$Tensor Product II$

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$Tensor Product II$

Intro to Tensor Products review.

Let $R$ be a comutative ring with 1 and $A$ an $R$-module with a bilinear operation $m:A×A\rightarrow A$ given by $m(a,b)=ab$ for $a,b∈A$.
We know that $m$ is associated to a $R$-module homomorphism
$μ:A\otimes_R A \rightarrow A$.
Let the abuse of notation $(A\otimes_RA)\otimes_R A=A\otimes_R(A\otimes_RA)$ be true and let $Id_A:A\rightarrow A$ be the identity function in $A$.

Show that $m$ is an associative operation if and only if $\mu \cdot ( \mu \otimes Id_A)=\mu \cdot (Id_A\otimes \mu)$

Abstract Algebra Algebra

Matchmaticians Alumnus

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We know that $\mu (a\otimes b)=m(a,b)=ab$. Also note that $\mu \circ (\mu\otimes Id),\mu \circ (Id\otimes \mu):A\otimes A\otimes A\to A. $ And $m$ is associative if $a(bc)=(ab)c$ for every $a,b,c\in A$. Now $$( \mu \circ (\mu\otimes Id))(a\otimes b\otimes c)=\mu ((\mu\otimes Id)(a\otimes b\otimes c)) \\ =\mu (\mu(a\otimes b)\otimes Id(c))=\mu (ab\otimes c)=(ab)c$$ and
$$( \mu \circ (Id\otimes \mu))(a\otimes b\otimes c)=\mu ((Id\otimes \mu)(a\otimes b\otimes c)) \\ =\mu (Id(a)\otimes \mu(b\otimes c))=\mu (a\otimes bc)=a(bc)$$ So $a(bc)=(ab)c$ if and only if $ ( \mu \circ (\mu\otimes Id))(a\otimes b\otimes c)=( \mu \circ (Id\otimes \mu))(a\otimes b\otimes c)$. But every element of $A\otimes A\otimes A$ is an $R$-linear combination of of finitely many terms of the form $a\otimes b\otimes c$ and $\mu \circ (Id\otimes \mu), \mu \circ (\mu\otimes Id)$ are $R$-module homomorphisms, so we get the desired.

Martin

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$Tensor Product II$

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