Show that $A \otimes B = A \odot B$ if $A$ is a finite-dimensional $C*$-algebra. Let $A$ and $B$ be $C^*$-algebras. If $A$ is finite-dimensional, then I want to show that the algebraic tensor product $A \odot B$ is complete for the minimal $C^*$-norm on it (in fact, the choice of norm does not matter since $A$ is nuclear).
Attempt:
I know that $$A \cong M_{n_1}(\mathbb{C}) \oplus \dots \oplus M_{n_k}(\mathbb{C})$$
And thus
$$A \odot B \cong (A \odot  M_{n_1}(\mathbb{C}))\oplus \dots \oplus (A\odot M_{n_k}(\mathbb{C})) $$
so I believe it suffices to show that $A \odot M_{n_1}(\mathbb{C})\cong A \odot \mathbb{C}^{n^ 2}$  is complete.
Then I'm stuck.
Maybe I'm overcomplicating this though....
 A: You wrote as if $B$ is the finite-dimensional one, so I'll stick with that.
Let $\gamma$ be a C$^*$-norm on $A\otimes M_n(\mathbb C)$ (one certainly exists, because we can represent $A\subset B(H)$ and then $A\otimes M_n(\mathbb C)$ can be represented in $B(H\otimes\mathbb C^n)$).
For any $k$, the map $a\longmapsto \gamma(a\otimes E_{kk})$ defines a C$^*$-norm on $A$. As a C$^*$-algebra admits a unique C$^*$-norm, we get that $\gamma(a\otimes E_{kk})=\|a\|$. Even if $k\ne j$,
$$
\gamma(a\otimes E_{kj})^2=\gamma(a^*a\otimes E_{jj})=\|a^*a\|=\|a\|^2.
$$
Now suppose that we have a Cauchy sequence  $\{x_n\}\subset A\otimes M_n(\mathbb C)$. We can write $$x_n=\sum_{k,j} a_n(k,j)\otimes E_{k,j}.$$
For any $k,j$,
\begin{align}
\|a_n(k,j)-a_m(k,j)\|^2&=\gamma((a_n(k,j)-a_m(k,j))\otimes E_{11})\\[0.3cm]
&=
\gamma((1\otimes E_{11})(x_n-x_m)(1\otimes E_{11}))\\[0.3cm]
&\leq\gamma(x_n-x_m).
\end{align}
So each sequence $\{a_n(k,j)\}_n$ is Cauchy on $A$. As $A$ is a C$^*$-algebra, there exist $a(k,j)\in A$ with $\lim_na_n(k,j)=a(k,j)$. We want to show that $x_n\to\sum_{k,j} a(k,j)\otimes E_{k,j}$. For this
\begin{align}
\gamma(x_n-\sum_{k,j} a(k,j)\otimes E_{k,j})
&=\gamma(\sum_{k,j} (a_n(k,j)-a(k,j))\otimes E_{k,j})\\[0.3cm]
&\leq \sum_{k,j} \gamma((a_n(k,j)-a(k,j))\otimes E_{k,j})\\[0.3cm]
&=\sum_{k,j} \|a_n(k,j)-a(k,j)\|\\[0.3cm]
&\to0,
\end{align}
where it is essential in this last step that the sum is finite. So $A\otimes M_n(\mathbb C)$ is complete with respect to $\gamma$, and so it is a C$^*$-algebra. This in particular shows that $\gamma$ is necessarily the norm described in the second paragraph.
A: The point is that $A \odot B$ is closed in $A \otimes B$. To see this, choose a basis $\{e_i\}_{i=1}^n$ for $A$ and suppose that $\{x_k\}_{k=1}^\infty$ is a sequence in $A \odot B$ that converges to $x \in A \otimes B$. Then we can write
$$x_k = \sum_{i=1}^n e_i \otimes b_i^k$$
and since $x_k \to x$ in $A \otimes B$, applying the slice map $\tau_j \otimes \iota: A \otimes B \to B$ where $\tau \in A^*$ satisfies $\tau(e_i) = \delta_{i,j}$, we see that
$$b_j^k = (\tau_j \otimes \iota)(x_k) \to (\tau_j \otimes \iota)(x).$$
Hence, $$x=\sum_{i=1}^n e_i \otimes (\tau_i \otimes \iota)(x) \in A \odot B.$$
