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

I think I have to prove if for any ai,bja_i, b_j, we have

((a1a2)a3⊗(b1b2)b3)=(a1(a2a3)⊗b1(b2b3))((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 (a1(a2a3)=((a1a2)a3) (a_1(a_2 a_3)=((a_1a_2)a_3) and also for the elements of B B .

Is my idea right?

If I assume b1=b2=b3=1b_ 1=b_2=b_3=1 then I get (a1(a2a3)⊗1)=((a1a2)a3)⊗1)(a_1(a_2 a_3) \otimes 1 )=((a_1a_2)a_3) \otimes 1), but how can I say (a1(a2a3)=((a1a2)a3) (a_1(a_2 a_3)=((a_1a_2)a_3) ?

Thank you for your help.

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Careful: you want AA and BB to be nonzero.

– darij grinberg

2 days ago

@darijgrinberg yes right.

– user115608

yesterday

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1 Answer

1

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You have 2 homomorphisms, ı:A→A⊗B\imath:A\to A\otimes B such that f(a)=a⊗1f(a)=a\otimes 1, and π:A⊗B→A\pi:A\otimes B\to A such that π(a⊗b)=a\pi(a\otimes b)=a, then you have π∘ı=id\pi\circ\imath = id

ı((a1a2)a2)=(a1a2)a3⊗1=a1(a2a3)⊗1\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.