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Prove that a set is a monoid, but not a ring.

  1. Apr 24, 2005 #1
    End n(k) is the set of all polynomial mappings: k^n->k^n. i have to prove that end n(k) is a monoid.

    k is a field of q elements and n is the number of variables.

    the composition of two mappings F G is: F o G = F o G(v) = [F1(G1(v),..Gn(v)), ... Fn(G1(v),...Gn(v))]

    i must prove that the composition is associative and that there is a unit element.

    id=X=(x1,...,xn) is obviously the unit.

    but how do i prove associativity? i was thinking of using commutativity somehow:(x^n)^m = x^(n*m) = x^(m*n) = (x^m)^n

    last, how do i show that End n(k) is not a ring? is it distributivity it fails? for all x,y,z E A ->
    (x+y)z=xz + yz og z(x+y)=zx + zy
     
  2. jcsd
  3. Apr 24, 2005 #2

    Hurkyl

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    You should already know that!


    I presume you mean with composition as the multiplicative operation?

    Well, if it's not a ring, and all of the other properties are satisfied, then it would have to be!
     
  4. Apr 24, 2005 #3
    i guess it is obvious that composition is associative, but i'm not sure how to prove it.

    and yes, composition is the multiplicative operation.
     
  5. Apr 24, 2005 #4

    Hurkyl

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    Associativity is a basic result about the composition of functions -- it shouldn't be something you have to prove.

    But if you really want to do it, simply use the definition of composition.
     
  6. Apr 24, 2005 #5
    hm, ok. so i guess then i have to show that if F and G is in End n(k), then so is F o G in End n(k). however, this seems pretty obvious to me. how do prove that? is there a elegant proof of this?
     
  7. Apr 24, 2005 #6

    Hurkyl

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    You always had to prove the composition of two things in End n(k) was again in End n(k) -- I had assumed you had already done that, or I would have said something.
     
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