(FD) Quotient by commutator ideal a subalgebra? A quotient of a $\mathrm{C}^*$-algebra is not generally a subalgebra. The counterexamples I have seen take an infinite dimensional commutative algebra and quotient by an ideal as per here.
My question concerns the quotient of a $\mathrm{C}^*$-algebra by its commutator ideal such that the quotient is non-zero finite dimensional aka there are only finitely many characters.
Is the quotient a subalgebra in this case?
 A: Look at
$$\mathcal A= \{ f: [0,1]\to M_2(\Bbb C) \mid f(0) = a\,e_{11}, f(1)= b\,e_{22}, \ \ a,b\in \Bbb C\}$$
where $e_{ij}$ is the usual basis of $M_2(\Bbb C)$.
Any commutator must vanish at $0$ and $1$. Outside of those two points note that any matrix in $M_2(\Bbb C)$ is in the commutator ideal of $M_2(\Bbb C)$. In particular you may write $e_{ij} = \sum_k A_k [B_k, C_k]$, further you may write any function $f$ on $[0,1]\to\Bbb C$ as a product of $3$ functions to get:
$$f(x) e_{ij} = f_{ij}(x) \sum_k A_k[B_k,C_k]= \sum_k h^{1}(x) A_k [h^2(x) B_k, h^{3}(x) C_k]$$
where if $f$ vanishes on $0$ and $1$ you may assume all the $h^k$ to do so also. So:

The commutator ideal of $\mathcal A$ is equal to all functions to $M_2(\Bbb C)$ vanishing on $0$ and $1$. The quotient of $\mathcal A$ by this ideal is then $\Bbb C^2$.

Now lets check that $\Bbb C^2$ is not a sub-algebra of $\mathcal A$. The simplest way to do this is to note that $\mathcal A$ doesnt have two disjoint self-adjoint projections:
If $p$ is a projection then it must take on either the values $0$ or $e_{11}$ (resp $e_{22}$) on at $0$ (resp $1$). If we have two disjoint projections then at least one of them needs to take on $0$ at one of the endpoints (else the product is not zero). If a projection $p$ has $p(x)=0$ for some $x$ then there must be some $x'$ with $0<\|p(x')\|<1$ (else its constant $0$). But the equation
$$p(x') \overset!= p(x')^2$$
cannot be satisfied for any self-adjoint matrix of norm greater than $0$ and less than $1$.
