Laurent polynomial algebra simple Assume any element of $A_q=\mathbb{C}\langle x^{\pm 1}, y^{\pm 1}\rangle/\langle xy-qyx\rangle$ (for $q$ not a root of unity) can be uniquely written in the form $\sum_{i=m}^nf_i(x)y^i$, where $m<n$ are integers (not necessarily positive), and $f_i(x)$ are Laurent polynomials in $x$ (i.e., $f_i(x)\in\mathbb{C}[x,x^{-1}]$). Show that the $\mathbb{C}$-algebra $A_q$ is simple.
My attempt:
To show that $A_q$ is simple, it would suffice to show that every two-sided ideal is trivial. Since $A_q$ is commutative, it suffices to show that for any $a\in A_q$, $\langle a\rangle=A_q$. Towards this end I thought if I could show that $\mathbb{C}[x,x^{-1}]$ is a division algebra, then I could show the case where $a=f_i(x)y^i$ for some $i$. But even in this case, I couldn't find a proof (Although I was able to show that if we replace $\mathbb{C}[x,x^{-1}]$ by $\mathbb{C}[[x,x^{-1}]]$ then everything would work).
Any suggestions on how to continue? After that avenue of thought I'm lost.
 A: We can show any two-sided ideal $I$ contains a nonzero constant, hence $I$ is the whole algebra.
To do this, start with an arbitrary (without loss of generality nonzero) element
$$ S=\sum_{i} f_i(x)y^i\in I.$$
One of its terms $f_j(x)y^j$ must be nonzero, in which case we may right-multiply by $y^{-j}$ in order to obtain a new element $R=Sy^{-j}\in I$, say given by
$$ R=\sum_i g_i(x)y^i\in I, $$
whose $g_0(x)y^0$ term is nonzero (it's just $f_j(x)$). If this is the only summand, then we've shown $I$ contains a nonzero polynomial $g(x)\in I$. Suppose it isn't the only summand.
Since the relation $xyx^{-1}=qy$ holds, if we conjugate by $x$ we get
$$ xRx^{-1} = \sum_i g_i(x)xy^ix^{-1}=\sum_i g_i(x) (xyx^{-1})^i=\sum_i q^i g_i(x)y^i.$$
Since $q^0=1$ and $q^i\ne 1$ for $i\ne0$ (as $q$ is not a root of unity), this has changed all but the constant coefficient. That means if we take the difference
$$xRx^{-1}-R=\sum_i (1-q^i)g_i(x)y^i \in I$$
we get an element of $I$ with exactly one less term (the $g_0(x)y^0$ term is deleted). We may proceed in this way to delete more and more terms until we end up with a nonzero polynomial $g(x)\in I$.
At this point, we may write $g(x)=\sum_i c_i x^i$, then do essentially the same thing: first multiply by a power of $x$ in order to assume without loss of generality that $c_0x^0$ is a nonzero term, then conjugate by $y$ to get
$$ yg(x)y^{-1}=\sum_i  q^{-i}c_ix^i, $$
in which case
$$ yg(x)y^{-1}-g(x)=\sum_i (1-q^{-i})c_ix^i\in I $$
is an element of $I$ with exactly one less term than $g(x)$. We may proceed in this way to delete term after term until we get a constant in $I$.
