Editing Tensor product (section)

== Tensor products of modules over a ring ==
{{main|Tensor product of modules}}
The tensor product of two [[module (mathematics)|modules]] {{math|''A''}} and {{math|''B''}} over a ''[[commutative ring|commutative]]'' [[ring (mathematics)|ring]] {{math|''R''}} is defined in exactly the same way as the tensor product of vector spaces over a field:
<math display="block">A \otimes_R B := F (A \times B) / G ,</math>
where now <math>F(A \times B)</math> is the [[Free module|free {{math|''R''}}-module]] generated by the cartesian product and {{math|''G''}} is the {{math|''R''}}-module generated by [[Tensor_product_of_modules#Balanced_product|these relations]].

More generally, the tensor product can be defined even if the ring is [[Noncommutative ring|non-commutative]]. In this case {{math|''A''}} has to be a right-{{math|''R''}}-module and {{math|''B''}} is a left-{{math|''R''}}-module, and instead of the last two relations above, the relation:
<math display="block">(ar,b)\sim (a,rb)</math>
is imposed. If {{math|''R''}} is non-commutative, this is no longer an {{math|''R''}}-module, but just an [[abelian group]].

The universal property also carries over, slightly modified: the map <math>\varphi : A \times B \to A \otimes_R B</math> defined by <math>(a, b) \mapsto a \otimes b</math> is a [[Tensor product of modules#Balanced product|middle linear map]] (referred to as "the canonical middle linear map"<ref>
{{cite book|last=Hungerford|first=Thomas W.|title=Algebra|
publisher=Springer|year=1974|isbn=0-387-90518-9}}</ref>); that is, it satisfies:<ref name=chen>
{{citation|last=Chen|first=Jungkai Alfred|title=Advanced Algebra II|chapter=Tensor product|chapter-url=http://www.math.ntu.edu.tw/~jkchen/S04AA/S04AAL10.pdf|type=lecture notes|date=Spring 2004|place=National Taiwan University|url-status=live|archive-url=https://web.archive.org/web/20160304040639/http://www.math.ntu.edu.tw/~jkchen/S04AA/S04AAL10.pdf|archive-date=2016-03-04}}</ref>
<math display="block">\begin{align}
  \varphi(a + a', b) &= \varphi(a, b) + \varphi(a', b) \\
  \varphi(a, b + b') &= \varphi(a, b) + \varphi(a, b') \\
      \varphi(ar, b) &= \varphi(a, rb)
\end{align}</math>

The first two properties make {{math|''φ''}} a bilinear map of the [[abelian group]] {{tmath|1= A \times B }}. For any middle linear map <math>\psi</math> of {{tmath|1= A \times B }}, a unique group homomorphism {{math|''f''}} of <math>A \otimes_R B</math> satisfies {{tmath|1= \psi = f \circ \varphi }}, and this property determines <math>\varphi</math> within group isomorphism. See the [[tensor product of modules|main article]] for details.

=== Tensor product of modules over a non-commutative ring ===

Let ''A'' be a right ''R''-module and ''B'' be a left ''R''-module. Then the tensor product of ''A'' and ''B'' is an abelian group defined by:
<math display="block">A \otimes_R B := F (A \times B) / G</math>
where <math>F (A \times B)</math> is a [[free abelian group]] over <math>A \times B</math> and G is the subgroup of <math>F (A \times B)</math> generated by relations:
<math display="block">\begin{align}
  &\forall a, a_1, a_2 \in A, \forall b, b_1, b_2 \in B, \text{ for all } r \in R:\\
  &(a_1,b) + (a_2,b) - (a_1 + a_2,b),\\
  &(a,b_1) + (a,b_2) - (a,b_1+b_2),\\
  &(ar,b) - (a,rb).\\
\end{align}</math>

The universal property can be stated as follows. Let ''G'' be an abelian group with a map <math>q:A\times B \to G</math> that is bilinear, in the sense that:
<math display="block">\begin{align}
  q(a_1 + a_2, b) &= q(a_1, b) + q(a_2, b),\\
  q(a, b_1 + b_2) &= q(a, b_1) + q(a, b_2),\\
         q(ar, b) &= q(a, rb).
\end{align}</math>

Then there is a unique map <math>\overline{q}:A\otimes B \to G</math> such that <math>\overline{q}(a\otimes b) = q(a,b)</math> for all <math>a \in A</math> and {{tmath|1= b \in B }}.

