Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Tensor product
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
=== 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.
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)