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Equality of two notions of tensor products over a commutative ring

Let $A$ be a commutative ring. There are two common definitions of tensor product of two $A$-modules $M$ and $N$ in terms of universal property. One definition defines it as a universal object of a map $$\tau:M\times N\rightarrow G,$$ where $G$ is an abelian group, and $\tau$ is $A$-balanced, i.e.,

  1. $\tau(m+m',n)=\tau(m,n)+\tau(m',n)$,
  2. $\tau(m,n+n')=\tau(m,n)+\tau(m,n')$,
  3. $\tau(a\cdot m,n)=\tau(m,a\cdot n)$.

Another definition defines it as a universal object of a map $$\sigma:M\times N\rightarrow L,$$ where $L$ is an $A$-module, and $\sigma$ is $A$-bilinear, i.e.,

  1. $\sigma(m+m',n)=\sigma(m,n)+\sigma(m',n)$,
  2. $\sigma(m,n+n')=\sigma(m,n)+\sigma(m,n')$,
  3. $\sigma(a\cdot m,n)=a\cdot\sigma(m,n)$,
  4. $\sigma(m,a\cdot n)=a\cdot\sigma(m,n)$.

I always thought they were equivalent definitions, but now I made a couple of observations:

  1. Let $\tau:M\times N\rightarrow G $ be a universal object in the first sense. If $G$ happens to be an $A$-module and $\tau$ happens to be $A$-bilinear, then $\tau$ is a universal object in the second sense.
  2. Let $\sigma:M\times N\rightarrow L$ be a universal object in the second sense. It doesn't seem to follow that $\sigma$ is a universal object in the first sense.

So intuitively, it seems that "the set of the first universal objects contains the set of the second universal objects". Am I correct? If so, then it seems necessary to distinguish the two definitions of tensor product.

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marked as duplicate by M Turgeon, ashpool, Zhen Lin, t.b., sdcvvc Aug 28 '12 at 1:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

@Ted, The first definition is certainly not "incorrect," since it is the definition most commonly found in textbooks, and also Wikipedia. –  ashpool Aug 24 '12 at 15:52
This question was raised recently:… –  M Turgeon Aug 24 '12 at 15:55
@Ted The reason why the first definition is not incorrect is because it is the only one which can be generalised to non-commutative rings –  M Turgeon Aug 24 '12 at 15:56
@M Turgeon, Thanks for the reference. Let me see if I can close down my question... –  ashpool Aug 24 '12 at 16:00
@Ted That's exactly my point :) –  M Turgeon Aug 24 '12 at 16:04

2 Answers 2

up vote 2 down vote accepted

[It is quite elementary. But if you try to keep things apart by using different symbols for different structures, it is harder to see it. I will stick to your notation, but use $UM$ for the underlying abelian group of a module, $U\sigma$ for $\sigma$ regarded as a morphism of abelian groups, $FG$ for a group turned into a module and $F\tau$ for a group morphism treated as a linear map or bilinear map of modules. You can disregard the $U$'s and $F$s' to get the basic idea.]

Consider the universal $A$-balanced map $\tau_0 : M \times N \to M \otimes_A N$. I show that it is a universal $A$-bilinear map. It is helpful to use the notation $m \otimes n$ for $\tau_0(m,n)$. Note that everything in $M \otimes_A N$ is of the form $m \otimes n$.

$M \otimes_A N$ is an abelian group. But, because $am \otimes n = m \otimes an$, it has already been set up to become an $A$-module. Just define scalar multiplication as $a \cdot (m \otimes n) = am \otimes n$.

  • Notice that $a \cdot (m \otimes n) = m \otimes an$. Just transitivity of equality at play here.
  • Notice that $\tau_0$ is now an $A$-bilinear map.

[If you want to check the types bureaucracy, we have defined $F(M \otimes_A N)$, by extending $M \otimes_A N$ with scalar multiplication, and $F\tau_0 : M \times N \to F(M \otimes_A N)$ which is the same as $\tau_0$ regarded as a bilinear map.]

Nxt we want to argue that $F\tau_0$ is universal among all $A$-bilinear maps. We know that all bilinear maps have underlying balanced maps . So, given any bilinear $\sigma : M \times N \to K$, we have an underlying balanced map $U\sigma : M \times N \to UK$. There is a unique abelian group morphism $\tau' : M \otimes_A N \to UK$ such that $U \sigma = \tau' \circ \tau_0$. It is in fact defined by $\tau'(m \otimes n) = U\sigma(m,n)$.

  • Notice that $\tau'$ is bilinear. $\tau'(am \otimes n) = \tau'(a \cdot (m \otimes n)) = a \cdot \tau'(m \otimes n)$ because $a \cdot (m \otimes n)$ is $am \otimes n$ by definition. Similarly, $\tau'(m \otimes an) = a \cdot \tau'(m \otimes n)$.

[To get the types bureaucracy right, we have shown $\sigma = FU\sigma = F\tau' \circ F\tau_0$. And, this $F\tau'$ is the unique factor for $FU\sigma$ just because $\tau'$ was the unique factor for $U\sigma$.]

So, we have shown that every bilinear map $\sigma : M \times N \to K$ uniquely factors through $F\tau_0 : M \times N \to F(M \otimes_A N)$, ergo $F(M \otimes_A N)$ is the tensor product (in the sense of being the codomain of the the universal bilinear map).

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These two tensor products are kind of different from each other in a categorical sense.

The first can be done with any ring $R$, commutative or not, and creates the abelian group $M\otimes_R N$ from the right module $M$ and left module $N$. This version is coming from the product category of right $R$ modules and left $R$ modules and going into the category of abelian groups.

If $S$ is another ring, and $M$ is an $S-R$ bimodule now, and $N$ is still a left $R$ module, then you can also form $M\otimes_R N$, but now the tensor product has a natural left $S$ module structure. So, this functor is most appropriately viewed as being from the product of the bimodule category with the left $R$ module category, into the category of left $S$ modules.

When you let $R$ be commutative all of these distinctions become blurred (as is often the case with commutativity.) The object you get with the first definition will be naturally compatible with the naive $R$ bimodule structure on each of the modules. So then, not only will the tensor product be an abelian group, it'll have an $R-R$ bimodule structure. It all depends on your initial point of view which categories you are shuttling between.

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