1
$\begingroup$

I am really struggling to understand several parts of the definition of tensor product given in my lecture notes:

Definition of the tensor product *Denote by L the free A-module with a basis consisting of elements l$_{m,n}$ indexed by elements of M x N. So an arbitrary element of L is a finite sum $\Sigma$ a$_i$l$_{m_i,n_i}$ with a$_i$$\in$ A, m$_i$$\in$ M and n$_i$$\in$ N. Let K be the A-submodule of L generated by elements: l$_{m_1+m_2,n}$ - l$_{m_1,n}$ - l$_{m_2,n}$; l$_{m,n_1+n_2}$ - l$_{m,n_1}$ - l$_{m,n_2}$ ; l$_{am,n}$ - al$_{m,n}$; l$_{m,an}$ - al$_{m,n}$ (for all a $\in$ A, m $\in$ M and n $\in$ N ). Denote T = L/K. The image of l$_{m,n}$ in T , i.e. the coset l$_{m,n}$ + K is usually denoted by m $\otimes$ n. Since L is generated by l$_{m,n}$, the module T is generated by m $\otimes$ n, i.e. T = {$\Sigma$ a$_i$m$_i$$\otimes$n$_i$ : a$_i$ $\in$ A;m$_i$ $\in$ M; n$_i$ $\in$ N }

These satisfy relations: (m$_1$ + m$_2$) $\otimes$ n = m$_1$ $\otimes$ n + m$_2$ $\otimes$ n; (am) $\otimes$ n = a(m $\otimes$ n); m $\otimes$ (n$_1$ + n$_2$) = m $\otimes$ n$_1$ + m $\otimes$ n$_2$;
m $\otimes$ (an) = a(m $\otimes$ n) (for all a $\in$ A, m $\in$ M and n $\in$ N ). The module T is denoted M $\bigotimes$$_A$ N and is called the tensor product of M and N over A.*


In the notes it is stated, as if obvious, that K is an A-submodule of L and that T = L/K is a module - but I just don't see how.

I also don't understand why the relations are satisfied - I try plugging in some arbitrary values l$_{m_1,n_1}$ etc. to try and prove the relations, but I have no idea how to manipulate these since to me they are just indexed l's which I can't do anything with.

Really appreciate any guidance you can give me

$\endgroup$

1 Answer 1

3
$\begingroup$

The definition of the tensor product is its universal property, which is quite simple. What you are struggling with is the construction of the tensor product - this is something different. If you want to see a construction of the tensor product which avoids free modules at all, see here.

$K$ is by definition a submodule, since it is defined as the submodule generated by some elements. Also, you should be already familiar with the general fact that a quotient of a module by a submodule is again a module. This is done exactly as in linear algebra with vector spaces. See quotient module. Recall that the basic idea of quotient is to make some elements zero. (The idea is not to compute with cosets - this is again just a construction which has no meaning at all.)

The idea behind the construction is that we want to compute with symbols $m \otimes n$ which become bilinear in each variable, for example $(m_1+m_2) \otimes n = m_1 \otimes n + m_2 \otimes n$, i.e. $(m_1 + m_2) \otimes n - m_1 \otimes n - m_2 \otimes n = 0$. For this, we consider the free module generated by symbols $(m,n)$ and just mod out the smallest submodule containing elements such as $(m_1+m_2,n)-(m_1,n)-(m_2,n)$ (and other ones). If $m \otimes n$ denotes the image of $(m,n)$, we have then $(m_1 + m_2) \otimes n - m_1 \otimes n - m_2 \otimes n = 0$ by construction.

$\endgroup$
2
  • $\begingroup$ Thanks for your reply. So let me see if I've understood correctly - the generators of K are chosen specially so that their image in L/K is the zero element of L/K (e.g. let k = l$_{m_1+m_2,n}$ - l$_{m_1,n}$ - l$_{m_2,n}$ $\in$ K. Then k + K $\in$ L/K equals K i.e. 0$_{L/K}$, so we can rearrange this element to make the relations satisfied)? $\endgroup$ Oct 14, 2014 at 15:48
  • $\begingroup$ Yes, that's right. $\endgroup$ Oct 14, 2014 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.