed subset ''S'' of ''R'',

::<math>S^{-1}(M \otimes_R N) = S^{-1}M \otimes_{S^{-1}R} S^{-1}N</math>

:as <math>S^{-1} R</math>-module. Since <math>S^{-1} R</math> is an ''R''-algebra and <math>S^{-1} - = S^{-1} R \otimes_R -</math>, this is a special case of:

*(commutes with base extension) If ''S'' is an ''R''-algebra, writing <math>-_{S} = S \otimes_R -</math>,

::<math>(M \otimes_R N)_S = M_S \otimes_S N_S;</math><ref>Proof: (using associativity in a general form) <math>(M \otimes_R N)_S = (S \otimes_R M) \otimes_R N = M_S \otimes_R N = M_S \otimes_S S \otimes_R N = M_S \otimes_S N_S</math></ref>

:cf. {{section link||Extension of scalars}}.

*(commutes with direct limit) for any direct system of ''R''-modules ''M''_''i'',

::<math>(\varinjlim M_i) \otimes_R N = \varinjlim (M_i \otimes_R N).</math>

*(tensoring is right exact) if

::<math>0 \to N' \overset{f}\to N \overset{g}\to N'' \to 0</math>

:is an exact sequence of ''R''-modules, then

::<math>M \otimes_R N' \overset{1 \otimes f}\to M \otimes_R N \overset{1 \otimes g}\to M \otimes_R N'' \to 0</math>

:is an exact sequence of ''R''-modules, where <math>(1 \otimes f)(x \otimes y) = x \otimes f(y).</math> This is a consequence of:

*([[tensor-hom adjunction|adjoint relation]]) <math>\operatorname{Hom}_R(M \otimes_R N, P) = \operatorname{Hom}_R(M, \operatorname{Hom}_R(N, P))</math>.

*(tensor-hom relation) there is a canonical ''R''-linear map: <math display="block">\operatorname{Hom}_R(M, N) \otimes P \to \operatorname{Hom}_R(M, N \otimes P),</math> which is an isomorphism if either ''M'' or ''P'' is a [[finitely generated projective module]] (see {{section link||As linearity-preserving maps}} for the non-commutative case);<ref>{{harvnb|Bourbaki|loc=ch. II §4.4}}</ref> more generally, there is a canonical ''R''-linear map: <math display="block">\operatorname{Hom}_R(M, N) \otimes \operatorname{Hom}_R(M', N') \to \operatorname{Hom}_R(M \otimes M', N \otimes N')</math> which is an isomorphism if either <math>(M, N)</math> or <math>(M, M')</math> is a pair of finitely generated projective modules.

*(tensor-hom relation) there is a canonical ''R''-linear map:

::<math>\operatorname{Hom}_R(M, N) \otimes P \to \operatorname{Hom}_R(M, N \otimes P),</math>

:which is an isomorphism if either ''M'' or ''P'' is a [[finitely generated projective module]] (see {{section link||As linearity-preserving maps}} for the non-commutative case);<ref>{{harvnb|Bourbaki|loc=ch. II §4.4}}</ref> more generally, there is a canonical ''R''-linear map:

::<math>\operatorname{Hom}_R(M, N) \otimes \operatorname{Hom}_R(M', N') \to \operatorname{Hom}_R(M \otimes M', N \otimes N')</math>

:which is an isomorphism if either <math>(M, N)</math> or <math>(M, M')</math> is a pair of finitely generated projective modules.

To give a practical example, suppose ''M'', ''N'' are free modules with bases <math>e_i, i \in I</math> and <math>f_j, j \in J</math>. Then ''M'' is the [[direct sum of modules|direct sum]] <math>M = \bigoplus_{i \in I} R e_i</math>

and the same for ''N''. By the distributive property, one has:

:<math display="block">M \otimes_R N = \bigoplus_{i, j} R(e_i \otimes f_j);</math>;

i.e., <math>e_i \otimes f_j, \, i \in I, j \in J</math> are the ''R''-basis of <math>M \otimes_R N</math>. Even if ''M'' is not free, a [[free presentation]] of ''M'' can be used to compute tensor products.

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The tensor product, in general, does not commute with [[inverse limit]]: on the one hand,

:<math display="block">\Q \otimes_{\Z} \Z /p^n = 0</math>

(cf. "examples"). On the other hand,

:<math display="block"> \left (\varprojlim \Z /p^n \right ) \otimes_{\Z} \Q = \Z_p \otimes_{\Z} \Q = \Z_p \left [p^{-1} \right ] = \Q_p</math>

where <math>\Z_p, \Q_p</math> are the [[ring of p-adic integers]] and the [[field of p-adic numbers]]. See also "[[profinite integer]]" for an example in the similar spirit.

