# A tensor product vs the tensor product.

I have been reading about tensor products in different books and struggle with the definition of tensor products in the following sense. There is a common construction of a tensor product as a quotient group. One often uses this construction to prove the existence of a tensor product an then denotes the quotient group with $$M \otimes_R N$$ (assuming the construction was done for $$R$$-modules $$M,N$$) and denotes its elements with $$m \otimes n$$. The question arises whether this tensor product is unique or not and one usually proves that this is unique up to unique isomorphism. This then leads to several authors saying that it is reasonable to speak of the tensor product of $$M$$ and $$N$$. My questions are the following.

$$(1)$$ Why is this a reason to speak of the tensor product? Can't this cause any harm in whatever sense possible (for example lead to contradictions)?

$$(2)$$ Let $$M \otimes_R N$$ denote another tensor product of $$M$$ and $$N$$, then what would the elements $$m \otimes n$$ be? Would they be the image of $$m \otimes n$$ (in the quotient group construction) under the unique isomorphism?

$$(3)$$ Would it be a possible approach to call the construction of the quotient group the tensor product of $$M$$ and $$N$$ and prove all properties with this definition? One could then call any other object satisfying the universal property a tensor product of $$M$$ and $$N$$ and that each of those objects is isomorphic (uniquely) to $$M \otimes_R N$$ (the quotient group construction). It should be the case that the proven properties then also hold for the other tensor products since there exists the unique isomorphism. If this would not cause any "harm" this would for now be the more elegant way to work with the definition for me, since this works around the non uniqueness of the tensor product.

Edit for clarification: What I mean with "possible" in $$(3)$$ isn't the technical or proof level but rather if proceeding this way would have any downsides or "harm" in any way.

Edit 2: I read lecture notes, Dummit & Foote and Atiyah Macdonald (perhaps some more, but I can't really remember those). As far as I can tell now, Atiyah actually defines the tensor product as precisely the quotient group, if I am not mistaken. This should then answer my 3rd question. It would be nice, however, to get confirmation by someone that I am not misunderstanding this.

• May I ask which books you're reading? – user20672 Apr 18 at 15:41
• @user20672 I have edited the post answering your question in edit 2. – GeraltHauser Apr 18 at 17:30

This issue goes back to the question of what 'equality' is in a given context. To stay in your example of modules, when would you call two $$R$$-modules equal? Of course, one can argue that we should call $$M$$ and $$N$$ equal only in the case that they have the exact same underlying sets and relations on them. But this would be a strange choice, because if there is a distinct isomorphism of modules $$\varphi: M \rightarrow N$$ (which is not the identity though), all the other $$R$$-modules cannot tell the difference between $$M$$ and $$N$$: There is a one to one correspondence between $$R$$-module homomorphisms $$P \rightarrow M$$ and homomorphisms $$P\rightarrow N$$ for all $$R$$-modules $$P$$! To illustrate that, take any $$R$$-module and just rename its elements. Then it would be a silly thing to call this new module different from the original one; the two look the same for all other modules.

Therefore, it does no harm at all to talk about the tensor product while meaning any representative of the isomorphism call of modules satisfying the universal property. To go even further, it does not make sense to talk about one particular realization of a solution of the universal property as 'the' tensor product, since the definition of the tensor product does not specify enough data to singe out one 'special' realization; it only tells us the isomorphism class we should look in.

In working with tensor products, we often give a concrete solution of their universal property, because that can make it easier to visualize them or get intuition for them if one encounters them for the first time. But you should make it clear to yourself that this concrete realization mathematically is by no means special in the isomorphism class of solutions.

Edit: Of course it is important to make sure that there is indeed a module satisfying the universal property. This is another way in which a concrete realization is useful.

