Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Extension of scalars ... D&F, Section 10.4: Tensor Products

  1. Apr 7, 2016 #1
    I am reading Dummit and Foote's book: Abstract Algebra ... ... and am currently focused on Section 10.4 Tensor Products of Modules ... ...

    I have a basic question regarding the extension of the scalars ...

    Dummit and Foote's (D&Fs) exposition regarding extension of the scalars reads as follows:



    ?temp_hash=91100eea6862f8bf7a811dd7da0380d4.png
    ?temp_hash=91100eea6862f8bf7a811dd7da0380d4.png
    ?temp_hash=91100eea6862f8bf7a811dd7da0380d4.png




    Question 1

    In the above text from D&F (towards the end of the quote) we read the following:

    "... ... Suppose now that ##\sum s_i \otimes n_i = \sum s'_i \otimes n'_i##

    are two representations for the same element in ##S \otimes_R N##. Then ##\sum (s_i, n_i) - \sum (s'_i, n'_i)## is an element of ##H## ... ... ... "


    Can someone please explain exactly why ##\sum s_i \otimes n_i = \sum s'_i \otimes n'_i## in ##S \otimes_R N## implies that ##\sum (s_i, n_i) - \sum (s'_i, n'_i)## is an element of ##H## ... ...



    [ ***Note*** I am a little unsure of the general nature of elements of ##H## ... and even more unsure of the nature of elements of ##S \otimes_R N## ... ... ]




    Question 2

    In the above text from D&F (towards the end of the quote) we read the following:

    "... ... for any ##s \in S## also ##\sum (ss_i, n_i) - \sum (ss'_i, n'_i)## is an element of ##H##. But this means that ##\sum ss_i \otimes n_i - \sum ss'_i \otimes n'_i)## in ##S \otimes_R N## ... ... "


    Can someone please explain exactly why if ##\sum (ss_i, n_i) - \sum (ss'_i, n'_i)## is an element of ##H## ... ... that we then have ##\sum ss_i \otimes n_i - \sum ss'_i \otimes n'_i)## in ##S \otimes_R N## ... ...

    ... ... although the above seems right, why exactly is it the case ... ?




    Hope someone can help ... I suspect my main problem is the general nature and characteristics of elements of ##H## and elements of ##S \otimes_R N##

    Peter
     

    Attached Files:

  2. jcsd
  3. Apr 8, 2016 #2

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It will help if we give a name, say ##B##, to the free ##\mathbb{Z}##-module on the set ##S\times N##. ##B## is the set of all maps from ##S\times N## to ##\mathbb{Z}## with finite support. This is directly analogous to the set that was constructed in your Cooperstein work, of maps from ##V_1\times V_2## to ##F##. It is easy to show that ##B## is a ##\mathbb{Z}##-module.

    Abelian groups can be considered as ##\mathbb{Z}##-modules, where the group operation is the module addition and multiplication of a group member by an integer ##z## corresponds to adding together ##z## copies of the integer. So we can use the terms Abelian Group and ##\mathbb{Z}##-module interchangeably.

    The author uses ##(s,n)## to denote an element of ##B##, and by that they actually mean the function ##\chi_{(s,n)}##, using the Cooperstein notation, ie the function that returns zero on all elements of ##S\times N## except for ##(s,n)##. The fact that ##(s,n)## is used both for elements of ##S\times N## and elements of ##B## can be resolved by the fact that the context will usually require it to mean specifically one of those two.

    ##H## is the submodule of ##B## generated by elements of the form set out in 10.3. These are the elements that would be in a loose sense equivalent to zero if we were to forget that ##H## elements are just formal sums and instead allow linear algebra type rules of addition and scalar multiplication to apply. The analogous construction is used when constructing tensor products of vector spaces using the quotient method.

    As is always the case with quotients, the subgroup over which we have taken the quotient - ##H## in this case - is the zero element of the quotient group.

    ##S\otimes_R N## (which I'll call ##T## for short) is then the quotient group (or ##\mathbb{Z}##-module) ##B/H##. So elements of ##T## are of the form ##\chi_{(s,n)}+H## where ##(s,n)\in S\times N##. That element can also be denoted ##s\otimes n##.

    If ##\sum s_i\otimes n_i=\sum s'_i\otimes n'_i##
    then ##\sum s_i\otimes n_i-\sum s'_i\otimes n'_i=0##
    and since the terms in this equation are elements of the tensor space ##T##, which is a quotient over ##H##, that means that
    ##\sum s_i\otimes n_i-\sum s'_i\otimes n'_i=H##.

    The LHS is
    $$\sum s_i\otimes n_i-\sum s'_i\otimes n'_i=
    \sum \left(\chi_{(s_i,n_i)}+H\right)-\sum \left(\chi_{(s'_i,n'_i)}+H\right)$$
    $$=\sum \left(\chi_{(s_i,n_i)}\right)-\sum\left(\chi_{(s'_i,n'_i)}\right)+H
    =\sum (s_i,n_i)-\sum(s'_i,n'_i)+H$$
    using the author's notation for elements of ##B## in that last step.

    And since this is equal to ##H## that means that ##\sum (s_i,n_i)-\sum(s'_i,n'_i)\in H##
     
  4. Apr 8, 2016 #3

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The answer to Q2 is similar to that to Q1, only in reverse.

    If ##\sum(ss_i,n_i)-\sum(ss'_i,n'_i)\in H## then, noting that the two sides of that equation are elements of ##B##, that means that, in ##T##:

    $$\left[\sum(ss_i,n_i)+H\right]-\left[\sum(ss'_i,n'_i)+H\right]=H=0_T$$

    Changing to the ##\otimes## notation for elements of ##T##, under which ##(s,n)+H\equiv s\otimes n##, we can rewrite this as

    $$\left[\sum(ss_i\otimes n_i)\right]-\left[\sum(ss'_i\otimes n'_i)\right]=0_T$$

    whence
    $$\left[\sum(ss_i\otimes n_i)\right]=\left[\sum(ss'_i\otimes n'_i)\right]$$

    as required.
     
