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I hope that my question yields some standard fact from (noncommutative) ring theory. In discussions with other graduate students, we have outlined some approaches to tackling the question, but haven't come up with an answer.

Notational caveats: I will work over some commutative ring that I will call $\mathbb Z$, but if you like, you may let $\mathbb Z$ denote your favorite algebraically closed field, or whatever — I am agnostic about that type of thing. I will denote by $\mathbb Z_q$ the Laurent polynomial ring $\mathbb Z_q = \mathbb Z[q^{\pm 1}]$ in a commuting parameter $q$, but again I am reasonably agnostic: if you need to use the field of rational functions, say, then so be it.

The quantum torus: I am interested in the noncommutative (but "$q$-mutative") Laurent polynomial ring of "functions on the quantum torus", defined by: $$ R = \mathbb Z_q\langle x^{\pm 1},y^{\pm 1}\rangle / (yx = qxy) $$ It is an "exponential" version of the Heisenberg ring $\mathbb Z[\hbar]\langle \xi,\upsilon\rangle / ([\upsilon,\xi] = \hbar)$, with $q = e^\hbar$, $x = e^\xi$, and $y = e^\upsilon$.

I am also interested in a distinguished left $R$-module: $$ M = \mathbb Z_q\langle x^{\pm 1}\rangle $$ $$ x \triangleright f(x) = x\ f(x), \quad y \triangleright f(x) = f(qx) $$ Again, this is an exponential version of the action of differential operators on polynomials, in which $\upsilon$ acts by $\frac{\partial}{\partial \xi}$.

Fix $f\in M$. It defines a left ideal $\operatorname{Ann}(f) \subseteq R$ consisting of those $r\in R$ such that $r\triangleright f = 0$.

My question: (When) is $\operatorname{Ann}(f)$ principal? I.e. (for which $f$) does there exist an element $a_f \in R$ such that $\operatorname{Ann}(f) = Ra_f$?

If it exists, then $a_f$ is determined up to units in $R$, which are precisely the monomials (unless I am mistaken). Is there an algorithm that computes $a_f$ from $f$?

For example, I believe that $\operatorname{Ann}(x^n)$ is generated by $y - q^n$. On the other hand, included within $\operatorname{Ann}(x-1)$ are the elements $(y-1)(y-q)$ and $x(y-q) - (y-1)$, and I don't see immediately a principal generator.

Not the main question, but an interesting topic for discussion: The classical limit $q \to 1$ gives the classical torus $R|_{q=1} = \mathbb Z[x^{\pm 1},y^{\pm 1}]$, with the symplectic structure $\omega = \frac{\mathrm dy}{y}\wedge \frac{\mathrm dx}{x}$ as the 1-jet of the noncommutativity. If I coarsely take the classical limit before computing the annihilation ideal, then for all non-zero $f$ I get the ideal generated by $(y-1)$, because $M$ is a domain. The more interesting thing to do is to compute $\operatorname{Ann}(f)$ in the quantum ring $R$, and then specialize $q \to 1$. This gives the ideal generated by $a_f(q=1)$, provided $\operatorname{Ann}(f)$ is principal in $R$. Can the vanishing locus of $\operatorname{Ann}(f)|_{q=1}$ be described in terms of the (symplectic) geometry of the function $f$?

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Oh please don't let $q$-mute catch on! –  Mariano Suárez-Alvarez Nov 27 '12 at 5:07
I know lots of people who use "q-mute" --- I certainly did not originate the phrase. –  Theo Johnson-Freyd Nov 27 '12 at 6:15

1 Answer 1

A longuish comment.

I'll work in $R=k\langle x^{\pm1},y^{\pm1}\rangle/(yx-qxy)$ with $q\in k\setminus0$ not a root of unity.

Notice that $M\cong R/R(y-1)$. If $f\in M$ is not zero, then $M/Rf$ is finite dimensional. As $R$ is infinite dimensional and simple (this is why I picked $q$ of infinite order), $M/Rf$ must be zero, that is, $Rf=M$.

Now we have short exact sequences $$0\to R(y-1)\to R\to M\to 0$$ with the last map being $r\mapsto r\triangleright 1_M$, and $$0\to\operatorname{ann}(f)\to R\to Rf\to 0$$ with the last map $r\mapsto r\triangleright f$, of course. The first one is a projective resolution of $M$, so $\operatorname{ann}(f)$ is projective, and Schanuel's lemma implies that there is an isomorphism of left $R$-modules $$R(y-1)\oplus R\cong \operatorname{ann}(f)\oplus R$$ We thus see that $\operatorname{ann}(f)$ is stably principal :-) (and that it can be generated by two elements)

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The $C^*$ variant of $R$ (the universal $C^*$-completion of the «normal» $R$) satisfies the cancellation property for projectives (this is a theorem of Rieffel) so modulo $C^*$-ization, we do get that the annihilator is principal. I cannot recall of cancellation holds also for our $R$, though. –  Mariano Suárez-Alvarez Nov 27 '12 at 6:09
This cancellation property doesn't hold in the algebraic case - in particular there are many non-isomorphic non-cyclic (left) ideals in $A_q$ (when $q \in \mathbb C^*$ isn't a root of unity). (By "non-isomorphic" I mean non-isomorphic as left modules.) See arxiv.org/abs/1010.3779 I would guess that typical annihilators are not principal, but I don't have a proof. –  Peter Samuelson Nov 27 '12 at 22:46

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