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Abstract Algebra: Non-trivial Rings Containing Only Zero-Divisors

  1. May 29, 2013 #1

    n+1

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    1. The problem statement, all variables and given/known data
    Is there a finite non-trivial ring such that for some a, b in R, ac = bc for all c in R?

    Does there exist finite non-trivial rings all of whose elements are zero-divisors or zero?

    2. The attempt at a solution
    Let a, b ≠ 0 in R such that ac=bc for all c in R. Subtracting the right from the left side, we see that (a-b)c=0, which since c is arbitrary, gives us that every element of R is a zero-divisor or zero.

    At this point, I opted to leave the initial problem to explore the second question: does there exist non-trivial rings all of whose elements are zero-divisors or zero? Seeing how I'm not so much trying to prove anything yet as I am just trying to gain some insight, my approach circles around identifying key information that's at play more than formally proving anything (I hope this approach doesn't make it too unreadable). I do this by assuming a ring of all zero-divisors and then seeing where I might be forced to make the ring trivial.

    Let a ≠ 0 in R, and let a0[\SUB] be a zero-divisor of a (we're guaranteed at least one). Since closure doesn't really play a factor here, there are only two properties of rings that involve multiplication (and thus zero-divisors): associativity (over multiplication) and distributivity. Exploring associativity reveals that
    0 = (aa0)b = a(a0b),
    and so we have that S={a0r | r in R} is a set of zero-divisors of a and zero. However, since a0r1 = a0r2 only implies that a0 and r1-r2 are zero-divisors, the map r to a0r is NOT necessarily one-to-one. In fact, we're guaranteed it's not one-to-one since both r=0 and r=a are sent to 0. This gives us hope that perhaps ar ≠ 0 for all r in R. Of course, it's also not sufficient proof that ar ≠ 0 for all r, for there could be more zero-divisors of a not in S (indeed we have done nothing to exclude trivial rings thus far, and so it would be absurd to claim that S necessarily contained all the zero-divisors of a).

    Exploring distributivity we see that
    0 = aa0 = naa0 = a(na0)
    for some n in Z (the integers), and so all members of R of form na0 are also zero-divisors of a (so clearly a non-trivial ring containing all zero-divisors could NOT be cyclic). Putting this information together with that of the last paragraph, we have the following set of zero-divisors of a: {na0r | n in the integers, r in R}. Say |S| = m, we can stretch our set of zero-divisors even further by giving the elements of S an order: s1, s2, ..., sm; S* = {n1s1 + ... + nmsm | n \in \Z}. In other words, S* is just the subgroup of R+ (I'm not sure if this is correct notation; I mean the 'group part' of the ring, R) generated by all the elements of S (all the different right multiples of our initial a0). ...


    Closing Remarks
    Anyway, this is where I am now. I'm looking for examples of such rings as described in the question or some proof of why such rings are impossible (the proof does not necessarily have to be connected to my work thus far).

    Thanks!
     
  2. jcsd
  3. May 29, 2013 #2

    Dick

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    Concerning your second question, I think you might want to think more about finding an example. Think about rings made of matrices. Matrices can be nilpotent. Then apply that to the first question.
     
  4. Jun 1, 2013 #3

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    Thanks for the response, Dick. So concerning nilpotent matrices, obviously any ring consisting of only nilpotent matrices would satisfy (2). Unfortunately the set of all nxn nilpotent matrices does not form a ring since it's not closed under multiplication (consider [[1,-1],[1,-1]]*[[1,-1/2],[2,-1]] to see this). Of course, the previous pseudo-ring wasn't closed under multiplication because it wasn't commutative; if we take the center of this pseudo-ring, then, we do have a ring. However, that this ring contains elements other than 0, I'm uncertain and even doubtful.

    As for any of this possibly applying to the first question, I'm skeptical. The additional constraint on a ring that would satisfy (1) is that there must be a non-zero element that when multiplied by ANY other element of the ring yields 0. While nilpotency is a nice way to guarantee zero-divisors, I don't see how it would help in producing an element as described in the previous sentence without getting seriously funky.
     
  5. Jun 1, 2013 #4

    Dick

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    Who said you had to take ALL nilpotent matrices? Take the ring generated by one, say [[0,0],[0,1]].
     
  6. Jun 1, 2013 #5

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    Hmmm. I'm not quite sure how the ring generated by [[0,0],[0,1]] would satisfy anything, especially since it has an identity. Also, wouldn't <[[0,0],[0,1]]> just be isomorphic to Z? It like the matrix part of all this is no longer important...

    Which is fine! Because all of this talk about nilpotency and single term generated rings got me going in the right direction. The example I've come up with is the subring of Z/2nZ (for n \in Z) generated by 2. This ring satisfies both parts of my question (it's straightforward to check; consider 2n-1) and is also relatively transparent/easy to work with.

    If this wasn't the example you were thinking of, I would love to hear yours. Once again, thanks for the help, Dick.
     
  7. Jun 1, 2013 #6
    If you have a commutative ring with at least one nilpotent element you can just take the ring generated by that element.
     
  8. Jun 1, 2013 #7

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    Yep. Because if an = 0, an-1 multiplied by any element in <a> can be rewritten as an*whocares=0. Cool cool.
     
  9. Jun 2, 2013 #8

    Dick

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    Ok, so also cool on the first part, right? Your ring can't contain an identity. If it does you can't do much with either part.
     
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