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Suppose $k$ is a field of characteristic zero and $P$ and $Q$ are commuting elements of the first Weyl algebra. Is it true that $P$ and $Q$ are polynomials in a some third element $H$?

I asked a similar question before but without the characteristic zero condition.

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No. This counterexample is due to Dixmier, Proposition 5.5 of:

Dixmier, J. Sur les algèbres de Weyl Bull. Soc. Math. France 96 (1968) p. 209–242.

I learned about it from the early pages of

Makar-Limanov, L. Centralizers in the quantum plane algebra Studies in Lie Theory, Progress in Mathematics Volume 243, (2006), p. 411-416.

Let $$U = (\partial^3 + x^2 - 1)^2 + 2 \partial$$ $$V = (\partial^3 + x^2 - 1)^3 + \frac{3}{2} \left( (\partial^3 + x^2 - 1)^2 \partial + \partial (\partial^3 + x^2 - 1)^2 \right).$$

Then $$UV=VU \quad \mbox{and} \quad V^2 = U^3 + 1.$$ Disclaimer: I have not checked these identities; I just copy-typed them.

Suppose, for the sake of contradiction, that $U=u(T)$ and $V=v(T)$ for some polynomials $U$ and $V$ and some differential operator $T$. Clearly $T$ is nonconstant. Then $T$ does not obey any polynomial relation within the Weyl algebra (exercise). So we must have $$v(t)^2 = u(t)^3+1$$ as an equality of polynomials. We claim there are no nonconstant polynomials $u$ and $v$ obeying $v(t)^2 = u(t)^3+1$ (in characteristics not $2$ or $3$).

Conceptual proof: $v^2=u^3+1$ is a genus one curve, so it can't be algebraically parametrized by a genus zero curve.

Direct proof: Since $0^2 \neq 0^3+1$, $u$ and $v$ have no common zeros. Taking derivatives of both sides, we have $$2 v v' = 3 u^2 u'.$$ So every zero of $v$ is also a zero of $u'$ (with multiplicity), and we deduce that $\deg v \leq \deg u -1$. But also, every zero of $u$ is a zero of $v'$ (twice, in fact) so $\deg u \leq \deg v-1$. Contradiction.

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Thank you, I had actually read these articles at an earlier point but clearly part of them had slipped my mind. –  Johan Oct 12 '12 at 8:28

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