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The associativity holds more generally for non-commutative rings: if ''M'' is a right ''R''-module, ''N'' a (''R'', ''S'')-module and ''P'' a left ''S''-module, then

:<math display="block">(M \otimes_R N) \otimes_S P = M \otimes_R (N \otimes_S P)</math>

as abelian group.

The general form of adjoint relation of tensor products says: if ''R'' is not necessarily commutative, ''M'' is a right ''R''-module, ''N'' is a (''R'', ''S'')-module, ''P'' is a right ''S''-module, then as abelian group<ref>{{harvnb|Bourbaki|loc=ch.II §4.1 Proposition 1}}</ref>

:<math display="block">\operatorname{Hom}_S(M \otimes_R N, P) = \operatorname{Hom}_R(M, \operatorname{Hom}_S(N, P)), \, f \mapsto f'</math><ref>{{harvnb|Bourbaki|loc=ch.II §4.1 Proposition 1}}</ref>

where <math>f'</math> is given by <math>f'(x)(y) = f(x \otimes y).</math> {{See also: [[tensor|Tensor-hom adjunction]].}}

===Tensor product of an ''R''-module with the fraction field===

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*For any ''R''-module ''M'', <math>K \otimes_R M \cong K \otimes_R (M / M_{\operatorname{tor}})</math> as ''R''-modules, where <math>M_{\operatorname{tor}}</math> is the torsion submodule of ''M''.

*If ''M'' is a torsion ''R''-module then <math>K \otimes_R M = 0</math> and if ''M'' is not a torsion module then <math>K \otimes_R M \ne 0</math>.

*If ''N'' is a submodule of ''M'' such that <math>M/N</math> is a torsion module then <math>K \otimes_R N \cong K \otimes_R M</math> as ''R''-modules by <math>x \otimes n \mapsto x \otimes n</math>.

*In <math>K \otimes_R M</math>, <math>x \otimes m = 0</math> if and only if <math>x = 0</math> or <math>m \in M_{\operatorname{tor}}</math>. In particular, <math>M_{\operatorname{tor}} = \operatorname{ker}(M \to K \otimes_R M)</math> where <math>m \mapsto 1 \otimes m</math>.

*<math>K \otimes_R M \cong M_{(0)}</math> where <math>M_{(0)}</math> is the [[localization of a module|localization of the module]] <math>M</math> at the prime ideal <math>(0)</math> (i.e., the localization with respect to the nonzero elements).

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The adjoint relation in the general form has an important special case: for any ''R''-algebra ''S'', ''M'' a right ''R''-module, ''P'' a right ''S''-module, using <math>\operatorname{Hom}_S (S, -) = -</math>, we have the natural isomorphism:

:<math display="block">\operatorname{Hom}_S (M \otimes_R S, P) = \operatorname{Hom}_R (M, \operatorname{Res}_R(P)).</math>

This says that the functor <math>-\otimes_R S</math> is a [[left adjoint]] to the forgetful functor <math>\operatorname{Res}_R</math>, which restricts an ''S''-action to an ''R''-action. Because of this, <math>- \otimes_R S</math> is often called the [[extension of scalars]] from ''R'' to ''S''. In the [[representation theory]], when ''R'', ''S'' are group algebras, the above relation becomes the [[Frobenius reciprocity]].

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*<math>R^n \otimes_R S = S^n,</math> for any ''R''-algebra ''S'' (i.e., a free module remains free after extending scalars.)

*For a commutative ring <math>R</math> and a commutative ''R''-algebra ''S'', we have: <math display="block">S \otimes_R R[x_1, \dots, x_n] = S[x_1, \dots, x_n];</math> in fact, more generally, <math display="block">S \otimes_R (R[x_1, \dots, x_n]/I) = S[x_1, \dots, x_n]/ IS[x_1, \dots, x_n],</math> where <math>I</math> is an ideal.

*Using <math>\Complex = \R [x]/(x^2 + 1),</math> the previous example and the [[Chinese remainder theorem]], we have as rings <math display="block">\Complex \otimes_{\R} \Complex = \Complex [x]/(x^2 + 1) = \Complex [x]/(x+i) \times \Complex[x]/(x-i) = \Complex^2.</math> This gives an example when a tensor product is a [[direct product]].

