Talk:Tensor product of algebras

Lifting

How does the lifting of bilinear maps work? AFAICT a bilinear map f:A×B → C should lift to an R-algebra homomorphism A⊗B → C iff f(aa', bb') = f(a,b) f(a',b'), but can someone confirm (and add) this? What's the general name for such a map? Jun

Canonical Mappings

You have mentioned that for Z-Algebras (or commutative rings) the canonical mapping A -> A @_R B and B -> A @_R B are injections, why is this so?

Sincerely Jose

Well, that may not be true, I think. I have made it 'homomorphisms'. Charles Matthews 15:16, 4 April 2007 (UTC)[reply]

No they aren't, there is an example.. for instance Q @_Z Z_p ... for the Z-algebras Q and Z_p,

1@x = 0@0 for any x in Z_p (since 1@x=1/p@xp=1/p@0 ), thus Z_p is not an injection into Q @_Z Z_p

But if A and B were R overrings, then I think this could lead into injections.. I need to look up some homological algebra which I have almost 0 knowledge in.

Sincerely, Jose

Unital R-algebras?

Why is the tensor product necessarily unital? unless it can be done only for unital R-algebras~but that's not mentioned. I checked out the references and it's not even as detailed as this article Noix07 (talk) 09:40, 4 August 2014 (UTC)[reply]

I think we take R-algebras to mean unital R-algebras in this article, by its original definition. I agree we should include something about the tensor product of nonunital, or even non-associative R-algebras, but it's true that if A and B are unital R-algebras, 1_A⊗1_B = 1_A⊗B. This can be shown by the universal property defining it - for any R-algebra homomorphisms f:A→C and g:B→C of commutative R-algebras, there is a unique R-algebra homomorphism f⊗g:A⊗B → C such that (f⊗g)∘i = f and (f⊗g)∘j = g for i and j the canonical injections of the coproduct diagram i:a ↦ a⊗1_B, etc. In fact it is true that (f⊗g)(a⊗b) = f(a)g(b), the product of f(a) and g(b) in C, for a∈A, b∈B. This shows that the tensor product of unital algebras is also a unital algebra. --Daviddwd (talk) 22:01, 27 August 2014 (UTC)[reply]

Tensor product and usual product

(I'm writing this so I don't forget). If A, B are subrings of some ring C and if they are all algebras over say R, then isn't tensor product of elements of A and B is the same as that of C? I'm not sure about the generality but in the special case (some sort of freeness) probably yes. (I found the answer myself: linearly disjoint.) In any case, this matter should be discussed in the article. -- Taku (talk) 22:01, 8 October 2015 (UTC)[reply]