TITLE: Smallest $C^*$-subalgebra in tensor product
QUESTION [2 upvotes]: I would like to prove that:

If $\mathfrak B$ and $\mathfrak C$ are two finite dimensional
$C^*$-algebras, and $\mathfrak A$ is a subalgebra of $\mathfrak B \otimes \mathfrak C$, then, there exists a $C^*$-subalgebra $\mathfrak B_0$ of $\mathfrak B$ such that it is the smallest $C^*$-algebra
satisfying  $$ \mathfrak A \subset \mathfrak B_0 \otimes \mathfrak C. $$

This is vaguely proved here in Lemma 7, pages (20-)21.
My attempt:
After fixing basis $\{f_1,\ldots,f_m\}$ of $\mathfrak B$ and $\{e_1,\ldots,e_n\}$ of $\mathfrak C$, every element $a$ of $\mathfrak A$ is an element of $\mathfrak B\otimes \mathfrak C$,
so that it can be written uniquely
in terms of the basis $\{f_j\otimes e_i : i=1,\ldots,n, j=1,\ldots,m\}$ of $\mathfrak B\otimes \mathfrak C$,
$$ a = \sum_{j=1}^m\sum_{i=1}^n \alpha_{ji} f_j\otimes e_i = \sum_{i=1}^n \Big(\sum_{j=1}^m \alpha_{ji} f_j \Big) \otimes e_i = \sum_{i=1}^n a_{i}\otimes e_i,$$
for some complex numbers $\{\alpha_{ji}:j=1,\ldots,m,i=1,\ldots,n\}$,
where we have denoted $a_i = \sum_{j=1}^m \alpha_{ji} f_j \in\mathfrak B$, for every $i\in\{1,\ldots,n\}$.
If $\mathfrak D$ is any $C^*$-subalgebra of $\mathfrak B$ satisfying
$\mathfrak A \subset \mathfrak D \otimes \mathfrak C$, then, necessarily,
$a_i\in \mathfrak D$ for each $i\in\{1,\ldots,n\}$, so that the $C^*$-subalgebra
generated by the set of elements $a_1,\ldots,a_n$ arising in this way, whenever the starting
$A\in\mathfrak A$ is, is necessarily the wanted $C^*$-subalgebra, $\mathfrak B_0$.
Edit: Why? Because if we consider the functionals $\omega_i:\mathfrak C\to\mathbb C$ such that $\omega_i(e_k) = \delta_{i,k}$, that is, the
corresponding dual basis for $\mathfrak C^*$ associated to $\{e_i\}$, then
$$ (\mathrm{Id}\otimes \omega_i)(a) = a_i $$
and since $(\mathrm{Id}\otimes \omega_i)$ maps $\mathfrak B\otimes \mathfrak C$ to $\mathfrak B$, then $a_i$, written as above, should belong to $\mathfrak B$.
If we change $\{f_j\}$ and $\{e_i\}$ this should not be a problem, and the $C^*$-algebra generated by the elements $a_i$ arising with the new basis should be the same, so that
$\mathfrak B_0$ is independent of this basis (I haven't proved rigourously this yet)
Is everything right?

REPLY [1 votes]: Your argument seems to depend on assuming that if $\sum_j a_j\otimes b_j\in A\otimes B$, then $a_j\in A$ for all $j$. That's not true.
For an easy example, take $A=D\otimes D\subset M_2(\mathbb C)\otimes M_2(\mathbb C)$, where $D$ is the diagonal algebra. Let $x$ be any non-diagonal matrix. Then $$0=x\otimes I+(-x)\otimes I\in D\otimes D,$$ while $x\not\in D$.
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