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.


1. Is the cardinality of the set of maximal ideals of $M\_{\mathscr U}$ known?
2. 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.



Maximal ideals of ultraproducts of full matrix algebras
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 $\sum w\_ixw\_i^\*\geq \frac{\varepsilon}{2}I.$


