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Let $M$ be an $R$-module, and let $T(M) = \bigoplus_{k \geq 0 } T^k(M)$ denote the tensor algebra of $M$.

By definition $T^0(M)=R$ . I am looking for a simple explanation to this problem.

I am looking for units in $T(M)$. I can figure out that the units in $R$ are also units in $T(M)$ but are there any other units in $T(M)$. Can anyone give few examples where units in $T(M)$ which are not contained in $T^0(M)$?

I am studying about tensor algebra and I am wondering about the existence of units in $T(M)$. Any such example would help me in finding how elements of $T(M)$ behave. I am in a misconception that elements of $T(M)$ of degree $m$ and $n$ will multiply and give me an element whose component is in $T^{m+n}(M)$ and hence it cannot be an unit.

azif00
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zero2infinity
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1 Answers1

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It's definitely true that if $a\in T^n(M)$ and $b\in T^m(M)$, then $ab\in T^{n+m}(M)$. But that doesn't preclude additional units other than those in $T^0$, it just shows that no element in $T^m$ has an inverse in $T^n$ for $m>0$.

Here's one way to produce a unit of $T(M)$: let $R=\Bbb Z$ and $M=\Bbb Q/\Bbb Z$. Then $M\otimes_R M=0$, so $T(M)=\Bbb Z\oplus \Bbb Q/\Bbb Z$ with the multiplication defined by $(a,b)(c,d)\mapsto (ac,ad+bc)$, so the units are precisely those elements of the form $(\pm 1, x)$ with inverse $(\pm 1, -x)$ for $x\in\Bbb Q/\Bbb Z$.

KReiser
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