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Let $k$ a field of characteristic zero and $A = k\langle x , y\rangle / (xy - yx -1)$ the first Weyl algebra. Let $A$ be given a $\mathbb{Z}$-grading by setting $\deg(x) = 1$ and $\deg(y) = -1$. Is this a strong $\mathbb{Z}$-grading? (i.e. does $A_1A_{-1} = A_0 = A_{-1}A_{1}$.)

Checking this by brute force has not been successful, but there are a number of conditions that seem useful. My intuition is that it cannot be strongly $\mathbb{Z}$-graded, because this would then mean that every graded (left/right) ideal would be generated by the degree zero component (the right ideal $yA$ feels like it should be a counterexample) and that every graded (left/right) $A$-module would be strongly graded as well.

YSB
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    Btw, very often the condition of strong gradedness is not only what you described but that $A_n\cdot A_m=A_{n+m}$ for all $n$ and $m$. – Mariano Suárez-Álvarez Oct 19 '22 at 14:05
  • Indeed! I probably should have mentioned I was using the result that if we have a group grading it suffices to check $1 \in A_{\sigma_i}A_{\sigma_i^{-1}}$ for ${\sigma_i}$ a set of generators of the group. – YSB Oct 19 '22 at 16:35

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It turns out that it is not strongly $\mathbb{Z}$-graded. It can be shown with a little effort that as sets, $A_1 = xA_0 = A_0x$ and $A_{-1} = yA_0 = A_0y$. We also can argue that $A_0 = k[xy] \cong k[t]$ via $xy \mapsto t$ as rings.

Then $A_1A_{-1}$ is the image of the $A_0,A_0$-bimodule map \begin{align} A_1 \otimes A_{-1} = A_0 x \otimes y A_0 \to A_0 \xrightarrow{\sim} k[t], \\ x \otimes y \mapsto xy \mapsto t. \end{align}

It follows that $1 \not \in A_1A_{-1}$, so $A$ is not strongly $\mathbb{Z}$-graded.

YSB
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