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Is there a (finite) ring with exactly 2 minimal left ideal and exactly 3 minimal right ideal? (rings are assumed to have identity)

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2 Answers 2

Let $M$ be the set of all $3\times 2$ matrices over a field $F_2$ with usual addition. Let $P$ be the $2\times 3$ matrix consisting of $1$'s. Define a multiplication: $A*B= APB$. Then we get a ring (this is so-called Munn algebra) with required properties (a minimal left ideal is generated by a column).

In such a way you can obtain a ring with $m$ minimal left ideals and $n$ minimal right ideals.

Addendum (Thanks to Manny Reyes) $M$ doesn't contain an identity element, but we can join an identity $e$ to $M$, i.e. consider the ring $M_1$ of pairs $(a,me)$ ($a\in M, m\in F_2$) with multiplication $(a,me)*(b,ne)=(a*b+na+mb,mne)$. Then minimal ideals of $M_1$ are the same of $M$.

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This is a very clever idea, but this ring doesn't seem to have an identity element. Such an element $X \in M$ would need to be a $3 \times 2$ matrix satisfying $PX = I_2$ and $XP = I_3$. The latter is impossible by rank considerations. Am I missing something? –  Manny Reyes Sep 26 '13 at 19:11
    
@Manny Reyes: Thank you very much, I forgot about the identity. I will correct the answer. –  Boris Novikov Sep 26 '13 at 19:43
    
+1 Nice. What can you tell about the form of maximal (left or right) and prime ideals of $M$ (or $M_1$)? –  user37834 Sep 26 '13 at 20:48
    
@Silvi: There is an analogy with usual matrix rings over a field: if we consider a minimal left ideal as the additive group, it is the direct sum of the additive groups of the field. –  Boris Novikov Sep 26 '13 at 21:05
    
So what are the maximal left ideals of $M$? –  user37834 Sep 28 '13 at 19:23

Consider the ring $R = \Bbb{Z}_2<a,b \; | \; a^3 = b^2 = ba = a^2b = 0>$ ($a,b$ are non-commuting indeterminate). Then every element of $R$ is a linear combination of $1,a,b,a^2,ab$. Therefore $|R| = 32$, also $R$ is a local ring and satisfies the asked conditions.

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