Let $M$ be an $m$-by-$n$ matrix, here are three definitions$^5$ that we could use for rank:

  1. $rk(M) = \min k$ such that for matrices $P$, and $Q$ with $P$ of size $m$-by-$k$ and $Q$ of size $k$-by-$n$ we have $M = PQ$.

  2. $rrk_+(M)$ (3. $crk_+(M)$) is the maximum number of linearly independent rows (columns) of $M$. Where a set $S$ of vectors is linearly independent$^1$ if for all $u \in S$ there is no non-zero $k$ such that $ku$ can be written as a linear combination of vectors in $S - \{u\}$.

Now, if the elements of $M$ come from a field then $rk(M) = rrk_+(M) = crk_+(M)$. However, if we work with modules over a (semi-)rings then the story is different.

If we are working over a commutative (semi-)ring with no zero divisors$^2$ then how far apart can $rk$, $rrk_+$, and $crk_+$ be? Can this be bounded in terms of easily computable properties of $M$ and the (semi-)ring? I am primarily interested in finite $M$ and "well-behaved"$^3$ (semi-)rings$^4$.


  1. This definition of linear independence is not necessarily equivalent to the 'make the $0$-vector' definition for arbitrary (semi-)rings because (semi-)rings lack minus. I added the there is no non-zero $k$ part (instead of just the $k = 1$ case) after @Manos' answer.

  2. I ask for no zero divisors, because it is easy to come up with separations for general rings. For instance, consider $\begin{pmatrix} 2 & 2 \\ 3 & 3 \end{pmatrix}$ over $\mathbb{Z}_6$ and the obvious generalizations. I am not interested in examples like this because they are in some way an artifact of just the ring, not of the module.

  3. By "well-behaved", I mean you can assume all the nice properties you want since I want the issue to come from the module-ness aspect, not just the the (semi-)ring itself. In fact, I will be more than happy with an answer just for finite (semi-)rings (that meet my other restrictions).

  4. I know that semirings are much less studied than rings, so I will be happy with an answer for integral domains if the "semi-" part makes the question too awkward.

  5. I realize that there are many other definitions of rank and dimension for modules and (semi-)rings. I am interested in these three specific definitions of rank not to generalize fields but because they come up in application and I want to learn how to play with them. I appreciate pointers to other definitions or rank/dimension if you point out how they relate to the definitions I provide.

  • $\begingroup$ Migrating to MO by OP request. $\endgroup$ Feb 20, 2014 at 8:32

1 Answer 1


Consider the ring of integers $\mathbb{Z}$. Then this is a principal ideal domain. Consider the $1 \times 2$ matrix $M = \left[ 2 \, \, \, 3 \right]$. Then $rk(M)=1=rrk_+(M)$, if we make the convention that any non-zero element of a $\mathbb{Z}$-module is linearly independent, but $crk_+(M)=2$ because $2 (3)$ can not be expressed as a linear combination of $3 (2)$ (maybe you can come up with more general examples using this idea).

I believe that in general these notions can be quite far apart, but even more importantly for algebraic purposes, they are inconsistent. Also, one can not go far with defining linear dependence in terms of expressing an element as a linear combination of others. A more general notion is that of algebraic dependence, where one requires the existence of not all zero coefficients such that a linear combination of the elements under examination yields zero.

Any any case, if you have a (finite) module $M$ over an integral domain $R$, then the rank of $M$ is defined as the dimension of the $K$-vector space $M \otimes_R K$, where $K$ is the field of fractions of the integral domain $R$. This turns out to be a very useful notion.

  • $\begingroup$ Thanks for the example in paragraph 1, it still rides in on just the ring, so I need to think about my question more. Paragraph 2: I specifically don't want to use that definition of algebraic dependence (see note 1) because in semi-rings we don't have minus and so it isn't clear that it is more general but just different (same with rings with zero-divisors). Paragraph 3: I am not looking for other ranks, but specifically the ones I defined in the question (this comes from an application in CS); is there a way in which the definition of rank you give here relates to the two I list? $\endgroup$ Feb 12, 2014 at 22:08
  • $\begingroup$ Your paragraph 1 example is very much like the one I tried to rule out by excluding zero-divisors, and can be generalized to give arbitrary separation by, for instance, taking the first $n$-primes with product $N = p_1\cdot...\cdot p_n$ and then taking a row matrix $M = [a_1 , ... , a_n]$ where $a_j = P/p_j$. $\endgroup$ Feb 12, 2014 at 22:18
  • $\begingroup$ I modified the definition of linear independence I use to better approximate the one you give in paragraph 2 but that will still work for what I need on semi-rings. Thanks for your counter-example in helping clarify my answer, I will give a +1 but won't accept in hopes of other answers. Sorry for the "chase the definition" edits, I am trying to formulate this precisely in my mind since it comes from specific applications. $\endgroup$ Feb 12, 2014 at 22:30
  • $\begingroup$ Regarding the last question of your first comment: As far as i know, the only way to make things consistent in terms of ranks is by passing to some related vector space. So if you tensor your definitions with the field of fractions (equivalently you localize at zero), then we find ourselves in the linear algebra case and your new definitions are equivalent to the one i gave. I understand this may sound like cheating, depending what you have in mind, but i really think that's the only way to do it. Finally regarding your comment "it still rides in on just the ring", the ring itself is a module $\endgroup$
    – Manos
    Feb 12, 2014 at 22:30
  • $\begingroup$ Yes, the ring is a module, but I was hoping for examples that really use the module-ness and not just the ring-ness. I know, I'm impossible to please! Sorry. $\endgroup$ Feb 12, 2014 at 22:32

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