# Help me understand the tensor product

I have several books and other literature that define the tensor product, but I understand none of them. Since this really concerns one topic, namely understanding the construction of and arithmetic of (with?) tensor products, I'll pose a few questions here instead of asking them separately. I don't feel that the specific case of tensoring vector spaces helps with understanding the general construction, so my questions will deal with tensoring a right $R$-module with a left $R$-module over the ring $R$, which we do not assume to be commutative.

1) Now then, the first step is to form the free abelian group $F(M \times N)$. I've never stumbled upon free abelian groups prior to reading about the tensor product, but it seems simple enough. Correct me if I'm wrong, but $F(M \times N) = \{ \sum a_{ij}(m,n) \mid a_{ij} \in \mathbb{Z} \}$, the sums ranging over all $(m,n) \in M \times N$. But some authors say something else, which I trust is the same thing but nevertheless confuses me. Keith Conrad (assuming $R$ is commutative) says $$R=\bigoplus_{(m,n) \in M \times N} R \delta_{(m,n)}$$ without any mention of what $\delta_{(m,n)}$ is, but it's probably similar to what I have in a compendium (which also assumes the ring is commutative), "notation tagging the component that corresponds to the element $(m,n) \in M \times N$". What does that even mean, and why is that particular notation only used when $R$ is assumed commutative?

2) Next define the subgroup $S \subset F(M \times N)$ generated by "all elements of the following three types" $(a,b+b') - (a,b) - (a,b')$ etc. Well, what does $-$ mean? Is an equivalence relation meant? If so, when are they considered equivalent? I've only seen this along the lines of "define $x,y$ to be equivalent if property $P(x,y)$ holds".

3) $M \otimes N$ is an abelian group. Fine, so how are elements "added"? No mention of this is made in any of my literature, except that $a \otimes (b+b') = a \otimes b + a \otimes b'$, but how about $a \otimes b + c \otimes d$?

4) So-called elementary or pure tensors. Conrad writes "Tensors in $M \otimes_R N$ that have the form $m \otimes n$ are called elementary tensors". The elementary tensors span the tensor product (right?), but what elements of $F(M \times N)$ wind up as elementary tensors? While my compendium (which, as Conrad, assumes the ring to be commutative) also makes note of elementary tensors, Rotman ("Introduction to Homological Algebra"), not assuming commutativity, makes no mention of them, but says: "Since $A \otimes_R B$ is generated by the elements of the form $a \otimes b$, every $u \in A \otimes_R B$ has the form $u = \sum_i a_i \otimes b_i$. This expression is not unique..." The other names for "elementary tensor" which Conrad lists are not in the index to Rotman's book, so are elementary tensors only relevant to tensoring over a commutative ring?

Thanks in advance for any help. I've had some homework regarding tensors and I'm unable to take even a first step in answering the questions, so I have to learn tensor products somehow.

• For your second question, have you come across the notation of a subgroup generated by a set? That's what's happening there. The set $S$ is now all elements of the form you described. This may be helpful too: en.wikipedia.org/wiki/Generating_set_of_a_group – user38268 Sep 27 '13 at 1:35
• Also TBH I have never ever used such a construction of a tensor product in my life (because at the end one cares about the universal property). Note that for finite dimensional spaces, the tensor product IIRC can also be constructed as the dual space of all bilinear maps $V \times W \to F$. – user38268 Sep 27 '13 at 1:36

2. The subgroup generated by those elements is precisely the set that consists of sums and inverses of elements of the form $(a,b+b')-(a,b)-(a,b')$, etc. The $-$ that goes with the element $(a,b)$, for example, means the additive inverse of $(a,b)$ in $F(M\times N)$. Since $F(M\times N)$ is abelian, and we take the subgroup generated by the elements above, then said subgroup is normal and so if we break up $F(M\times N)$ into cosets, this gives an equivalence relation on $F(M\times N)$. Explicitly, if $H$ is the subgroup $H=\langle (a,b+b')-(a,b)-(a,b'),\mbox{ etc...}:a,b,b'\in M\times N\rangle$ (the smallest subgroup that contains the elements you defined), then if $x,y\in F(M\times N)$, $x\sim y$ if $x-y\in H$. When we mod out by this equivalence relation (which is the same as taking the quotient $F(M\times N)/H$) we basically declare the elements you defined to be 0.

3. If there is no obvious relation between $a,b,c,d$, then there is no obvious way to add $a\otimes b+c\otimes d$. It's just left like that, and we see it as an element of the tensor product anyways.

