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

1. Apr 24, 2005

grimster

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. Apr 24, 2005

Hurkyl

Staff Emeritus

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!

3. Apr 24, 2005

grimster

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. Apr 24, 2005

Hurkyl

Staff Emeritus
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. Apr 24, 2005

grimster

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. Apr 24, 2005

Hurkyl

Staff Emeritus
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.