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Let $R=k\langle x,y\rangle$ be the free algebra on two variables. We can think of it as an algebra of non-commutative polynomials. Consider elements of the form $p[x,y]q$, where $p,q\in R$ are some monic monomials, and $[x,y]=xy-yx$.

My question is: how to check that elements $p[x,y]q$ are linearly independent?

I have tried to define some linear map from the subspace of $R$, spanned by all $p[x,y]q$ to some other linear space that will separate the elements, but it didn't really work. I was trying to order monomials and do some induction on the order, but it also didn't work.

I don't know the answer, so it may not be true. I might be missing something obvious, and I would be happy if you point that out for me.

Thank you very much!

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1 Answer 1

up vote 2 down vote accepted

Suppose $\sum_{i=1}^n a_ip_i[x,y]q_i=0$ is a dependence relation with $n$ minimal. This gives the equation:

$$\sum_{i=1}^na_ip_ixyq_i=\sum_{i=1}^na_ip_iyxq_i$$

Because $n$ is minimal, the $n$ monomials that make up each side of this equation must be in bijection with each other. That is, we can rearrange our indices so that:

$$a_1p_1xyq_1=a_2p_2yxq_2\\a_2p_2xyq_2=a_3p_3yxq_3\\\vdots\\a_np_nxyq_n=a_1p_1yxq_1\\$$

Now, notice that it follows that all of the $p_i$'s start with the same string (the $p_j$ of minimal length). Getting rid of this common string, and relabeling cyclically if necessary, we can assume that $p_1=1$. From the first equation, it follows that $p_2$ begins with an $x$, which then implies by the second equation that $p_3$ also begins with an $x$. Continuing with this reasoning, we find that $p_n$ begins with an $x$, which contradicts the last equation (remember that $p_1=1$).

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