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

  • Thread starter grimster
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  • #1
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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
 

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  • #2
Hurkyl
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i must prove that the composition is associative
You should already know that!


last, how do i show that End n(k) is not a ring?
I presume you mean with composition as the multiplicative operation?

is it distributivity it fails?
Well, if it's not a ring, and all of the other properties are satisfied, then it would have to be!
 
  • #3
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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.
 
  • #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.
 
  • #5
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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?
 
  • #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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