Rational rotational algebra (noncommutative torus) is not simple I would like to show that the rational rotational algebra $A_\theta$ is not simple where $A_\theta= C^{*}(u,v : u,v$ are unitaries and $uv=e^{i2\pi\theta}vu$) and $\theta$ is a rational number.
The hint is given that I need to show the existence of unital $*$-homomorphisms $\phi : A_\theta → B $ and $\psi : A_\theta  → D $ such that $\phi (v^{q})=1$ and $\psi(v^{q})\neq 1$.
And I do not know how to show the existences of such homomorphisms.
Any help would be appreciated.
 A: It is sufficient to find some $C^*$-algebra $B$ containing unitaries $U, V$ satisfying $UV = e^{2 \pi i \theta} VU$ such that there is a non-injective *-homomorphism $\pi : A_\theta \rightarrow B$, as then $\ker \pi $ is a nonzero closed ideal in $A_\theta$. Continuing, if $\pi $ is chosen such that $\ker \pi \neq A_\theta$, then this would show that $A_\theta$ is not simple. 
One way to construct $\pi$ is to first define $\pi$ by the *-homomorphism satisfying $\pi (u) = U$ and $\pi (v) = V$ as the mapping $\mathrm{span}(v^m u^n : m,n\in \mathbb{Z}) \rightarrow B$, in noting that the closure of this span is $A_\theta$. If you can then establish that this mapping is continuous, then it would extend to a *-homomorphism $A_\theta \rightarrow B$. 
For what $B$ is, since $\theta$ is rational, choose $k \in \mathbb{Z}^+$ such that $k \theta \in \mathbb{Z}$. Let $B = M_k(\mathbb{C})$. Let 
$U = \begin{pmatrix}0 & 1 & 0 & \cdots & 0\\
0 & 0 & 1 & \cdots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & \cdots & 1\\
1 & 0 & 0 & \cdots & 0
\end{pmatrix}$ and $V = \text{diag}\left(1,e^{2\pi i\theta},e^{4\pi i\theta},\ldots,e^{2\left(k-1\right)\pi i\theta}\right)$
where diag stands for diagonal matrix. It is immediate to show that $U, V$ are unitaries satisfying $UV = e^{2 \pi i \theta} VU$ in $B$. Using the commuting relation for $u, v$ in $A_\theta$, and that $\{e^{2 \pi i n \theta} : n \in \mathbb{Z} \}$ is dense in $\mathbb{T}$, it is not hard to show that $\sigma(u) = \mathbb{T}$. However, since $B = M_k(\mathbb{C})$, $\sigma(U) \neq \mathbb{T}$. So if $\pi$ were injective, and possibly using some additional properties for $\pi$, a contradiction would be derived using $\pi (u) = U$ and these differing spectrums. Moreover $\pi (u) = U $ with $u, U \neq 0$ showing $\ker \pi \neq A_\theta$. 