• (+1) but I think the last paragraph is written in a misleading way. I does matter that there are realizations as otherwise we could be talking the whole time about a non-existing objects. But the construction itself only really matters for showing precisely this: There are some objects satisfying the characteristic universal property (i.e. the isomorphism class is non-empty). – mrtaurho Apr 18 at 17:13
• First of all, thank you for your answer! I am not sure if I would completely agree here though. Perhaps I am off here, but for me two algebraic objects (e.g. groups) would be equal if and only if they are the same set with the same underlying operation. I see that it often does not matter to distinguish between isomorphic objects, however there seem to be cases in which it may be problematic: mathoverflow.net/questions/24318/… (if I understand it correctly). – GeraltHauser Apr 18 at 17:41
• @GeraltHauserb Yes, there definitely are contexts where isomorphic objects should not be regarded as 'the same'. It can be difficult to determine in which situations problems arise from this, but often times one is on the safe side when something is unique up to unique isomorphism, as is the case here. – S.Farr Apr 18 at 17:49
• @GeraltHauser That depends on the property you want to show. If it is a property you expect to be true for all realizations of the tensor product, you will always be able to show it using only the universal property. I often prefer the abstract definition for proves as such proves generally involve less 'calculations'. – S.Farr Apr 18 at 18:25
• @GeraltHauser: that $M\otimes_R N$ is generated by the elements $m\otimes n$ does not require the concrete construction. It can be proved using only the universal property. Essentially every useful property of a tensor product can be shown using only the universal property, because satisfying the universal property is what makes tensor products useful. – Vercassivelaunos Apr 18 at 21:43

You have the same question in almost every part of mathematics. Frequently there are two ways to introduce a new mathematical "object":

1. Constructive approach: Simply give an explicit construction of the object.

2. Conceptual approach: Specify a characteristic property determining the desired object up to unique isomorphism and then prove existence via giving an explicit construction.

In my opinion the conceptual approach should be preferred because it explains the purpose of what is introduced. This approach is of course more abstract (and perhaps too abstract for beginners).

In the first case one certainly uses the definite article "the" (e.g the tensor product), but in the second case I personally would tend to use the indefinite article "a" (e.g a tensor product) because there are many objects having the characteristic property, i.e. there exist many concrete models realizing the abstract concept. However, in practice one commonly uses the article "the" without caring about a concrete construction.

Here are some examples.

• The Cartesian product of sets. Everybody says the Cartesian product although there are many ways to introduce it formally. On the level of basic set theory there is nothing like a pre-existent Cartesian product. One can define the concept of an ordered pair $$(a,b)$$ by the property $$(a,b) = (a',b')$$ iff $$a = a'$$ and $$b = b'$$, and this seems to be completely intuitive - but what is $$(a,b)$$? One can define it as the set $$(a,b) = \{\{a\},\{a,b\}\}$$ and check that it has the desired property. However, there are other definitions which perform equally well.
In other words, there are various models for the Cartesian product of sets, but nobody cares about it because it is completely irrelevant for all practical purposes which model we take. The essence is the above characteristic property.

• The set of integers $$\mathbb N$$. Conceptually they are characterized by the Peano axioms. Again there are various models, e.g. we can recursively define sets $$0 := \emptyset$$ and $$n+1 = \{n\} \cup n$$. Again nobody cares about the concrete model, the essence is that the Peano axioms are satisfied.

• The set or real numbers $$\mathbb R$$. Conceptually they can be characterized as a Dedekind-complete ordered field. As you know there are various models: Starting with $$\mathbb Q$$, one can define real numbers e.g. as Dedekind cuts or as equivalence classes of Cauchy sequences. But again nobody cares.

These examples show that it is a philosophical question whether to use the definite or indefinite article. In fact it is irrelevant if you visualize a certain model or take any model without being specific.

Let us now come to your original question. As you know, the concept of tensor product can be defined via a universal property:

Given two $$R$$-modules $$M, N$$, we say that an $$R$$-bilinear map $$\tau : A \times B \to C$$ to an $$R$$-module $$C$$ has the tensor product property if for each $$R$$-bilinear map $$\beta : A \times B \to D$$ to an $$R$$-module $$D$$ there exists a unique $$R$$-linear $$\phi : C \to D$$ such that $$\phi \circ \tau = \beta$$.

Clearly for any two $$R$$-bilinear maps $$\tau_i : A \times B \to C_i$$ having the tensor product property the unique $$\phi$$ such that $$\phi \circ \tau_1 = \tau_2$$ is an isomorphism. This is what you mean by uniqueness up to unique isomorphism. It remains to prove the existence of a map with the tensor product property. Usually this is done by the quotient group construction mentioned in your question. The result is denoted by $$\tau : A \times B \to A \otimes_R B$$. It must be emphasized that the object $$A \otimes_R B$$ on its own does not have any universal property, the map $$\tau$$ is an essential constituent. If we look at the construction of the object $$A \otimes_R B$$, it may be fairly obvious what this map looks like, but nevertheless it belongs to the complete specification.

