Talk:Tensor product of modules

message: (I changed most of the HTML at the beginning to Tex.) - Sorry, I didn't realize this was such an apparently charged issue. The bilinearity axioms still render as non-Tex on my settings though. I'll let you decide what to do now.
Also, I added the universal property because the article was referring to an universal property "above", which wasn't there. Then I found the universal property below, so It's duplicated now. But I don't want to delete it; I think I already messed with the article enough :)
Functor salad 13:16, 25 September 2007 (UTC)
thread_title:  Tex vs. HTML 

message: Hello all.  There is an implicit claim here which is along the lines of "if the tensor functor is not exact, minimal generating sets may not be sent to minimal generating sets, but when the functor is exact minimal generating sets DO go to minimal generating sets". Vector spaces (the functor is always exact in this case) are given as an example, but I've only seen this proven in an ad hoc way which doesn't mention exact functors.  The claim seems plausible, but I would like to ask a more senior category theorist to double check what is written and provide confirmation.  Thanks! Rschwieb (talk) 20:19, 22 February 2011 (UTC)
thread_title:  Verification questions 

message: I've learned to be cautious with mathematical texts: when something obvious is not written it often means that it's wrong. But here, what's the point of doing a tensor product if we don't get what we want. Why do they specifically say the image is an abelian group when one naturally wants to say it's automatically a moduleNoix07 (talk) 14:25, 4 August 2014 (UTC)
replies: text: It says in the lead that the result is a module ("resulting in a third module") – provided that we start with modules over a commutative ring. In the case where we start with a left- and right-module and the ring is non-commutative, your observation is accurate: we do not end up with a module, only an abelian group, which is to say, addition is defined, but there is no scalar multiplication for the resulting object, which would be necessary to make it a module. This is explained under Tensor_product_of_modules#Multilinear_mappings.  Can you suggest where changes in wording would distinguish the two cases more clearly? —Quondum 14:54, 4 August 2014 (UTC)
thread_title: Why isn't it written that the tensor product is again an R-module? 

message: One has the other universal property: Let  R  be a ring and let M,N be left R-modules.
If X is a left R-module and M \to X is an R-module homomorphism then there is a unique map M \otimes_R N \to X making the obvious diagram commute. This can be useful (e.g. in commutative algebra) when one does not want to deal with cartesian products, which a priori have no real structure. — Preceding unsigned comment added by Mobius stripe (talk • contribs) 2014-11-23T15:05:41
:I think something (i.e., namely N) is missing in the input. There is
::\operatorname{Hom}_S (S \otimes_R M, X) = \operatorname{Hom}_R(M, X)
:when S is an R-algebra. (This stuff is perhaps worth mentioning.) -- Taku (talk) 15:42, 6 April 2015 (UTC)
thread_title:  Other universal property? 

message: The article says in Tensor product of modules#Definition:
:It can be shown that  and  are completely different from each other.
Question: could this be made a bit more specific?
Certainly R ⊗Z R = R, because the bilinear map of R × R → R is the multiplication in R.
But also R ⊗R R = R for the vector space R/R.
So certainly the two are different in algebraic structure, i.e.  as abelian group is furnished only with the trivial scalar product. But the two are the same as sets and as abelian groups. Aren't they ?
--Nomen4Omen (talk) 16:32, 25 January 2015 (UTC)
replies: text: (Disclaimer: not my area, but this may give some pointers; it may help to make this clearer in the article.)  See Tensor product § Definition – this allows us to think of the tensor product as a quotient of the Cartesian product, considered as a free Z-vector space.  The scalar multiplication is the same under all cases under the definition given in this article (only multiplication by Z is defined), and thus even  is a Z-vector space, not an R-vector space).  What is significant is that the equivalences are different:  only when  or  under this quotient AFAICT.  You can say that , the additive group of the real numbers (and is the same as real multiplication, but for the lack of scalar multiplication on the resulting algebraic object), but  is not the same as multiplication (Z-bilinearity and R-bilinearity are not the same thing), since we get for example that . —Quondum 00:08, 26 January 2015 (UTC)
thread_title:  '''R''' ⊗<sub>'''Z'''</sub> '''R'''  ≠  '''R''' ⊗<sub>'''R'''</sub> '''R''' ? 

