Are matrices of continuous functions on the circle non-simple?

Question

Let $C(\mathbb{T})$ denote the continuous function on the circle. I believe that $M_n(\mathbb{C}) \otimes C(\mathbb{T})$ is non-simple. More precisely, I suspect that $C(\mathbb{T})$ sits as an ideal inside of $M_n(\mathbb{C}) \otimes C(\mathbb{T})$, but all I want to determine is whether or not said $C^*$-algebra is simple. Are any of you aware of the answer to this question?

Answer 1

For $R$ a commutative ring, the two sided ideals of $M_n(R)$ are in correspondence with the ideals of $R$.

The correspondence is as follows: Given an ideal $I$ of $R$, we have $M_n(I)$, of matrices with entries in $I$. On the other hand, given an ideal $J$ of $M_n(R)$, we can from the ideal $E(J)$ generated by all entries of matrices from $J$. It is clear that $E(M_N(I)) = I$. It is clear that $M_N(E(J)) \supset J$. To see the other inclusion, we use a series of projections -- let $E_i$ be the projection onto the $ith$ basis vector. Then $E_i M E_j$ is the matrix with $m_{ij}$ in its originalspot, and zeros everywhere else. Then you can use permutations to move its position around.

$C(T)$ has many ideals. For instance, any point $p \in T$ determines a maximal ideal by $\{ f | f(p) = 0 \}$. The qoutient of this ideal is the evaluation at $p$ function onto $\mathbb{C}$.