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Intro abstract algebra along with basic set theory

  1. Sep 20, 2013 #1
    1. The problem statement, all variables and given/known data

    An interesting example of a ring:

    Begin with a nonempty set X and form the power set of X, P(X), which is the set of all subsets of X. On P(X), define addition + and multiplication × as follows:

    For A, B in P(X):

    A × B = A ∩ B

    A + B = (A\B) ∪ (B\A), where as usual A\B = {a in A | a not in B}

    I have to answer the following. I have bolded the questions i am having trouble on. the rest i have answered and just want to make sure i'm right.

    (a)Show there exists an element 0 in P(X) such that A + 0 = A for all A in P(X).

    (b)Show that for all A in P(X), there exists an element -A in P(X) so that A + (-A) = 0 (0 is from part (a)).

    (c)Show A×(B + C) = A×B + A×C for all A, B, C in P(X).

    (d) In fact this is a commutative ring with (multiplicative) identity. What element in P(X), not the zero 0 element, when multiplied by any A in P(X), gives you A itself?

    (e)What are the zero divisors in this ring, if any? What are the units in this ring, if any?

    2. Relevant equations

    --

    3. The attempt at a solution

    (a) The 0 element is the empty set. The empty set is a subset of all sets, therefore given any set X, the empty set is an element of the power set P(X).

    A + ∅ = (A\∅) ∪ (∅\A) = A ∪ ∅ = A.

    (b) This one troubles me.

    Suppose there exists such an additive inverse, let's find it. For an A in P(X), let the inverse of A be B. It exists if

    (A\B) ∪ (B\A) = ∅ implies we must have (A\B) = (B\A) = ∅.
    (A\B) = ∅ implies A is a subset of B.
    (B\A) = ∅ implies B is a subset of A.

    These two previous statements imply A = B. So this guarantees that for every A, there exists an additive inverse of A, which actually happens to be itself, which is always in P(X) by definition. Can a ring element be its own additive inverse?

    (c)Not sure where to start.

    (d) The set X itself, which generates the power set P(X), is the mult. identity. The question doesn't ask that we prove this. It just asks what is the the identity.

    (e) every element in P(X) except X and ∅ is a zero divisor. This is because there will always exist an element so that their intersection is disjoint. This element is simply A\X. X is not a zero divisor because X\X = ∅ which cannot be a zero divisor. Again an explanation isn't needed here but i thought i would throw it in on this forum. There is only one unit, which is X itself.
     
  2. jcsd
  3. Sep 20, 2013 #2

    haruspex

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    Yes
    You could start by translating both sides of the equation into usual set operations, according to the definitions of + and x.
     
  4. Sep 20, 2013 #3
    Just for part b) where your write (A\B) = (B\A) = ∅ it should read (A\B) or (B\A) = ∅.
    This changes the wording of what's to follow but still the same idea.

    As for c) equality of sets is shown through showing that the LHS is a subset of the RHS and then showing that the RHS is a subset of the LHS.

    To start you might say something like:
    Let x [itex]\in[/itex] A×(B + C)
    Then by definition of multiplication x [itex]\in[/itex] A∩(B+C)
    So x[itex]\in[/itex] A and x [itex]\in[/itex] (B+C)
    .....
    .....
    .....
    Does this help?
     
  5. Sep 20, 2013 #4
    I think for the union of two sets to be empty, we must have that both sets are the empty set. E.g., {1,2} U ∅ = {1,2} ≠ ∅, which would be required for additive inverse.
     
  6. Sep 20, 2013 #5
    Oh yes, I was thinking about sets in general and forgot things get a little different when you mention the empty set sometimes.

    Part c remains the same,
     
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