4. The elementary tensors come from elements of the form $(a,b)\in M\times N\subseteq F(M\times N)$. In the quotient these turn into $a\times b$. The term elementary tensor just means that these are basically the "building blocks" for the tensor product. You are right to say that these generate the tensor product.

A bit of motivation for the tensor product:

If we have the product $M\times N$, we would somehow like a module $T$ such that a bilinear map $\phi:M\times N\to P$ turns into an $R$-module morphism $T\to P$. The problem with $M\times N$ is that if $(m,n),(m',n')\in M\times N$ then $$\phi((m,n)+(m',n'))=\phi(m+m',n+n')=\phi(m,n+n')+\phi(m',n+n')=\phi(m,n)+\phi(m',n)+\phi(m,n')+\phi(m',n'),$$ which isn't the same as $\phi(m,n)+\phi(m',n')$. The way to fix this is by adding the elements of $M\times N$ a different way and modding out by others. For example, instead of $(m,n)+(m',n')=(m+m',n+n')$, we want $(m,n)+(m',n)"="(m+m',n)$, etc. If we had this then $\phi$ would be a morphism of $R$-modules. The way to get this relation is to declare that $(m,n)+(m',n)-(m+m',n)"="0$, that is, we mod out by the relations you put above. This is how we obtain the tensor product.

Remark The motivation to define tensor product I put above basically characterizes the tensor product (this is the universal property that defines the tensor product).

• I don't know how much this will help or further confuse, but there is no 'the tensor product' but rather 'a tensor product'. The notion of tensor product is not uniquely defined. There are plenty of different constructions that will produce tensor products for $M$ and $N$. Thus, writing $M\otimes N$ assumes you also made some choice of constructing this particular tensor product. What is guaranteed is that all constructions yield isomorphic objects. But that does not mean you can pretend they are all equal. It is the universal property of the tensor product that guarantees this uniqueness. – Ittay Weiss Sep 27 '13 at 1:10
• it is also almost always much easier to work with the universal property rather than with a particular construction of a tensor product. The sooner you'll get used to reasoning about tensors using the universal property the better. – Ittay Weiss Sep 27 '13 at 1:11
• Yes, the universal property characterizes the tensor product and everyone works with that. And I say "the" as an abuse of terminology by what you say. Everyone says "the", however, in the same sense that there is only "one" group of order $p$ for every prime $p$. – rfauffar Sep 27 '13 at 1:37

1) What he means when he writes $F_R(M\times N) = \bigoplus R \delta_{(m,n)}$ is that you're taking a direct sum of as many copies of $R$ as there are elements $(m,n) \in M \times N$. By definition, an element in the direct sum of an infinity of modules is a finite $R$-linear combination of the modules involved (here, they are all copies of $R$), so a generic element in $F_R(M\times N)$ would look like $r_1 \delta_{(m_1,n_1)} + \cdots + r_k \delta_{(m_k,n_k)}$. If you want, you can view such an element as an $R$-valued sequence indexed by $M \times N$ with finitely many non-zero elements.

2) The $-$ is the symbol for the additive inverse in the free module. Thinking in terms of sequences above, the addition is componentwise addition. So what you do is construct the submodule generated by elements of the given form so that you get exactly the bilinearity relations that you want in the quotient module.

3) The tensor product is constructed as a quotient module $F_R(M\times N)/D$, so the operation is the usual coset addition: $(a + D) + (b + D) := (a + b) + D$. If we want to be explicit, like above, we can write $$(r_1 \delta_{(m_1,n_1)} + \cdots + r_k \delta_{(m_k,n_k)} + D) + (r'_1 \delta_{(m'_1,n'_1)} + \cdots + r'_l \delta_{(m'_l,n'_l)} + D) = (r_1 \delta_{(m_1,n_1)} + \cdots + r_k \delta_{(m_k,n_k)} + r'_1 \delta_{(m'_1,n'_1)} + \cdots + r'_l \delta_{(m'_l,n'_l)}) + D.$$

4) The elementary tensors are those which come from the generators $\delta_{(m,n)} \in F_R(M\times N)$. Elements of the form $r \delta_{(m,n)}$ also reduce to elementary tensors in the quotient, because the identifications give $r(m \otimes n) = (rm) \otimes n = m' \otimes n$.

It is okay to struggle with the construction of the tensor product at first, but keep in mind that the construction is not very important; what really matters is the universal property of the tensor product, which in essence states that bilinear mappings $M\times N \rightarrow L$ (or, equivalently, linear mappings $M \rightarrow \operatorname{Hom}(N,L)$) are naturally the same thing as linear mappings $M \otimes N \rightarrow L$.