# Understanding Tensor Product Construction

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?

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.
• 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 $z_1(s_1\otimes n_1)+\cdots + z_n(s_n\otimes n_n)$? Apr 7, 2018 at 19:57
• Further, how would one add two elements of $\mathbb{Z}(S\times N)$? Apr 7, 2018 at 21:37
• 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 $n_1 = n_2$. Apr 7, 2018 at 23:41