Is there a relation between $End(M)$ and $M$ under tensor products?

Question

Let $R$ be a commutative ring and $M$ be an $R$-module (If necessary, you can assume more conditions)

Let $\phi$ be an $R$-endomorphism on $M$ and $\overline{\phi}:M^{\otimes n}\rightarrow M^{\otimes n}$ be the tensor product of $\phi$. (I used the notation $\overline{\phi}$ to distinguish the one that will come below)

Now let consider the tensor power of $End(M)$. That is, $End(M)^{\otimes n}$. And consider a tensor of it, namely, $\phi^{\otimes n}$.

Is there a relation between $\overline{\phi}$ and $\phi^{\otimes n}$?

I think the reason why I'm asking this is I don't actually have a picture of the concept tensor..

Answer 1

Consider the case $n=2$. The only general answer I know is this: by the universal property of tensor products, given $R$-modules $M_1,M_2,N_1, N_2$ there is a canonical homomorphism: $$\DeclareMathOperator{\Hom}{Hom}\Hom_R(M_1,N_1)\otimes_R\Hom_R(M_2,N_2)\to\Hom_R(M_1\otimes_RM_2, N_1\otimes_R N_2)$$ and this homomorphism is an isomorphism if one of the pairs $(M_1,M_2)$, $(M_1,N_1)$, $(M_2,N_2)$ consists in finitely generated projective $R$-modules (Bourbaki, Algebra, Ch. II, Linear Algebra, §4 no4, prop.4). So in your the answer is they're isomorphic for finite projective $R$-modules.

Note: This covers the case of finite dimensional vector spaces.

Answer 2

$\require{enclose}$ I just reformulate the rusults form vector spaces that i know.

For any modules $M_1,N_1,M_2,N_2$ there is an $\enclose{horizontalstrike}{\text{injective}}$ (EDIT. Apparently injectivity doesn't hold for modules over rings. Modules require additional conditions.) mapping $$j:Hom(M_1,N_1)\otimes Hom(M_2,N_2)\rightarrow Hom(M_1\otimes M_2,N_1\otimes N_2),$$ such that $$j(\phi\otimes\psi)(m_1\otimes m_2)=\phi(m_1)\otimes \psi (m_2).$$ So we can identify elements in $Hom(M_1,N_1)\otimes Hom(M_2,N_2)$ with elements in $Hom(M_1\otimes M_2,N_1\otimes N_2)$ (if this additional conditions hold). Hence given simple $\phi\in End(M)^{\otimes n}$ we can think of this as an element of $End(M^{\otimes n}).$ So $$\phi^{\otimes n}(m_1\otimes\dots\otimes m_n)=\phi(m_1)\otimes\dots\otimes \phi(m_n)=\bar\phi(m_1\otimes\dots\otimes m_n).$$

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If you want to read more, here is the reference: Greub, Multilinear Algebra, Springer 1967.