Commutative algebra 20 Tensor products review

[@annaclarafenyo8185]

Regarding the word "unique" in mathematics, if you think about it more deeply, I think you will come to see that it always means 'unique up to canonical isomorphism'. The process of Platonic elevation occurs when we internalize the canonical isomorphism classes. Once this happens, once objects elevate themselves into our Platonic realm, we use the word 'unique' thinking it means something other than canonically isomorphic.

[@pianostein7590]

4:40 The commutativity of the diagram only gives you, that the composition of the map from MxN to the tensor product with the composition of the two maps is the same as the map from MxN to the tensor product. But by the definition of the tensor product we then know that the composition of the two maps is unique with this property. Since the identity also satisfies this property, they has to be the same.

[@hausdorffm]

21:40 Tensor product of Z/mZ and Z/nZ equals 0, if m and n are coprime. In fact, if integers m,n are coprime, then there are some integers a,b such that am+bn=1. For any integers x, y, the tensor product of x and y = x*y = x(am+bn)*y = m*xay + xby*n = 0 in tensor product of Z/mZ and Z/nZ. I do not understand proofs of 15:43 and isomorphisms 22:30, 21:14.

[@zy9662]

Did he define the operation of the tensor product of two elements in a previous lecture?

[@zy9662]

19:53 (m,n) is suppose to be the minimum common multiple of m and n?

[@lucmerci2284]

It was correct, the greatest common divisor.

[@robinbalean958]

As Luc says, it is the gcd and the tensor product (m,n)(1 \otimes 1)=0 because we can always find integers a,b such that am + bn = (m,n) .