Two-sided closed ideals of $C(X,M_2(\mathbb C))$

Question

Let $X$ be compact and Hausdorff space. I know all closed ideals of $C(X)$. I want to substitute $\mathbb C$ by $M_2(\mathbb C)$. What can we say about two-sided closed ideals of $C(X,M_2(\mathbb C))$?

Answer 1

You should study Morita equivalence of C*-algebras (in the sense of Rieffel): the fact is that $C(X,M_2)$ is Morita equivalent to $C(X)$. And it is a general fact that Morita equivalent C*-algebras have same closed (two-sided) ideals. It follows that the closed ideals of $C(X,M_2)$ are all of the form $C_0(U,M_2)$ (functions vanishing outside $U$) for $U\subseteq X$.

More generally, you can replace $M_2$ by $M_n$ or even the algebra $K(H)$ of compact operators on some Hilbert space $H$. You can also replace $X$ by a locally compact space (and use $C_0(X)$ instead).

Answer 2

Perhaps an easier approach is as follows: Let $R$ be any unital commutative ring, then there is a one-to-one correspondence between ideals in $R$ and ideals in $M_2(R)$. In fact, if $J \subset M_2(R)$ is an ideal, then $$ I = \{a \in R : \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} \in J\} $$ is the corresponding ideal in $R$ such that $J = M_2(I)$. Thus, if $R$ is also a Banach algebra, one can go further and show that if the ideal in $M_2(R)$ is closed, then the corresponding ideal in $R$ must also be closed and vice-versa.

Now you just need to observe that $$ C(X)\otimes M_2(\mathbb{C}) \cong M_2(C(X)) $$ Thus, every ideal in $C(X)\otimes M_2(\mathbb{C})$ must be of the form $$ C_0(U)\otimes M_2(\mathbb{C}) $$ for some open set $U \subset X$.