*For a commutative ring <math>R</math> and a commutative ''R''-algebra ''S'', we have:

::<math>S \otimes_R R[x_1, \dots, x_n] = S[x_1, \dots, x_n];</math>

:in fact, more generally,

::<math>S \otimes_R (R[x_1, \dots, x_n]/I) = S[x_1, \dots, x_n]/ IS[x_1, \dots, x_n],</math>

:where <math>I</math> is an ideal.

*Using <math>\Complex = \R [x]/(x^2 + 1),</math> the previous example and the [[Chinese remainder theorem]], we have as rings

::<math>\Complex \otimes_{\R} \Complex = \Complex [x]/(x^2 + 1) = \Complex [x]/(x+i) \times \Complex[x]/(x-i) = \Complex^2.</math>

:This gives an example when a tensor product is a [[direct product]].

*<math>\R \otimes_{\Z} \Z[i] = \R[i] = \Complex.</math>

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Let ''G'' be an abelian group in which every element has finite order (that is ''G'' is a [[torsion abelian group]]; for example ''G'' can be a finite abelian group or <math>\Q/\Z</math>). Then:<ref>Example 3.6 of http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf</ref>

<math display="block">\Q \otimes_{\Z} G = 0.</math>

:<math>\Q \otimes_{\Z} G = 0.</math>

Indeed, any <math>x \in \Q \otimes_{\Z} G</math> is of the form

<math display="block">x = \sum_i r_i \otimes g_i, \qquad r_i \in \Q , g_i \in G.</math>

:<math>x = \sum_i r_i \otimes g_i, \qquad r_i \in \Q , g_i \in G.</math>

If <math>n_i</math> is the order of <math>g_i</math>, then we compute:

<math display="block">x = \sum (r_i / n_i )n_i \otimes g_i = \sum r_i / n_i \otimes n_i g_i = 0.</math>

:<math>x = \sum (r_i / n_i )n_i \otimes g_i = \sum r_i / n_i \otimes n_i g_i = 0.</math>

Similarly, one sees

<math display="block">\Q /\Z \otimes_{\Z} \Q /\Z = 0.</math><!--

:<math>\Q /\Z \otimes_{\Z} \Q /\Z = 0.</math><!--

Consider the [[rational number]]s, '''Q''', and the [[modular arithmetic|integers modulo ''n'']], '''Z'''_''n''. As with any abelian group, both can be considered as modules over the [[integer]]s, '''Z'''.

Let {{nowrapmath|''B'' : '''Q''' × '''Z'''_''n'' → ''M''}} be a '''Z'''-bilinear operator. Then {{nowrapmath|1=''B''(''q'', ''k'') = ''B''(''q''/''n'', ''nk'') = ''B''(''q''/''n'', 0) = 0}}, so every bilinear operator is identically zero. Therefore, if we define {{nowrapmath|'''Q''' ⊗_'''Z''' '''Z'''_''n''}} to be the trivial module, and ⊗ to be the zero bilinear function, then we see that the properties for the tensor product are satisfied. Therefore, the tensor product of '''Q''' and '''Z'''_''n'' is {0}.<ref>Hazewinkel, et al. (2004), [https://books.google.com/books?id=AibpdVNkFDYC&pg=PA97 p. 97], Ex. 4.5.1.</ref>

Since an [[abelian group]] is the same thing as a '''Z'''-module,<ref>{{cite book |first=Nathan |last=Jacobson |title=Basic Algebra |volume=I |edition=2nd |publisher=Dover |year=2009 |page=164 }}</ref> the tensor product of '''Z'''-modules is the same thing as the '''tensor product of abelian groups'''.-->

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Here are some identities useful for calculation: Let ''R'' be a commutative ring, ''I'', ''J'' ideals, ''M'', ''N'' ''R''-modules. Then

#<math>R/I \otimes_R M = M/IM</math>. If ''M'' is [[flat module|flat]], <math>IM = I \otimes_R M</math>.<ref group="proof">Tensoring with ''M'' the exact sequence <math>0 \to I \to R \to R/I \to 0</math> gives

:<math display="block">I \otimes_R M \overset{f}\to R \otimes_R M = M \to R/I \otimes_R M \to 0</math>

where ''f'' is given by <math>i \otimes x \mapsto ix</math>. Since the image of ''f'' is ''IM'', we get the first part of 1. If ''M'' is flat, ''f'' is injective and so is an isomorphism onto its image.</ref>

#<math>M/IM \otimes_{R/I} N/IN = M \otimes_R N \otimes_R R/I</math> (because tensoring commutes with base extensions)

#<math>R/I \otimes_R R/J = R/(I + J)</math>.<ref group="proof">

:<math display="block">R/I \otimes_R R/J = {R/J \over I(R/J) }= {R/J \over (I + J)/J} = R/(I+J).</math> Q.E.D. <math>\square</math></ref>

'''Example:''' If ''G'' is an abelian group, <math>G \otimes_{\Z } \Z /n = G/nG</math>; this follows from 1.

