0
$\begingroup$

I am reading about tensor products of modules in Dummit and Foote, and I am a little confused about what the elements of the tensor product actually look like.

The construction I am referring to starts with a ring $S$ with unity, and a subring $R$ with $1_S=1_R$. We then take an abelian group $N$ that is an $R$-module and consider the a map $S\times N$ into $N$ where the image of $(s,n)$ is denoted $sn$. We then consider the free $\mathbb{Z}$-module (the free abelian group) on the set $S\times N$, which is the collection of all finite commuting sums of elements of the form $(s_i,n_i)$, where $s_i\in S, n_i\in N$.

1) Does this mean that in this free abelian group, denoted $\mathbb{Z}(S\times N$), elements are of the form $r_1(s_1,n_1)+\cdots +r_n(s_n,n_n)$ where $r_i\in R$? Can we combine the summands into one element of the form $(s,n)$? The next line states that there are "no relations between any distinct pairs $(s,n)$ and $(s',n')$. What does this mean?

Next, we consider the subgroup $H$ of the free abelian group above that is generated by all elements of the form $(s_1+s_2,n)-(s_1,n)-(s_2,n), (s,n_1+n_2)-(s,n_1)-(s,n_2)$, and $(sr,n)-(s,rn)$ for $s,s_1,s_2\in S$, $n,n_1,n_2\in N$, and $r\in R$. We then consider the quotient of $\mathbb{Z}(S\times N)$ by $H$ to obtain the tensor product of $S$ and $N$ over $R$, denoted $S\otimes_RN$.

Now, elements of the tensor product are denoted $s\otimes n$, and this denotes the coset containing $(s,n)$. However, the book then says that elements of the tensor product can be written as finite sums of "finite tensors" which are of the form $s\otimes n$. This is confusing to me. An element of the tensor product is both a simple tensor and a finite sum of simple tensors?

$\endgroup$

1 Answer 1

1
$\begingroup$

1) You cannot combine summands into one element. This is what they mean when they say that there are no relations between the pairs. You could say that all of the pairs are linearly independent.

The elements of $\Bbb{Z}(S \times N)$ are each of the form $z_1(s_1,n_1)+\cdots +z_n(s_n,n_n)$ (where $z_i \in \Bbb{Z}$) by definition. So these represent the cosets. A "simple tensor" is a coset represented by one pair. But not all cosets are represented by only one pair. The coset containing $z_1(s_1,n_1)+\cdots +z_n(s_n,n_n)$ could be written as $z_1(s_1 \otimes n_1)+\cdots +z_n(s_n \otimes n_n)$. The reason is this: let $\overline{v}$ be denote the coset containing $v$. Then $\overline{z_1(s_1,n_1)+\cdots +z_n(s_n,n_n)}=z_1\overline{(s_1,n_1)}+\cdots +z_n\overline{(s_n,n_n)}=z_1(s_1 \otimes n_1)+\cdots +z_n(s_n \otimes n_n)$, since this is how cosets are added in a quotient module.

$\endgroup$
3
  • $\begingroup$ How do you justify the last statement, namely that the coset containing $z_1(s_1,n_1)+⋯+z_n(s_n,n_n)$ can be written as $z_1(s_1\otimes n_1)+\cdots + z_n(s_n\otimes n_n)$? $\endgroup$
    – ponchan
    Apr 7, 2018 at 19:57
  • $\begingroup$ Further, how would one add two elements of $\mathbb{Z}(S\times N)$? $\endgroup$
    – ponchan
    Apr 7, 2018 at 21:37
  • 1
    $\begingroup$ Ok, I updated the solution. The way that you add elements of $\Bbb{Z}$ is by just leaving your sums as is. For example, $(s_1,n_1) + (s_2,n_2)$ will not simplify further unless $s_1 = s_2$ and $n_1 = n_2$. $\endgroup$ Apr 7, 2018 at 23:41

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.