# Algebriac Structures

1. Jan 25, 2007

### happyg1

1. The problem statement, all variables and given/known data

HI, I'm working on this:
If n>m, prove that there is a homomorphism of F^(n) onto F^(m) with a kernel W which is isomorphic to F^(n-m).

2. Relevant equations

Def: If U and V are vector spaces over F (a field) then the mapping T of U onto V is said to be a homomorphism if:
a) (u1+u2)T=u1T+u2T
b) (a u1)T=a(u1)T

If T in addition is one-to-one, we call it and isomorphism. The Kernel ot T is defined as {u in U|uT=0} where o is the identity element of the addition in V.

3. The attempt at a solution
These are my thoughts:
It seems trivial to me that there is a homomorphism from F^(n) onto F^(m) since n>m. I just don't know how to formalize that argument. I can't get a picture in my mind to write it down.
The part about the kernel of the homomorphism isomorphic to F^(n-m) also seems to be intuitively simple..the kernel will have n-m elements in it, so there's got to be an isomorphism between the kernel and F^(n-m).

thanks
cc

2. Jan 25, 2007

### StatusX

what is F^n explicitly? Write down a general element in F^n and it should be obvious where to send this in F^m to get a homomorphism (usually called a linear map in this case) with the desired properties. Think about the case F=R, the real numbers.

3. Jan 25, 2007

### Kreizhn

Without working out the details explicitly, I would imagine the multivariate projection map from $$F^n \rightarrow F^m$$ should satisfy the homomorphism conditions.

Furthermore, have you considered using the first isomorphism theorem for rings to help formally show that $$F^{n-m}$$ is isomorphic to the kernel?

It shouldn't be too hard to show that the projection map $$\pi (x)$$ is surjective, and so that $$Image(\pi(x))=F^m$$ is a subring of $$F^n$$. Then $$F^m$$ is isomorphic to $$F^n \setminus_{Ker(\pi)}$$. Consider the natural map between these two sets and then see if you can get anything from there.

This might be a bit over the top, but was the first thing that came to my mind.

4. Jan 27, 2007

### happyg1

Thanks for the input. It's been a year since I studied rings and I'm all rusty. I'm pretty sure I understand this one and I think I got it.

My next question is this:Prove that there exists an isomorphism from $$F^n$$ into $$Hom(Hom(F^n , F),F).$$
Again, I'm all rusty on this stuff, so any input will be helpful.
My confusion here lies in that I just finished proving (by contradiction) that $$F^1$$ is not isomorphic to $$F^n$$ for n>1.
So I'm not sure what $$F$$ is or what $$Hom(F^n,F)$$ looks like. It seems to me that $$F^n$$ is finite and $$F$$ is infinite and then my mind just starts going in circles. Please help me understand this.
thanks
CC

Last edited: Jan 27, 2007