'''Example:''' <math>\Z /n \otimes_{\Z } \Z /m = \Z /{\gcd(n, m)}</math>; this follows from 3. In particular, for distinct prime numbers ''p'', ''q'',

:<math display="block">\Z / p\Z \otimes \Z / q\Z=0.</math>

Tensor products can be applied to control the order of elements of groups. Let G be an abelian group. Then the multiples of 2 in

:<math display="block">G \otimes \Z / 2\Z</math>

are zero.

'''Example:''' Let <math>\mu_n</math> be the group of ''n''-th roots of unity. It is a [[cyclic group]] and cyclic groups are classified by orders. Thus, non-canonically, <math>\mu_n \approx \Z /n</math> and thus, when ''g'' is the gcd of ''n'' and ''m'',

:<math display="block">\mu_n \otimes_{\Z } \mu_m \approx \mu_g.</math>

'''Example:''' Consider <math>\Q \otimes_{\Z} \Q.</math> Since <math>\Q \otimes_{\Q } \Q</math> is obtained from <math>\Q \otimes_{\Z } \Q </math> by imposing <math>\Q </math>-linearity on the middle, we have the surjection

:<math display="block">\Q \otimes_{\Z } \Q \to \Q \otimes_{\Q } \Q </math>

whose kernel is generated by elements of the form <math>{r \over s} x \otimes y - x \otimes {r \over s} y</math>

where ''r'', ''s'', ''x'', ''u'' are integers and ''s'' is nonzero. Since

:<math display="block">{r \over s} x \otimes y = {r \over s} x \otimes {s \over s} y = x \otimes {r \over s} y,</math>

the kernel actually vanishes; hence, <math>\Q \otimes_{\Z } \Q = \Q \otimes_{\Q } \Q = \Q .</math>

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'''Example:''' We propose to compare <math>\R \otimes_{\Z} \R </math> and <math>\R \otimes_{\R } \R </math>. Like in the previous example, we have: <math>\R \otimes_{\Z} \R = \R \otimes_{\Q} \R </math> as abelian group and thus as <math>\Q</math>-vector space (any <math>\Z</math>-linear map between <math>\Q</math>-vector spaces is <math>\Q</math>-linear). As <math>\Q</math>-vector space, <math>\R </math> has dimension (cardinality of a basis) of [[Cardinality of the continuum|continuum]]. Hence, <math>\R \otimes_{\Q } \R </math> has a <math>\Q</math>-basis indexed by a product of continuums; thus its <math>\Q</math>-dimension is continuum. Hence, for dimension reason, there is a non-canonical isomorphism of <math>\Q</math>-vector spaces:

:<math display="block">\R \otimes_{\Z } \R \approx \R \otimes_{\R } \R </math>.

<!-- but they are not isomorphic as rings since the ring on the left is not even a field. -->

Consider the modules <math>M=\Complex [x,y,z]/(f),N=\Complex [x,y,z]/(g)</math> for <math>f,g\in \Complex[x,y,z]</math> irreducible polynomials such that <math>\gcd(f,g)=1.</math> Then,

:<math display="block">\frac{\Complex [x,y,z]}{(f)}\otimes_{\Complex [x,y,z]}\frac{\Complex [x,y,z]}{(g)} \cong \frac{\Complex [x,y,z]}{(f,g)}</math>

Another useful family of examples comes from changing the scalars. Notice that

:<math display="block">\frac{\Z [x_1,\ldots,x_n]}{(f_1,\ldots,f_k)} \otimes_\Z R \cong \frac{R[x_1,\ldots,x_n]}{(f_1,\ldots,f_k)}</math>

Good examples of this phenomenon to look at are when <math>R = \Q, \Complex, \Z/(p^k), \Z_p, \Q_p.</math>