There are certainly other models of maps having the tensor product property, for example when $$A = B = R$$ we may take $$\tau: R \times R \to R$$ to be the multiplication map on the ring $$R$$. In this case your question could be read as

Do we have $$R \otimes_R R = R$$ or only $$R \otimes_R R \approx R$$?

Actually it is irrelevant, we could take $$R \otimes_R R = R$$. But the price would be to work with a case-by-case construction of tensor products. You can do that, but it has a serious problem. In fact one wants to have a functor

$$\otimes_R : \textbf{Mod}_R \times \textbf{Mod}_R \to \textbf{Mod}_R .$$

It is not expedient to say "pick any model of the tensor product" - doing so would involve a variant of the axiom of choice for proper classes. Concerning classes have a look at Can we say," The set of all compact metric spaces"?

This problem can easily be avoided by taking a concrete tensor product construction as that via quotients. We do not need any choice here.

Remark:

A related question is Why define simplicial chain groups as functions rather than free Abelian groups?

• (1) Thank you for your answer! Especially the cartesian product seems to be a perfect analogy to my question. However I do have some remaining questions. If I would define a tensor product to be anything that satisfies the universal property, then what is meant with $M \otimes_R N$? – GeraltHauser Apr 19 at 12:12
• (2) Is $M \otimes_R N$ the concrete construction, or is it the entire class of isomorphisms. In the first case, its clear how to prove results about $M \otimes_R N$ and once proven, they should hold for all other tensor products as well. In the second case however, one would have to pick an arbitrary tensor product and then prove the property. Are both ways common to do? In my opinion the first case is way more comfortable. – GeraltHauser Apr 19 at 12:13
• If I imagine writing notes introducing the tensor product, I think this is the way it is done: One states the universal property, then proves the existence and then that they are essentially unique. Afterwards, $M \otimes_R N$ is defined to be the concrete construction and one speaks of the tensor product since all the others are isomorphic to $M \otimes N$, meaning that the properties proven for $M \otimes N$ (the concrete construction) also hold for any other tensor product. I think my problem now is "what exactly is $M \otimes N$. If its the construction, then I think I understood it. – GeraltHauser Apr 19 at 12:18
• @GeraltHauser The symbol $M \otimes_R N$ does not denote the whole isomorphism class. If you see $M \otimes_R N$ in a text, then you can interpret it either as the standard quotient group construct or as any other model of the tensor product - it will not make a real difference. Nevertheless working with the concrete construction may be very helpful to prove results. I agree that this is more comfortable than always recurring to a universal property. But certainly some people have a different opinion. So the question what $M \otimes_R N$ "really is" means which model we should chose. – Paul Frost Apr 19 at 14:16
• Perhaps this is a similar question as what a point in Euclidean geometry "really is". In Euclid's Elements one finds the definition A point is that which has no part. More than 2000 years later Hilbert said that the names of the objects of geometry, such as point, line and plane could be substituted by tables, chairs and glasses of beer - it is their defined relationships that are discussed. – Paul Frost Apr 19 at 14:17

For what it's worth I think there are genuine unsolved and/or disagreed-upon philosophy of math issues that this question touches on. It's a bit like the question of whether $$1 \in 2$$ is meaningful in ZFC. See the nice discussion here:

https://mathoverflow.net/questions/336191/cauchy-reals-and-dedekind-reals-satisfy-the-same-mathematical-theorems

One belief that you can find in the comments here, which perhaps best fits with what mathematicians really do in practice: when we construct an object like $$M \otimes N$$ satisfying a universal property, we should then add an axiom to our system saying that the object exists. The construction can then be re-interpreted as a proof that this axiom is consistent with our system, and reference to $$M \otimes N$$ (and its simple tensors $$m \otimes n$$ are references to "the one guaranteed by the axioms.")

This is a bit of a "software engineering" framing (make your code match your intuitive perception of reality), and the "computer science" framing might be that $$M \otimes N$$ "is" the quotient but that it (prove-ably!) doesn't matter what it "is."

• Thank you for your answer! Throughout studying mathematics I seem to always struggle with problems that seem to be quite philosophical, however I will make sure to check out your link. I think its also weird that one never gets taught why this is an okay thing to do in lectures, its just being done without any comment whatsoever (I noticed this especially when identifying isomorphic/homeomorphic objects). – GeraltHauser Apr 18 at 18:25