The tensor product of algebras is associative if both of the algebras are associative.

Question

Let $A,B$ be nonzero algebras over a field with identity elements, but not necessarily associative. Prove that if the tensor product of $A$ and $B$ is associative, then so are $ A,B $.

I think I have to prove if for any $a_i, b_j$, we have $((a_1a_2)a_3 \otimes (b_1 b_2)b_3)= (a_1(a_2 a_3) \otimes b_1 (b_2 b_3))$,

then we have $ (a_1(a_2 a_3)=((a_1a_2)a_3) $and also for the elements of $ B $.

Is my idea right?

If I assume $b_ 1=b_2=b_3=1$ then I get $(a_1(a_2 a_3) \otimes 1 )=((a_1a_2)a_3) \otimes 1)$, but how can I say $ (a_1(a_2 a_3)=((a_1a_2)a_3) $ ?

Thank you for your help.

Answer 1

You have 2 homomorphisms, $\imath:A\to A\otimes B$ such that $f(a)=a\otimes 1$, and $\pi:A\otimes B\to A$ such that $\pi(a\otimes b)=a$, then you have $\pi\circ\imath = id$ $$\imath((a_1 a_2) a_2)=(a_1a_2)a_3\otimes 1 = a_1(a_2a_3)\otimes 1$$ and applying the oother homomorphism we get the desired result.