==Construction==

The construction of {{nowrapmath|''M'' ⊗ ''N''}} takes a quotient of a [[free abelian group]] with basis the symbols {{nowrapmath|''m'' ∗ ''n''}}, used here to denote the [[ordered pair]] {{nowrapmath|(''m'', ''n'')}}, for ''m'' in ''M'' and ''n'' in ''N'' by the subgroup generated by all elements of the form

# −''m'' ∗ (''n'' + ''n''′) + ''m'' ∗ ''n'' + ''m'' ∗ ''n''′

# −(''m'' + ''m''′) ∗ ''n'' + ''m'' ∗ ''n'' + ''m''′ ∗ ''n''

# (''m'' · ''r'') ∗ ''n'' − ''m'' ∗ (''r'' · ''n'')

where ''m'', ''m''′ in ''M'', ''n'', ''n''′ in ''N'', and ''r'' in ''R''. The quotient map which takes {{nowrapmath|1=''m'' ∗ ''n'' {{=}} (''m'', ''n'')}} to the coset containing {{nowrapmath|''m'' ∗ ''n''}}; that is,

:<math display="block">\otimes: M \times N \to M \otimes_R N, \, (m, n) \mapsto [m * n]</math>

is balanced, and the subgroup has been chosen minimally so that this map is balanced. The universal property of ⊗ follows from the universal properties of a free abelian group and a quotient.

More category-theoretically, let σ be the given right action of ''R'' on ''M''; i.e., σ(''m'', ''r'') = ''m'' · ''r'' and τ the left action of ''R'' of ''N''. Then the tensor product of ''M'' and ''N'' over ''R'' can be defined as the [[coequalizer]]:

:<math display="block">M \times R \times N {{{} \atop \overset{\sigma \times 1}\to}\atop{\underset{1 \times \tau} \to \atop {}}} M \times N \overset{\otimes}\to M \otimes_R N,</math>

together with the requirements

:<math display="block">m \otimes (n + n') = m \otimes n + m \otimes n',</math>

:<math display="block">(m + m') \otimes n = m \otimes n + m' \otimes n.</math>

If ''S'' is a subring of a ring ''R'', then <math>M \otimes_R N</math> is the quotient group of <math>M \otimes_S N</math> by the subgroup generated by <math>xr \otimes_S y - x \otimes_S ry, \, r \in R, x \in M, y \in N</math>, where <math>x \otimes_S y</math> is the image of <math>(x, y)</math> under <math>\otimes: M \times N \to M \otimes_{S} N.</math> In particular, any tensor product of ''R''-modules can be constructed, if so desired, as a quotient of a tensor product of abelian groups by imposing the ''R''-balanced product property.

In the construction of the tensor product over a commutative ring ''R'', the ''R''-module structure can be built in from the start by forming the quotient of a free ''R''-module by the submodule generated by the elements given above for the general construction, augmented by the elements {{nowrapmath|''r'' ⋅ (''m'' ∗ ''n'') − ''m'' ∗ (''r'' ⋅ ''n'')}}. Alternately, the general construction can be given a Z(''R'')-module structure by defining the scalar action by {{nowrapmath|1=''r'' ⋅ (''m'' ⊗ ''n'') = ''m'' ⊗ (''r'' ⋅ ''n'')}} when this is well-defined, which is precisely when ''r'' ∈ Z(''R''), the [[Center (ring theory)|centre]] of ''R''.

The [[direct product]] of ''M'' and ''N'' is rarely isomorphic to the tensor product of ''M'' and ''N''. When ''R'' is not commutative, then the tensor product requires that ''M'' and ''N'' be modules on opposite sides, while the direct product requires they be modules on the same side. In all cases the only function from {{nowrapmath|''M'' × ''N''}} to ''G'' that is both linear and bilinear is the zero map.

==As linear maps==

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===Dual module===

{{see also|Duality (mathematics)#Dual objects}}

The '''dual module''' of a right ''R''-module ''E'', is defined as {{nowrapmath|Hom_''R''(''E'', ''R'')}} with the canonical left ''R''-module structure, and is denoted ''E''^∗.<ref>{{harvnb|Bourbaki|loc=ch. II §2.3}}</ref> The canonical structure is the [[pointwise]] operations of addition and scalar multiplication. Thus, ''E''^∗ is the set of all ''R''-linear maps {{nowrapmath|''E'' → ''R''}} (also called ''linear forms''), with operations

:<math display="block">(\phi + \psi)(u) = \phi(u) + \psi(u), \quad \phi, \psi \in E^*, u \in E</math>

:<math display="block">(r \cdot \phi) (u) = r \cdot \phi(u), \quad \phi \in E^*, u \in E, r \in R,</math>

The dual of a left ''R''-module is defined analogously, with the same notation.

