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The polynomial ring $k[x,y]$ has graded $k$-dimension (aka. Poincaré-series) $\frac{1}{(1-q)^2}$. This is clear since the polynomial ring in one indeterminate has $\dim_q k[x]=1+q+q^2+\cdots=\frac{1}{1-q}$ the harmonic series.

Now I may regard $k[x,y]$ as the quotient $k\langle x,y\rangle/(xy-yx)$ of the free (non-commutative) algebra on two generators by the ideal encoding commutativity. I want to use this approach to compute $\dim_q k[x,y]$.

In $R:=k\langle x,y\rangle$, I have $2^l$ possibilities two form a monomial of degree $l$ from $x$ and $y$. Thus $k\langle x,y\rangle$ has graded dimension $d:= \sum_{l=0}^{\infty} (2q)^{l}=\frac{1}{1-2q}$. The (two-sided) ideal $R(xy-yx)R$ has dimension $q^2d^2$. The quotient thus should have graded dimension $$\frac{1}{1-2q}\Big(1-\frac{q^2}{1-2q}\Big) = \frac{1-2q-q^2}{(1-2q)^2}. $$ This does not equal $\frac{1}{(1-q)^2}$.

Question: Where is the conceptual mistake in my approach?

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You are assuming, without justification, that $R\otimes_k R\to R(xy-yx)R$ given by $f\otimes g\mapsto f(xy-yx)g$, is an isomorphism. It isn't.

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  • $\begingroup$ You are right that this assumption lacks justification. However, I do not see why this is no isomorphic, or even a base for its kernel. Would you mind helping me? $\endgroup$ – Bubaya Aug 14 '17 at 20:04
  • $\begingroup$ An element of its kernel is $(xy-yx)\otimes1-1\otimes(xy-yx)$. $\endgroup$ – Lord Shark the Unknown Aug 14 '17 at 20:04
  • $\begingroup$ Too obvious, thank you. $\endgroup$ – Bubaya Aug 14 '17 at 20:42

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