Maximal ideals of ultraproducts of full matrix algebras Let $\mathscr U$ be a non-principal ultrafilter over the natural numbers. Let $M_{\mathscr U}$ be the ultraproduct of all full matrix algebras $M_n$ along $\mathscr U$. This is a C*-algebra that is not simple as it contains a non-zero proper ideal, for example $\{[(x_n)]\colon \lim_{n, \mathscr U} \|x_n\|_{\rm HS} = 0\}$, where $\|\cdot\|_{\rm HS}$ stands for the Hilbert–Schmidt norm.

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*Is the cardinality of the set of maximal ideals of $M_{\mathscr U}$ known?

*Does $M_{\mathscr U}$ have an ideal of finite-codimension?

I anticipate that for Q1 the answer should be $2^{\mathfrak{c}}$ and for Q2 it should be no but I am somehow stuck.
 A: I think Nik Weaver is right that the ideal mentioned is the unique maximal ideal.
This simultaneously answers both questions (since the quotient is clearly infinite dimensional). Let $\tau$ be the trace on $M_\mathcal{U}$ defined as $\tau(x_n)=\lim_{n\rightarrow \mathcal{U}}\tau_n(x_n)$ where $\tau_n$ is the normalized trace on $M_n.$  As Nik Weaver mentioned in the comments, the ideal $\{ x\in M_\mathcal{U}:\tau(x^*x)=0 \}$ is maximal.  I claim this is the only maximal ideal. First we need a lemma from linear algebra
Claim: Let $a\in M_N$ be positive norm 1 and set $\varepsilon=\tau_N(A)>0.$  Then there are
$k=\frac{2}{\varepsilon}$ partial isometries $v_1,...,v_k\in M_N$ such that $\sum v_i^*av_i\geq \frac{\varepsilon}{2}I.$
Proof of Claim: Order the eigenvalues of $a$ as $a_1\geq a_2\geq\cdots \geq a_N$ and if $a_{\frac{N\varepsilon}{2}+1}<\frac{\varepsilon}{2}$ the trace is strictly less than $\varepsilon$ hence $a_i\geq \frac{\varepsilon}{2}$ for $1\leq i\leq \frac{N\varepsilon}{2}.$  Let $v_1$ project onto the $\frac{N\varepsilon}{2}$-dimensional subspace spanned by the first $\frac{N\varepsilon}{2}$ eigenvectors (corresponding to the ordering of the eigenvalues $a_i$).  Then twist this projection down the line with appropriate partial isometries to obtain the claim.
Back to the Answer: Let $I$ be an ideal that contains a positive, norm 1 element $x=(x_n)$ such that $\tau(x)>0.$  We will show that the ideal generated by $x$ contains the identity.  By replacing $(x_n)$ with an equivalent sequence we can assume each $x_n$ is positive, norm 1 and has $\tau_n(x_n)\geq\varepsilon$ for some $\varepsilon>0.$ Now just apply the above claim coordinate wise to produce $k=\frac{2}{\varepsilon}$ partial isometries $w_1,...,w_k\in M_\mathcal{U}$ so $\sum w_ixw_i^*\geq \frac{\varepsilon}{2}I.$