There is always a canonical homomorphism {{nowrapmath|''E'' → ''E''^∗∗}} from ''E'' to its second dual. It is an isomorphism if ''E'' is a free module of finite rank. In general, ''E'' is called a [[reflexive module]] if the canonical homomorphism is an isomorphism.

===Duality pairing===

We denote the [[natural pairing]] of its dual ''E''^∗ and a right ''R''-module ''E'', or of a left ''R''-module ''F'' and its dual ''F''^∗ as

:<math display="block"> \langle \cdot , \cdot \rangle : E^* \times E \to R : (e', e) \mapsto \langle e', e \rangle = e'(e) </math>

:<math display="block"> \langle \cdot , \cdot \rangle : F \times F^* \to R : (f, f') \mapsto \langle f, f' \rangle = f'(f) .</math>

The pairing is left ''R''-linear in its left argument, and right ''R''-linear in its right argument:

:<math display="block"> \langle r \cdot g, h \cdot s \rangle = r \cdot \langle g, h \rangle \cdot s, \quad r, s \in R .</math>

===An element as a (bi)linear map===

In the general case, each element of the tensor product of modules gives rise to a left ''R''-linear map, to a right ''R''-linear map, and to an ''R''-bilinear form. Unlike the commutative case, in the general case the tensor product is not an ''R''-module, and thus does not support scalar multiplication.

* Given right ''R''-module ''E'' and right ''R''-module ''F'', there is a canonical homomorphism {{nowrapmath|''θ'' : ''F'' ⊗_''R'' ''E''^∗ → Hom_''R''(''E'', ''F'')}} such that {{nowrapmath|''θ''(''f'' ⊗ ''e''′)}} is the map {{nowrapmath|''e'' ↦ ''f'' ⋅ {{langle}}''e''′, ''e''{{rangle}}}}.<ref>{{harvnb|Bourbaki|loc=ch. II §4.2 eq. (11)}}</ref><!-- Thus, an element of a tensor product ''η'' ∈ ''F'' ⊗_''R'' ''E''^∗ acts as a right ''R''-linear map ''η'' : ''E'' → ''F''.-->

* Given left ''R''-module ''E'' and right ''R''-module ''F'', there is a canonical homomorphism {{nowrapmath|''θ'' : ''F'' ⊗_''R'' ''E'' → Hom_''R''(''E''^∗, ''F'')}} such that {{nowrapmath|''θ''(''f'' ⊗ ''e'')}} is the map {{nowrapmath|''e''′ ↦ ''f'' ⋅ {{langle}}''e'', ''e''′{{rangle}}}}.<ref>{{harvnb|Bourbaki|loc=ch. II §4.2 eq. (15)}}</ref><!--Thus, an element of a tensor product ''ξ'' ∈ ''F'' ⊗_''R'' ''E'' acts as a right ''R''-linear map ''E''^∗ → ''F'', and by similarity, as a left ''R''-linear map ''F''^∗ → ''E''. -->

Both cases hold for general modules, and become isomorphisms if the modules ''E'' and ''F'' are restricted to being [[finitely generated projective module]]s (in particular free modules of finite ranks). Thus, an element of a tensor product of modules over a ring ''R'' maps canonically onto an ''R''-linear map, though as with vector spaces, constraints apply to the modules for this to be equivalent to the full space of such linear maps.

* Given right ''R''-module ''E'' and left ''R''-module ''F'', there is a canonical homomorphism {{nowrapmath|''θ'' : ''F''^∗ ⊗_''R'' ''E''^∗ → L_''R''(''F'' × ''E'', ''R'')}} such that {{nowrapmath|''θ''(''f''′ ⊗ ''e''′)}} is the map {{nowrapmath|(''f'', ''e'') ↦ {{langle}}⟨''f'', ''f''′{{rangle}}′⟩ ⋅ {{langle}}⟨''e''′, ''e''{{rangle}}⟩}}.{{citation needed|date=April 2015}} Thus, an element of a tensor product ''ξ'' ∈ ''F''^∗ ⊗_''R'' ''E''^∗ may be thought of giving rise to or acting as an ''R''-bilinear map {{nowrapmath|''F'' × ''E'' → ''R''}}.

