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Given a vector space $V=\mathbb{F}^{d}$, the free algebra, or tensor algebra, of $V$ is $T\left(V\right)=\oplus_{n\geq0}V^{\otimes n}$. Now, it is stated everywhere, that this is exactly the algebra of non-commutative polynomials over $d$ indeterminates.

I'm probably missing something really basic, but can someone please show me why this is true? I would really appreciate an intuitive explanation with a fromal one, as i'm having a really hard time grasping the concepts in this subject. Thanks alot!

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    If we fix a basis $e_1,\ldots, e_d$ for $V$, then $V^{\otimes n}$ has a basis given by "words" of length $n$ in the letters $e_1,\ldots, e_d$. If this doesn't help, how do you think about non-commutative polynomials? Are you aware of the (basis independent) construction of commutative polynomials as the symmetric algebra of a vector space?2011-11-17
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    yes, that helps, i knew it was something small i need to think about and everything will fall into place. thanks!2011-11-17
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    @Aaron Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-10
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    @JulianKuelshammer: I have converted my comment into an answer. In the future, I will make sure my comments aren't useful enough to stop people from posting answers.2013-06-10

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