Re: R-homomorphisms


rod2 wrote:
Just as a linear transformation can be defined by specifying its action on the basis vectors, I am trying to show that an R-homomorphism can be defined by assigning arbitrary values on the elements of a basis (assuming it exists) and extending by linearity.

I'm not really sure what I am supposed to show. But this is what I came up with: Let B = {v_i:i in I} be a basis for module M. Specify T(v_i) for all v_i in B. Then extend the domain of T to M using linearity:

T(v) = T(a1*v1+...+an*vn) = a1*T(v1)+...+an*T(vn)

I then show T is indeed a homomorphism (basically linear).

Then I suppose T(v_i) = Q(v_i)

Let u = a1*v1 + ... + an*vn

T(u) = a1*T(v1)+...+an*T(vn)
Q(u) = a1*Q(v1)+...+an*Q(vn)

Thus T = Q, so it is unique. Is that correct/all I need to do?

One thing that bugs me is that I wrote an element as a finite linear combination, but I don't think this is right for an infinite basis. E.g., functions are written as infinite linear combinations of the trigonometric basis vectors in Fourier series.

Yes, it is right. For infinite sums to make sense you have to define
some notion of convergence. If a set is a basis in the algebraic sense,
then every element is a finite linear combination of some members of
the set.

Your argument is fine.

.

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