Rings 10 Tensor products of abelian groups

[@weierstra5326]

amazing lecture.

[@adityaekbote8498]

These + serge lang = fun

[@jovanagiovanna7749]

Also Hungerford.

[@birdboat5647]

wow~

[@migarsormrapophis2755]

yeeeeeeeee

[@xiaohuwang4173]

23:40 not sure if I'm right, but shouldn't the sequence of homomorphisms be ×2 ×3 ×4 ... instead of ×2 ×6 ×24 ...?

[@shahinhashemi5799]

it is to do with factorials.

[@stephenbigelow5449]

Yes, I think you are right.

[@analander9222]

The answers of the two exercises should be 0 and Qp (the p-adic numbers)

[@didierfelbacq7208]

Since Q=2Q, then Z/2Z \otimes Q=Z/2Z \otimes 2Q. By linearity: Z/2Z \otimes 2Q=2*Z/2Z \otimes Q=0 \otimes Q=0. Idem for Q\otimes Q/Z=0. I like the proof with exact sequence though (but it does not show the why's, more like "abstract non sense" proof).

[@annaclarafenyo8185]

There is a problem with your notation for bilinear maps. You write "bilinear: A x B -> C " using a product symbol x which usually means product in the category, but here the 'x' symbol must be interpreted as the set-theoretic product making pairs of elements without any abelian group structure, because if the x means a product in abelian groups, it's the sum, so any linear map in the sum obeys f(a+a', b+b') = f(a,b) + f(a',b') rather than f(a+a',b+b') = f(a,b) + f(a',b) + f(a,b') + f(b,b') as we would like. This genuinely confused me.

[@davidliu7246]

The notation is fine. The confusion is because bilinear maps are (usually) not linear. By saying that the map is bilinear, that means f(a+a',b+b') = f(a,b)+f(a,b')+f(a',b)+f(a'+b') , and in particular, f is usually not linear

[@anthonymurphy5689]

@@davidliu7246 Said slightly differently, a bilinear map between abelian groups (aka Z-modules) AxB->C is NOT the same as a homomorphism AxB->C in the category of abelian groups (aka Z-modules). In addition to the example above, we also have: Bilinear map of Z-modules: (1) F[r(a,b)] = F[(ra,rb)] (definition of Z-module scalar multiplication) = r.F[(a,rb)] (bilinearity) = r.r.F[(a,b)] (bilinearity) = r^2 . F[(a,b)] Homomorphism of Z-modules: (2) F[r(a,b)] = r. F[(a,b)] The bilinear map AxB->C does however give rise to a Z-module homomorphism (ie linear map) from the tensor product A⊗B->C (3) F[r(a⊗b)] = F[(ra)⊗b] or F[(a⊗(rb)] = r.F[(a,b)] ... this is the main point of the tensor product construction. And of course (ra)⊗(rb) = r^2 a⊗b, consistent wit (1) above and the definition of the tensor product. Having said all that, I completely sympathize with Anna's confusion and thank her for highlighting the need to make the distinction.

[@annaclarafenyo8185]

@@anthonymurphy5689 No, I think you miss my issue. Once you write "A x B -> X" you are working with the object AxB and morphisms from that object (AxB). The object AxB DOESN'T KNOW it's got A and B inside! It's just an abelian group, it has no idea that it's made up of A and B elements. You cany say "Bilinear : Ab -> X " for an arbitary abelian group, because bilinear needs an additional structure, an explicit splitting into factors. While we can see the factorization on the page, when he writes AxB, it is obvious to the eye, to the formal rules usually defined in category theory, the notation is inconsistent. I still figured it out after a minute or so, I just remember noting that this is an inconsistency. Perhaps it's a universal abuse of notation for bilinear maps before the tensor product structure has been introduced.

[@gunhasirac]

It’s very interesting because I didn’t even notice this until reading this comment. But i don’t think there’s any ambiguity in the context. As we define bilinear maps, we always think of the map being linear in each component while the other is fixed and structure of the product of the domain doesn’t come into play. It does coincide with notation of product of group but from the context I think it’s clear enough to not using separate notation here.

[@jovanagiovanna7749]

@@annaclarafenyo8185 In Hungerford's book that I follow, there is always a little remark that internal products are not equal to external products, but they are isomorphic (with external products containing monomorphic copies of substructures).