Furthermore, we can give <math>A \otimes_R B</math> a module structure under some extra conditions:
# If ''A'' is a (''S'',''R'')-bimodule, then <math>A \otimes_R B</math> is a left ''S''-module, where {{tmath|1= s(a\otimes b):=(sa)\otimes b }}.
# If ''B'' is a (''R'',''S'')-bimodule, then <math>A \otimes_R B</math> is a right ''S''-module, where {{tmath|1= (a\otimes b)s:=a\otimes (bs) }}.
# If ''A'' is a (''S'',''R'')-bimodule and ''B'' is a (''R'',''T'')-bimodule, then <math>A \otimes_R B</math> is a (''S'',''T'')-bimodule, where the left and right actions are defined in the same way as the previous two examples.
# If ''R'' is a commutative ring, then ''A'' and ''B'' are (''R'',''R'')-bimodules where <math>ra:=ar</math> and {{tmath|1= br:=rb }}. By 3), we can conclude <math>A \otimes_R B</math> is a (''R'',''R'')-bimodule.

=== Computing the tensor product ===
For vector spaces, the tensor product <math>V \otimes W</math> is quickly computed since bases of {{math|''V''}} of {{math|''W''}} immediately determine a basis of {{tmath|1= V \otimes W }}, as was mentioned above. For modules over a general (commutative) ring, not every module is free. For example, {{math|'''Z'''/''n'''''Z'''}} is not a free abelian group ({{math|'''Z'''}}-module). The tensor product with {{math|'''Z'''/''n'''''Z'''}} is given by:
<math display="block">M \otimes_\mathbf{Z} \mathbf{Z}/n\mathbf{Z} = M/nM.</math>

More generally, given a [[presentation of a module|presentation]] of some {{math|''R''}}-module {{math|''M''}}, that is, a number of generators <math>m_i \in M, i \in I</math> together with relations:
<math display="block">\sum_{j \in J} a_{ji} m_i = 0,\qquad a_{ij} \in R,</math>
the tensor product can be computed as the following [[cokernel]]:
<math display="block">M \otimes_R N = \operatorname{coker} \left(N^J \to N^I\right)</math>

Here {{tmath|1= N^J = \oplus_{j \in J} N }}, and the map <math>N^J \to N^I</math> is determined by sending some <math>n \in N</math> in the {{math|''j''}}th copy of <math>N^J</math> to <math>a_{ij} n</math> (in {{tmath|1= N^I }}). Colloquially, this may be rephrased by saying that a presentation of {{math|''M''}} gives rise to a presentation of {{tmath|1= M \otimes_R N }}. This is referred to by saying that the tensor product is a [[right exact functor]]. It is not in general left exact, that is, given an injective map of {{math|''R''}}-modules {{tmath|1= M_1 \to M_2 }}, the tensor product:
<math display="block">M_1 \otimes_R N \to M_2 \otimes_R N</math>
is not usually injective. For example, tensoring the (injective) map given by multiplication with {{math|''n''}}, {{math|''n'' : '''Z''' → '''Z'''}} with {{math|'''Z'''/''n'''''Z'''}} yields the zero map {{math|0 : '''Z'''/''n'''''Z''' → '''Z'''/''n'''''Z'''}}, which is not injective. Higher [[Tor functor]]s measure the defect of the tensor product being not left exact. All higher Tor functors are assembled in the [[derived tensor product]].