===Trace===

Let ''R'' be a commutative ring<!-- the non-commutative case seems unclear; any source anyone? --> and ''E'' an ''R''-module. Then there is a canonical ''R''-linear map:

:<math display="block">E^* \otimes_R E \to R</math>

induced through linearity by <math>\phi \otimes x \mapsto \phi(x)</math>; it is the unique ''R''-linear map corresponding to the natural pairing.

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If ''E'' is a finitely generated projective ''R''-module, then one can identify <math>E^* \otimes_R E = \operatorname{End}_R(E)</math> through the canonical homomorphism mentioned above and then the above is the '''trace map''':

:<math display="block">\operatorname{tr}: \operatorname{End}_R(E) \to R.</math>

When ''R'' is a field, this is the usual [[Trace (linear algebra)|trace]] of a linear transformation.

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The most prominent example of a tensor product of modules in differential geometry is the tensor product of the spaces of vector fields and differential forms. More precisely, if ''R'' is the (commutative) ring of smooth functions on a smooth manifold ''M'', then one puts

:<math display="block">\mathfrak{T}^p_q = \Gamma(M, T M)^{\otimes p} \otimes_R \Gamma(M, T^* M)^{\otimes q}</math>

where Γ means the [[space of sections]] and the superscript <math>\otimes p</math> means tensoring ''p'' times over ''R''. By definition, an element of <math>\mathfrak{T}^p_q</math> is a [[tensor field]] of type (''p'', ''q'').

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To lighten the notation, put <math>E = \Gamma(M, T M)</math> and so <math>E^* = \Gamma(M, T^* M)</math>.<ref>This is actually the ''definition'' of differential one-forms, global sections of <math>T^*M</math>, in Helgason, but is equivalent to the usual definition that does not use module theory.</ref> When ''p'', ''q'' ≥ 1, for each (''k'', ''l'') with 1 ≤ ''k'' ≤ ''p'', 1 ≤ ''l'' ≤ ''q'', there is an ''R''-multilinear map:

:<math display="block">E^p \times {E^*}^q \to \mathfrak{T}^{p-1}_{q-1}, \, (X_1, \dots, X_p, \omega_1, \dots, \omega_q) \mapsto \langle X_k, \omega_l \rangle X_1\otimes \cdots\otimes \widehat{X_l}\otimes \cdots\otimes X_p \otimes \omega_1\otimes \cdots \widehat{\omega_l}\otimes \cdots\otimes \omega_q</math>

where <math>E^p</math> means <math>\prod_1^p E</math> and the hat means a term is omitted. By the universal property, it corresponds to a unique ''R''-linear map:

:<math display="block">C^k_l: \mathfrak{T}^p_q \to \mathfrak{T}^{p-1}_{q-1}.</math>

It is called the [[tensor contraction|contraction]] of tensors in the index (''k'', ''l''). Unwinding what the universal property says one sees:

:<math display="block">C^k_l(X_1 \otimes \cdots \otimes X_p \otimes \omega_1 \otimes \cdots \otimes \omega_q) = \langle X_k, \omega_l \rangle X_1 \otimes \cdots \widehat{X_l} \cdots \otimes X_p \otimes \omega_1 \otimes \cdots \widehat{\omega_l} \cdots \otimes \omega_q.</math>

'''Remark''': The preceding discussion is standard in textbooks on differential geometry (e.g., Helgason<!-- perhaps Kobayashi-Nomizu? -->). In a way, the sheaf-theoretic construction (i.e., the language of [[sheaf of modules]]) is more natural and increasingly more common; for that, see the section {{section link||Tensor product of sheaves of modules}}.

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In general,

:<math display="block">-\otimes_R-:\text{Mod-}R\times R\text{-Mod}\longrightarrow \mathrm{Ab}</math>

is a [[bifunctor]] which accepts a right and a left ''R'' module pair as input, and assigns them to the tensor product in the [[category of abelian groups]].

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By fixing a right ''R'' module ''M'', a functor

:<math display="block">M\otimes_R-:R\text{-Mod} \longrightarrow \mathrm{Ab}</math>

arises, and symmetrically a left ''R'' module ''N'' could be fixed to create a functor

:<math display="block">-\otimes_R N:\text{Mod-}R \longrightarrow \mathrm{Ab}.</math>

Unlike the [[Hom bifunctor]] <math>\mathrm{Hom}_R(-,-),</math> the tensor functor is [[covariant functor|covariant]] in both inputs.