message: The trace, defined via E^* \otimes_R E \to R, \, \phi \otimes x \mapsto \phi(x), needs clarification.  In particular, not all tensors in E^* \otimes_R E can be written as simple products of elements of the respective modules.  A few words explaining why (and how) this extends to the whole tensor product space are needed.
As to (not) extending the trace to the noncommutative case, a demonstration of why this does not work is straightforward.  IMO, we can put this statement in without much contention, even if we cannot find a source for the nonworkingness of it now.  Take the simple case from of the (intended) canonical map (assume that E is a right R-module):
:E \otimes_R E^* \to R, \, x \otimes \phi \mapsto \phi(x).
We have that
:r \cdot \phi(x) = (r \cdot \phi)(x) = tr(x \otimes (r \cdot \phi)) = tr((x \cdot r) \otimes \phi) = \phi(x \cdot r) = \phi(x) \cdot r.
This, assuming a noncentral regular element r of R and that the trace can be nonzero, is a contradiction. Conclusion: the trace, as would fit the canonical map, does not exist if R contains a regular noncentral element. (Maybe I missed a corner case, but that's the idea.) The determinant will have similar issues in the noncommutative case, as should the exterior algebra of a module. —Quondum 17:41, 9 April 2015 (UTC)
replies: text: On second thoughts, this does not stop us defining the trace "up to conjugation". I don't know how useful this would be, though. —Quondum 17:53, 9 April 2015 (UTC)

text: First of all, it is standard and not problematic to define a linear map on the generating set, instead of all elements. Since pure tensors (x \otimes y) do generate the whole tensor product, it is enough to just map the generators (pure tensors in this case). It is clear (and in fact I just did actually check with a paper and pencil) to see the map extends by linearity. This is mentioned in the important proposition in the definition section (maybe that need to be clarified). As for the example, I have trouble following it (\phi is right R-linear, am I right?). To me the main issue is the module structure: since M \otimes_R N is just an abelian group, R-linearity doesn't make sense.. The example makes sense; I don't have a good feeling about the non-commutative case :) -- Taku (talk) 18:33, 9 April 2015 (UTC)
replies: text: You just preempted a discourse by me with that last edit ;). I'm not sure about what you're trying to say with the rest; I fully agree, and distinctly prefer the "induced by" approach (from memory, even Bourbaki does it this way). I hope you didn't interpret my remark as sarcasm. —Quondum 19:13, 9 April 2015 (UTC)
thread_title:  Trace of a tensor 

message: The edit summary (not quite unless the canonical homorphism is injective) for this edit suggests that I've run into a problem with my understanding of the meaning of a "right R-module homomorphism", or  for right R-modules E and F.  As I understand it,  is the set of all maps  that preserve the module structure ( implies that  (additivity) and  (homogeneity)). I see no problem so far: we know (from Bourbaki) that the canonical homomorphism  exists, and this just says that θ maps an element ξ of the tensor product to such a φ, .  If the canonical homomorphism is injective, this means there do no exist distinct ξ and ξ′ such that  and  for any φ.  How come injectivity of θ is required for the statement "Thus, an element of a tensor product  may be thought of as a right R-linear map " to hold? If it is not injective, all this means is that more than one element ξ may map to the same right R-linear map φ, but this seems irrelevant to the statement. —Quondum 15:25, 12 April 2015 (UTC)
replies: text: I think this is the matter of attitude; if there is a linear transformation from a vector space V to a vector space W, I don't interpret a vector in V as a vector in W. Each vector in V uniquely determines a vector in W; for me, this is not the same thing as a vector in V is a vector in W; that would require the transformation to be injective. -- Taku (talk) 16:49, 12 April 2015 (UTC)
replies: text: Ah, I see what you're saying – fair enough. So we need to find a different wording. The phrase "may be thought of as" is problematic.  Maybe "Thus, an element of a tensor product F ⊗R E  a right R-linear map E∗ → F "? —Quondum 17:23, 12 April 2015 (UTC)

text: Yeah, that would be definitely the better wording. -- Taku (talk) 19:15, 12 April 2015 (UTC)
thread_title:  Canonical map required to be injective for ''R''-linearity? 

message: Bourbaki, Algebra, ch. II §3.8 deals with this topic.  It is clearly a generalization of §§ Several modules, since it deals with the noncommutative case, which specializes to exactly the multilinear case over a single commutative ring.  Essentially, given a right R-module L, a R,S-bimodule M and a left S-module N, we get the isomorphisms .  My inclination would be to present the general case as Bourbaki does, and then to specialize it. —Quondum 19:11, 12 April 2015 (UTC)
replies: text: That section is not general enough (and requires updates). The general form of the associativity is already noted in the "properties" section. -- Taku (talk) 19:16, 12 April 2015 (UTC)
replies: text: Oops, yes, it already does. I'd prefer to have the properties present the general case first, and then the changes that result from specialization to a commutative ring.  Does this make sense? —Quondum 00:03, 13 April 2015 (UTC)
replies: text: It does but yes as you suggested it's probably not an optimal approach from the expository aspect. What is needed is to discuss the universal properties for multilinear maps and tensor product of several modules. I'm lazy busy and no one is doing the job for a moment :)  -- Taku (talk) 01:20, 13 April 2015 (UTC)

