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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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    @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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If we fix a basis $e_1,…,e_d$ for $V$, then $V^{\otimes n}$ has a basis given by "words" of length $n$ in the letters $e_1,…,e_d$. This allows you to construct an isomorphism between $T(V)$ and the free non-commutative algebra in $n$ variables.

This construction is analagous to the (basis independent) construction of commutative polynomial rings as the symmetric algebra of a vector space. (How helpful this analogy is depends on what experience one has with the aforementioned construction).