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==Additional structure==

If ''S'' and ''T'' are commutative ''R''-algebras, then {{nowrapmath|''S'' ⊗_''R'' ''T''}} will be a commutative ''R''-algebra as well, with the multiplication map defined by {{nowrapmath|1=(''m''₁ ⊗ ''m''₂) (''n''₁ ⊗ ''n''₂) = (''m''₁''n''₁ ⊗ ''m''₂''n''₂)}} and extended by linearity. In this setting, the tensor product become a [[fibered coproduct]] in the category of ''R''-algebras.<!--Note that any ring is a '''Z'''-algebra, so we may always take {{nowrapmath|''M'' ⊗_'''Z''' ''N''}}.-->

If ''M'' and ''N'' are both ''R''-modules over a commutative ring, then their tensor product is again an ''R''-module. If ''R'' is a ring, ''_RM'' is a left ''R''-module, and the [[commutator]]

{{block indent|em=1.5|text=''rs'' − ''sr''}}

:''rs'' − ''sr''

of any two elements ''r'' and ''s'' of ''R'' is in the [[Annihilator (ring theory)|annihilator]] of ''M'', then we can make ''M'' into a right ''R'' module by setting

{{block indent|em=1.5|text=''mr'' = ''rm''.}}

:''mr'' = ''rm''.

The action of ''R'' on ''M'' factors through an action of a quotient commutative ring. In this case the tensor product of ''M'' with itself over ''R'' is again an ''R''-module. This is a very common technique in commutative algebra.

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===Tensor product of complexes of modules===

If ''X'', ''Y'' are complexes of ''R''-modules (''R'' a commutative ring), then their tensor product is the complex given by

<math display="block">(X \otimes_R Y)_n = \sum_{i + j = n} X_i \otimes_R Y_j,</math>

:<math>(X \otimes_R Y)_n = \sum_{i + j = n} X_i \otimes_R Y_j,</math>

with the differential given by: for ''x'' in ''X''_''i'' and ''y'' in ''Y''_''j'',

<math display="block">d_{X \otimes Y} (x \otimes y) = d_X(x) \otimes y + (-1)^i x \otimes d_Y(y).</math><ref>{{harvnb|May|ch. 12 §3}}</ref>

:<math>d_{X \otimes Y} (x \otimes y) = d_X(x) \otimes y + (-1)^i x \otimes d_Y(y).</math><ref>{{harvnb|May|ch. 12 §3}}</ref>

For example, if ''C'' is a chain complex of flat abelian groups and if ''G'' is an abelian group, then the homology group of <math>C \otimes_{\Z } G</math> is the homology group of ''C'' with coefficients in ''G'' (see also: [[universal coefficient theorem]].)

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In this setup, for example, one can define a [[tensor field]] on a smooth manifold ''M'' as a (global or local) section of the tensor product (called '''tensor bundle''')

<math display="block">(T M)^{\otimes p} \otimes_{O} (T^* M)^{\otimes q}</math>

:<math>(T M)^{\otimes p} \otimes_{O} (T^* M)^{\otimes q}</math>

where ''O'' is the [[sheaf of rings]] of smooth functions on ''M'' and the bundles <math>TM, T^*M</math> are viewed as [[locally free sheaf|locally free sheaves]] on ''M''.<ref>See also [https://www.encyclopediaofmath.org/index.php/Tensor_bundle Encyclopedia of Mathematics - Tensor bundle]</ref>

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==Notes==

{{reflist|group=proof}}

<references/>

==References==

<references/>

* Bourbaki, ''Algebra''

*{{citation|first=Sigurdur|last=Helgason|title=Differential geometry, Lie groups and symmetric spaces|year=1978| publisher=Academic Press| isbn=0-12-338460-5}}

*{{citation|first1=D.G.|last1=Northcott|authorlink1=Douglas Northcott|title=Multilinear Algebra|publisher=Cambridge University Press| year=1984|isbn=613-0-04808-4}}.

*{{citation|first1=Michiel|last1=Hazewinkel|authorlink1=Michiel Hazewinkel|first2=Nadezhda Mikhaĭlovna| last2=Gubareni |authorlink2 =Nadezhda Mikhaĭlovna|first3=Nadiya|last3=Gubareni|authorlink3=Nadiya Gubareni|first4=Vladimir V.| last4=Kirichenko |authorlink4= Vladimir V. Kirichenko|title=Algebras, rings and modules|publisher=Springer|year=2004|isbn=978-1-4020-2690-4}}.