closed ideals of $C(X,\mathcal{B})$ Question:- Let $X$ be a compact Hausdroff space and $\mathcal{B}$ be a simple C*-algebra. Prove that the closed ideals of $\mathcal{B}\otimes_{min} C(X) \cong^* C(X,\mathcal{B})$ are of the form $\mathcal{B}\otimes J$ where $J$ is a closed ideal of $C(X)$.
Proof:- Let $I$ be a proper closed ideal of $C(X,\mathcal{B})$ and set 
$$E = \{x\in X : f(x) = 0 \text{ for all } f\in I\}.$$
Then $E$ is a closed subset of $X$ and $I\subseteq I(E) := \{f\in C(X,\mathcal{B}) : f(x) = 0 \text{ for all } x\in E\}$. Indeed, let $x$ be a limit point of $E$. Suppose there exist $f\in I$ such that $f(x)\neq 0$. Then, by continuity, there exists an open set $U$ containing $x$, such that $f(y)\neq 0$ for all $y\in U$. Since $U$ is neighborhood of $x$, there is $q\in E$ and $q\neq x$, for which $f(q)\neq 0$, which is a contradiction. Therefore, $x\in E$.
claim : $I(E)\subseteq I$.
Assume that $X\cap E = \emptyset$. Then, for every point $p\in X$, there exists a continuous function $f\in I$ such that $f(p)\neq 0$. Then, by continuity, there must exist an open set $U$ containing $p$ so that $f(q)\neq 0$ for all $q\in U$. Thus, we may assign to each point $p\in X$ a continuous function $f\in I$ and an open set $U$ of $X$ such that $f(q)\neq 0$ for all $q\in U$. Since this collection of open sets covers $X$, which is compact, there must exists a finite subcover which also covers $X$. Call this subcover $U_1,\cdots, U_n$ and the corresponding functions $f_1, \cdots, f_n$. Consider the function $g$ defined as 
$$g(x) = f_1(x) f^*_1(x)+\cdots+f_n(x) f^*_n(x).$$
Since $I$ is an ideal, $g\in I$. For every point $p\in X$, there exists an integer $i$ between $1$ and $n$ such that $f_i(p)\neq 0$. This implies that $g(p) \neq 0$. Also, $g$ is positive element of $\mathcal{B}$.
how to proceed further......
if I prove that $g^{-1}$ exist, then $g.g^{-1}\in I$, which implies, $I = C(X,\mathcal{B})$, a contradiction (as $I$ is proper). 
 A: I will post an answer for a more general case, since I couldn't see how to proceed in your approach. 
In my proof I'll need Kirchberg's slice lemma. The proof can be found in RØRDAM's book: "Classification of Nuclear $C^*$-Algebras.  Entropy in Operator Algebras"- Lemma 4.1.9.:

Kirchberg's Slice Lemma: Let $A$ and $B$ be unital $C^*$ algebras and let $D$ be a hereditary $C^*$-subalgebra of $A\otimes B$. Then there exists a non-zero element $z\in A\otimes B$ such that $zz^*\in D$ and $z^*z=a\otimes b$ for some $a\in A_+$ and $b\in B_+$.

Observation: Let $A$ and $B$ be unital $C^*$ algebras. If $I\subseteq A\otimes B$ is a non-zero ideal, then $I$ contains a non-zero elementary tensor $a\otimes b \neq 0$.
Proof: Let $z$ be the element we get by applying Kirchberg's slice lemma with $D=I$. Then $z^*zz^*z=a^2\otimes b^2$ is a non-zero element in $I$ as required.
Claim: Let $A$ be a unital, simple $C^*$-algebra and let $B$ be a unital, nuclear $C^*$-algebra. Then ideals in $A\otimes B$ have the form $A\otimes J$ where $J$ is an ideal in $B$.
Proof: Let $I$ be an ideal in $A\otimes B$. Since $A$ is a assumed to be unital, $(\Bbb{C}1_A\otimes B)\cap I$ is an ideal in $\Bbb{C}1_A\otimes B\cong B$, denote it $J$. Since we're working with the minimal tensor product, we have $A\otimes J\subseteq A\otimes B$. We hope to show that $I=A\otimes J$. The inclusion $A\otimes J\subseteq I$ is easy.
Let's show $I\subseteq A\otimes J$. 
Let $q: A\otimes B\to \frac{A\otimes B} {A\otimes J}\cong A\otimes (B/J)$ (using nuclearity of $B/J$ and corollary 3.7.4. in Brown&Ozawa's book) be the quotient map.
If $q(I)=0$ we're done. Assume that this is not the case. So $q(I)$ is a non-zero closed ideal in $A\otimes (B/J)$ and by the above observation there exists a non-zero element $a\otimes (b+J)\in q(I)$ $(*)$
Let $x\in I$ be such that $q(x)=a\otimes (b+J)=q(a\otimes b)$, then $x-(a\otimes b)\in ker(q)=A\otimes J\subseteq I$, and since $x\in I$ also $a\otimes b\in I$ (non-zero).
Since $I$ is an ideal in $A\otimes B$, we conclude that $<a>\otimes <b>\subseteq I $ (the notation $<.>$ is used for "the ideal generated by"). But $A$ is assumed to be simple, thus $A\otimes <b>\subseteq I$.
Therefore, $b\in J=I\cap B$, which contradicts assumption $(*)$, i.e. $a\otimes (b+J)\neq 0$.
Remark: My $A$ is your $B$ and my $B$ is your $C(X)$. I assumed twice in my proof that $A$ is unital. At least, to get a copy of $B$ in the tensor product I need a non-zero projection in $A$. I don't know how to generalize my proof to the non-unital case.
