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The following question is related to this post What elements in a tensor algebra are invertible? and mainly I am confused if the answer in the above post applies to showing that the tensor algebra of a vector space (finite dimensional) does not contain any zero divisors.

Let $T(V) = \oplus_{k=0}^{\infty} T^k(V)$ be the tensor algebra of a finite vector space $V$.

How do you $T(V)$ does not contain any zero divisors?

2 Answers 2

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One way to see this is as follows.

Pick a basis $\{x_1,\dots,x_n\}$ for $V$. Then the set $B$ of all non-commutative monomials $x_{i_1}x_{i_2}\cdots x_{i_r}$ of all lengths $r\geq0$ (including the zero length) and with $i_1,\dots,i_r\in\{1,\dots,n\}$, is a basis of $TV$. Moreover, this elements are multiplied by juxtaposition.

Now suppose $a$ and $b$ are non-zero elements of $TV$. Then we can write both of them as a linear combination of the elements of $B$. Suppose that $x_{i_1}x_{i_2}\cdots x_{i_r}$ is one of the monomials of $B$ which appears in $a$ with a non-zero coefficient and has maximal length, and suppose that $x_{j_1}x_{j_2}\cdots x_{j_s}$ is one of the monomials of $B$ which appears in $b$ with a non-zero coefficient and has maximal length. Then you can show that the monomial $x_{i_1}x_{i_2}\cdots x_{i_r}x_{j_1}x_{j_2}\cdots x_{j_s}$ is a monomials of $B$ which appears in $ab$ with a non-zero coefficient: it follows then that $ab\neq0$.

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Here is a philosophical reason: the tensor algebra is the free algebra on some vector space. That is, it only satisfies the relations it has to in order to be an associative algebra, and nothing more. If it had zero-divisors, that would be one more relation it satisfies that it doesn't have to. Also, it would no longer satisfies the universal property, those elements that were zero divisors can not be mapped to elements that are not zero divisors. But any map out of the tensor algebra is determined by a map out of the given vector space, and the vector space has no way of knowing that those elements would have to go to zero divisors.

(please excuse the personification)

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    You have to try and see! :)2011-10-24