Maximal ideal that annihilates entire ring Does there exist a ring $R$ with a nonzero maximal ideal $M$ such that $R^2=R$ and $MR = RM = 0$?
Here $R$ is associative but does not have an identity (obviously).  It seems a simple enough question but I'm not even sure of the answer if I insist that $M$ be the only proper ideal.  I know $R$ can't be commutative and that the conditions can't be relaxed (much); for example, there is a commutative ring satisfying these conditions if $M$ need not be maximal.
 A: Ok, I think I worked out the bugs in my previous answer.
Let $F$ be the field with two elements (for simplicity).  Let $T=F< a_{i}, x_{k,i},y_{k,i}:i\in \mathbb{N}, k\in K>$ be the free algebra over $F$, generated by the non-commuting variables $a_{i}$, $x_{k,i}$, and $y_{k,i}$ (where $k$ runs over an indexing set $K$ we will describe later, and $i$ runs over the non-negative integers), with no constant term.  Let $J$ be the ideal generated by the relations $a_{i}=a_{2i+1}a_{0}a_{2i+2}$, $x_{k,i}=x_{k,2i+1}a_{0}x_{2i+2}$, $y_{k,i}=y_{k,2i+1}a_{0}y_{k,2i+2}$ and $a_{0}^{2}a_{i}=a_{0}^{2}x_{k,i}=a_{0}^{2}y_{k,i}=a_{i}a_{0}^{2}=x_{k,i}a_{0}^{2}=y_{k,i}a_{0}^{2}=0$ for each $i$.  Let $S=T/J$ and identify each variable with its image in the quotient (and continue this practice below).  Notice that $S$ is generated by $a_{0}$ and $S^{2}=S$.  Also note that $a_{0}^{2}$ is a universal zero-divisor.  The ring $R$ we want will be a quotient of $S$, so will still be generated by $a_{0}$, we will still have $R^{2}=R$, and $a_{0}^{2}$ will still be a universal zero-divisor.  We will construct $R$ so that $a_{0}^{2}$ remains nonzero and generates a maximal ideal.
Let $I'$ be the set of words $w\in T$, $w\notin a_{0}^{2}+J$,  such that there is a word $w_{1}\in T$ of length 1 (i.e. a variable) with $w_{1}w-a_{0}^{2}\in J$ or $ww_{1}-a_{0}^{2}\in J$ but no words $w_{2},w_{3}\in T$ with $w_{2}w-a_{0}\in J$ or $ww_{3}-a_{0}\in J$ or $w_{2}ww_{3}-a_{0}\in J$. [For example, $a_{0}a_{1}a_{0}a_{5}a_{0}$ is such a word.  Multiplying on the right by $a_{6}$ it reduces to $a_{0}^{2}$, but it does not equal $a_{0}^{2}$ modulo $J$, and can never be multiplied to $a_{0}$.]  If we make words in $I'$ zero (or even zero divisors) that will possibly make $a_{0}^{2}$ zero.  So, let $I$ be the ideal generated by the following relations: $x_{w,0}wy_{w,0}=a_{0}$ if $w\in I''$, and take $S_{1}=S/I$.
At this point, we repeat the argument in the previous paragraph on $S_{1}$ (our new set $I_{1}'$ will have new words in it).  We do this countably many times, the resulting ring is $S_{\infty}$.
Next, mod out by the ideal generated by words $w\neq a_{0}^{2}$ such that there exist no words $w_{1}, w_{2}$ so that $w_{1}w=a_{0}^{2}$ or $ww_{2}=a_{0}^{2}$ or $w_{1}ww_{2}=a_{0}^{2}$ (in $S_{\infty}$).
The resulting quotient ring should be the structure you are looking for.  (The index set $K$ could initially be all words in $T$, then cut down to words that appeared in any of the $I'$'s above.)
This is complicated enough that I might have made a mistake somewhere--this is just a sketch of my ideals.  One of the main points that should be noted is that (besides $a_{0}$) each variable appears in exactly two relations (except those when a word is made equal to 0): one where it is used to reduce the size of a word, and one where the size expands.
A: This is not an answer, but maybe a useful reformulation. An equivalent question is: 
Does there exist a (nonunital) associative algebra $I$ over a field $k$, such that $I^2=I$, and such that $0$ and $I$ are the only ideals, and such that the sequence $I^{\otimes 3}\to I^{\otimes 2}\to I$ is not exact? Here the first map is $a\otimes b\otimes c\mapsto ab\otimes c-a\otimes bc$ and the second is $a\otimes b\mapsto ab$.
The main point here is that if $R$ and $M$ are as in the question then you can (1) take $I$ to be $R/M$ (2) make the cokernel of $a\otimes b\otimes c\mapsto ab\otimes c-a\otimes bc$ into a ring by $(a\otimes b)\times (c\otimes d)=ab\otimes cd$, and (3) define a surjective ring map from this to $R$. 
