12
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ if K=kR ,who r(k)=T .T is maximal ideal . $\endgroup$
    – user10564
    Nov 4, 2010 at 12:43

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ I'm confused; why is $S$ generated by $a_0$? $\endgroup$ Nov 5, 2010 at 2:40
  • $\begingroup$ Every variable is in the ideal generated by $a_{0}$. For example, $a_{i}=a_{2i+1}a_{0}a_{2i+2}$. An arbitrary element of $S$ can be written as a sum of words in the variables defining $T$, and each word contains at least one variable (since we taking the free algebra without constant terms). $\endgroup$ Nov 5, 2010 at 16:08
  • $\begingroup$ Oh, I thought you mean S is generated as an algebra by $a_0$. $\endgroup$ Nov 6, 2010 at 9:50
  • $\begingroup$ Nope, just as an ideal. $\endgroup$ Nov 8, 2010 at 15:36
1
$\begingroup$

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 $R$.

$\endgroup$
1
  • $\begingroup$ The last $R$ should probably be $I$, right? $\endgroup$ Nov 4, 2010 at 19:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.