Timeline for Separations between notions of rank for modules over commutative (semi-)rings with no zero divisors.
Current License: CC BY-SA 3.0
9 events
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| Oct 8, 2015 at 3:28 | comment | added | Fernando Martin | Why are $2$ and $3$ linearly independent? Unless I'm missing something, two elements $x,y$ in a commutative ring can't be linearly independent, since $x(y)=y(x)$. | |
| Feb 20, 2014 at 8:32 | history | migrated | from math.stackexchange.com (revisions) | ||
| Feb 12, 2014 at 22:35 | comment | added | Artem Kaznatcheev | I don't need the three definitions to be consistent, I am happy with them being different, I just want to see if I can bound the gap that separates them by something reasonable. Your examples in P1 shows that for my initial flawed definition of linear-independence, I can't bound it at all (better than 1 versus max of $n$, $m$) if I allow arbitrarily large coefficients, and if I count the number of bits to represent coefficients, I still can't hope for a small gap. | |
| Feb 12, 2014 at 22:32 | comment | added | Artem Kaznatcheev | 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. | |
| Feb 12, 2014 at 22:30 | comment | added | Manos | 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 | |
| Feb 12, 2014 at 22:30 | comment | added | Artem Kaznatcheev | 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. | |
| Feb 12, 2014 at 22:18 | comment | added | Artem Kaznatcheev | 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 $a_j = P/p_j$. | |
| Feb 12, 2014 at 22:08 | comment | added | Artem Kaznatcheev | 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? | |
| Feb 12, 2014 at 21:53 | history | answered | Manos | CC BY-SA 3.0 |