text: Bourbaki, who know tensors the best, use the notion of multimodule (more than two ring actions on a module, get it?); that's how they handle tensor products of several modules over non-commutative rings. I'm just not sure if we want to use this notion; there is probably a better way. -- Taku (talk) 01:43, 13 April 2015 (UTC)
replies: text: I still need to wrap my head around the multimodule concept.  This may be getting very deep for a WP article, other than the mention of what can be achieved this way. It is way over my head, but tantalizing. I'll keep my edits to what can be shown using bimodules, for now. —Quondum 02:48, 13 April 2015 (UTC)

text: Yargh.  This one keeps playing with my brain, and Bourbaki is a bit cryptic for me on this.  Am I correct in saying that w.r.t. modules, the left/right distinction for scalar multiplication is mathematically spurious (in the sense that every right scalar multiplication can be considered to be equivalent to a left scalar multiplication of the opposite ring)? This brings to mind a picture of a multimodule that simultaneously supports left, right, top, bottom, (whatever) scalar multiplication, and that the tensor product over one or more of these simultaneously would be possible. There is presumably some compatibility constraint on these scalar multiplications, even if only Z-linearity.  Or am I completely missing the intention? —Quondum 18:57, 13 April 2015 (UTC)
replies: text: I had exactly the same mental image as you: top action, bottom action, maybe 45-degree action, etc. But actually it's not as bad as it might appear first. The idea is simple: we simply allow more than one left or right action; what is a ring action anyway: it is a group homomorphism \pi: R \to \operatorname{End}_{\mathbb{Z}}(M). This means each element r in R determines a linear transformation \pi(r) of M; a right action is precisely a left action for the opposite ring of R, as you said. Then it is not much of a leap to consider a family of left actions \pi_i: R \to \operatorname{End}_{\mathbb{Z}}(M). When an "obvious" compatibility condition is met, a module becomes a left multimodule. Simiarly, one can consider a right multimodule structure, and then, in an "obvious" way, a module that has both left and right multimodule (or just multimodule). At least this is how I understood. -- Taku (talk) 20:59, 13 April 2015 (UTC)
replies: text: This is intriguing because it brings back the possibility of higher-order tensor products, and with it, exterior algebras; I had sort of assumed that went out of the window with noncommutativity. But now I have this picture of building tensor products out of multimodules like one builds covalently bonded molecules out of atoms ... but this may bring with it the possibility of nonisomorphic structural isomers, with consequences to the associativity of the tensor product. —Quondum 21:33, 13 April 2015 (UTC)

text: On second thoughts, perhaps the exterior algebra needs nothing more than an (R,R)-bimodule. So noncommutative geometry might still be good using this. —Quondum 21:37, 13 April 2015 (UTC)
thread_title:  Associativity of the tensor product 

message: The section 'Balanced product' defines a bilinear map as a synonym of balanced form. I see that Bourbaki does not refer to a balanced product (although it uses the concept without naming it), and only seems to use use bilinear map in the sense that <au,yb> = ab, where u and v are elements of modules, and a and b are their respective scalars. EOM defines it likewise. I have not located other references. Should we should remove this unfortunate use of bilinear map? —Quondum 05:31, 24 April 2015 (UTC)
replies: text: I don't know a good solution to this terminology issue. A "bilinear" could be very misleading in the case R is non-commutative; I personally like "balanced product". Some text doesn't even bother to introduce a terminology (they refer instead to a Z-bilinear map with the middle linearity). Bourbaki is s bit old so that might be why they don't use the word "balanced product". Dummit-Foote, Abstract Algebra, a fairly standard and reliable text, uses (if I remember) "balanced product" -- Taku (talk) 14:18, 24 April 2015 (UTC)
replies: text: Well, I think that simply dropping the use of "bilinear" in this sense is a good solution, then: it neither sensible, notable nor referenced; it is also in direct conflict with Bourbaki's sensible use of the term. .  I'll remove it.  I see there is the same problem at Bilinear map, which I'll remove too. —Quondum 15:36, 24 April 2015 (UTC)
thread_title:  Bilinear map 

message: I think the correct modern (post-Bourbaki) definition of the tensor product of several modules is that as the colimit of the diagram that generalize a coequalizer for two ring actions to several ring actions. Did I miss anything?  -- Taku (talk) 17:46, 8 June 2015 (UTC)
thread_title:  Tensor products of several modules 

message: In ,  seems to be missing some of the point.  At first glance, one could argue that only a one-sided module definition has been given, but in fact the natural bimodule extension (namely with r⋅x = x⋅r) must be assumed to make the result valid; this should really be stated upfront in the text.  If one does not have this assumption, the stated result falls apart.  Take, for example, a bimodule over the commutative field C with general element x and r ∈ C, defined such that r⋅x = x⋅r (where the  denotes the complex conjugate).  We can still form the tensor product of the module with itself, but it clearly violates the stated result.  The tensor product does match the more general result stated just below that though.  —Quondum 16:08, 16 May 2020 (UTC)
thread_title:  Sidedness of a module