  5. Apr 10, 2016 #4

    Thanks so much Andrew ... finally, after some work, I have followed that ... or most of it ... :smile:

    Just one issue that you may be able to clarify ...

    You write:

    "... ... It will help if we give a name, say ##B##, to the free ##\mathbb{Z}##-module on the set ##S\times N##. ##B## is the set of all maps from ##S\times N## to ##\mathbb{Z}## with finite support. This is directly analogous to the set that was constructed in your Cooperstein work, of maps from ##V_1\times V_2## to ##F##. It is easy to show that ##B## is a ##\mathbb{Z}##-module. ... "


    My question is ... why is the free ##\mathbb{Z}##-module on the set ##S\times N## the set of all maps from ##S\times N## to ##\mathbb{Z}## with finite support. ... ... ? ... why/how is this true ... ...?


    Hope you can explain/clarify ...

    Peter
     
  6. Apr 11, 2016 #5

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    For any type ##T## of algebraic object and any set ##S##, such as ring, ##\mathbb{Z}##-module or group, the 'free ##T## generated by ##S##' is the ##T##-object generated by considering the elements of ##S## to be distinct elements of the ##T##-object, and not allowing any equalities other than those that are specified by the axioms of the type ##T## (eg the Group axioms if ##T##=Group). We contrast this with a common way of specifying ('presenting') a group, which is by specifying generators and relations. For example, the additive Abelian group ##\mathbb{Z}_3## has generators {1,2} and relations 1+1=2,1+2=0,2+2=1. A Free object is one that has no relations. So the free Abelian group on {1,2} is the (infinite) set of all items of the form 1+1+......+1+2+2+.....+2, containing any finite number of 1s and of 2s.
    So you can see that an arbitrary element of a free Abelian group on ##n## generators is a sum like that that includes ##k_1## copies of the first generator through to ##k_{n}## copies of the ##n##th generator.
    That sum can be regarded as a function ##f## from the generating set ##\{g_1,....,g_n\}## to ##\mathbb{Z}## such that ##f(g_r)=k_r##. Those functions can be multiplied by scalars and added together in the usual way that applies to any functions with shared domains and shared numerical ranges. A little work shows that the object thus defined is an Abelian group under addition, and hence a ##\mathbb{Z}##-module.
    This extends naturally to the case where the generating set ##S## is infinite, as long as we require the sums to be of finite length, which means that only a finite number of different generators can be included in a sum, which means that the corresponding function must have finite support.
     
  7. Apr 11, 2016 #6


    Thanks Andrew ... most informative ...

    Just a point ... you write:

    " ... ... For any type ##T## of algebraic object and any set ##S##, such as ring, ##\mathbb{Z}##-module or group, the 'free ##T## generated by ##S##' is the ##T##-object generated by considering the elements of ##S## to be distinct elements of the ##T##-object, and not allowing any equalities other than those that are specified by the axioms of the type ##T## (eg the Group axioms if ##T##=Group). ... ...


    Where does such theory come from ... is it category theory?

    Can you suggest any algebra or category theory texts that cover the material you use in your post ...?

    Peter






    "
     
  8. Apr 11, 2016 #7

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I've never got very far into Category Theory, so in my case it more comes from the bottom up than top down. One comes across free groups when doing group theory and free modules when studying modules. Then one notices the similarities: that they are both objects of the given type (group or module) that have as little structure - in terms of relationships between the different elements - as possible. In fact, they have no structure other than what is imposed by the axioms of the algebraic type of object. One surmises from that that perhaps most algebraic objects could have a 'free' sub-class.

    Vector spaces are all 'free', because the entire algebraic structure of a vector space with basis of a given cardinality is determined by that cardinality and the overarching field, so it would not be possible to have both free and non-free vector spaces over the same set of generators and the same field. That is not the case with modules though, because they have more freedom in defining the result of multiplying a module element by a scalar (an element of the overarching ring).

    I mentioned rings above as a guess but on a quick search, it looks like there may not be such a thing as a free ring. I haven't thought through why that is yet. There is a free ideal ring though.
    Apparently there are free algebras, but I have not had anything to do with them.

    There is a concept in Category Theory of a free object. I expect that is a generalisation of the concepts of free modules, groups and algebras.

    The book from which I got most of my understanding of free objects - via free modules - is Hartley and Hawkes 'Rings Modules and Linear Algebra', which I think is an absolute classic! We studied it in second year uni pure maths in 1984 and I revisited it a couple of years ago and really enjoyed working through it again. It doesn't seem to have dated at all. Two thirds of the book is working towards a single result which is that, given a few regularity conditions, any module with a finite generating set can be expressed as a direct sum of a bunch of (identical) free modules and a bunch of non-free modules (which are called cyclical, for reasons that become apparent). Then the last third makes some really useful applications of the theory, to Abelian groups, deriving Jordan canonical forms of matrices, and vector space automorphisms. It's one of the very few of my texts that I have read from cover to cover.

    The wiki page on free groups is also quite good.
     
  9. Apr 21, 2016 #8
    Thanks Andrew ... most helpful as usual ...

    ... ... will definitely look up Hartley and Hawkes 'Rings Modules and Linear Algebra'

    Peter

    (Sorry for the late reply ... had to leave my island state of Tasmania and was travelling through regional Victoria ... a southern mainland state ... )
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted