Non-commutative algebra $A$ with a non-trivial maximal ideal $M$ such that $A/M$ is not a division algebra

Question

By "ideal", I mean "two-sided ideal".

I'm looking for an example of a real/complex algebra $A$ which is non-commutative and has some maximal ideal $M$ with $\{0\}\subsetneq M\subsetneq A$ and $A/M$ is not a division algebra.

The counterexamples for

$$ M \text{ maximal ideal}\implies A/M \text{ division algebra} $$

that I know of are $\mathbb{F}^{n\times n}$ (with $\mathbb F$ some field) and the Weyl algebra, but both of these are simple rings.

Answer 1

In other words you are looking for a ring with unique maximum ideal that is not simple and not local.

Take any noncommutative simple ring $R$ that is not a division ring. You said you knew a couple: you could take the two by two matrix ring over the reals or complex numbers.

Form the trivial extension $S=R\times R$ where the addition operation is coordinatewise and multiplication is $(a,b)(c,d)=(ac,ad+bc)$.

Since $I=\{0\}\times R$ is a nonzero nilpotent ideal, $S$ is not simple. Furthermore the maximal ideals of $S$ correspond to those of $S/I$ which is simple, so $I$ is the unique maximal ideal, but not the unique maximal right ideal (since $S/I$ lacks a unique maximal right ideal.)

Answer 2

Let $F$ be a field and let $V$ be a vector space over $F$ having a countable basis $\{e_0,e_1,\dots,e_n,\dotsc\}$.

Let $R$ be the endomorphism ring of $V$. Then this ring has a unique nontrivial two-sided ideal, namely the set $K$ of endomorphisms with finite rank (in other words, such that the image is finitely generated). That this is a two-sided ideal is easy to see.

Suppose that the ideal $I$ contains an endomorphism $f$ having infinite rank. Then we can select $\{e_{k_0},e_{k_1},\dotsc\}$ such that $\mathscr{B}=\{f(e_{k_0},f(e_{k_1}),\dotsc\}$ is a basis for the image of $f$. Extend it to a basis $\mathscr{B}\cup\mathscr{C}$ (disjoint union) of $V$.

Now consider $g\in R$ defined by $g(f(e_{k_i}))=e_i$ and $g(v)=0$ for $v\in\mathscr{C}$ and $h\in R$ defined by $h(e_i)=e_{k_i}$.

Then we have $$ gfh(e_i)=gf(e_{k_i})=e_i $$ so $gfh$ is the identity and $gfh\in I$ and $I=R$.

Thus $K$ is indeed a maximal two-sided ideal, but $R/K$ is not a division ring, because $R$ has proper left ideals that properly contain $K$ (find an example).

As an aside, if $V$ is a vector space of dimension $\beta$ (an infinite cardinal, choice assumed), the proper nonzero ideals of the endomorphism ring $R$ of $V$ are of the form $$ K_\alpha=\{f\in R:\dim\operatorname{im}f<\alpha\} $$ where $\alpha$ runs through the infinite cardinals less than or equal to $\beta$. So we have several other examples, because the same idea as before shows that $K_\beta$ is the unique maximal two-sided ideal, but the quotient ring is not a division ring.