Tensor Product of two Algebras is an Algebra

[sami]

Hi,

maybe I didn't look long enough, but I can't find a detailed proof for the proposition:

"If \(\displaystyle A, B\) are unital algebras over a field \(\displaystyle k\), \(\displaystyle A\otimes B\) with multiplication defined as \(\displaystyle (a_1 \otimes b_1)\cdot (a_2 \otimes b_2) = (a_1 a_2) \otimes (b_1 b_2)\) is a unital algebra."

Mostly the well-definition of the product is missing.

Thanks for any help!

 

[daon2]

Hi,

maybe I didn't look long enough, but I can't find a detailed proof for the proposition:

"If \(\displaystyle A, B\) are unital algebras over a field \(\displaystyle k\), \(\displaystyle A\otimes B\) with multiplication defined as \(\displaystyle (a_1 \otimes b_1)\cdot (a_2 \otimes b_2) = (a_1 a_2) \otimes (b_1 b_2)\) is a unital algebra."

Mostly the well-definition of the product is missing.

Thanks for any help!

I haven't looked at tensor products in over a year, but I believe you just use the universal property (most proofs will do this). I think this is more or less the solution (just fill in the little details):


For \(\displaystyle x\in A, y\in B\) define a map \(\displaystyle f:A\times B \to A\otimes B\) by \(\displaystyle (a,b)\mapsto ax\otimes by. \) This map is bilinear and so there is a unique (linear) map \(\displaystyle g:A\otimes B\to A\otimes B\) such that \(\displaystyle f=g\circ \otimes\). Thus if \(\displaystyle a\otimes b = a'\otimes b'\) then \(\displaystyle (a\otimes b)(x\otimes y)=f(a,b)=g(a\otimes b) = g(a'\otimes b') = f(a',b') = (a'\otimes b')(x\otimes y)\)