Quotient Rings and Homomorphic images in F(R)

"let g be the function from F(R) to RxR defined by g(f)=( f(0), f(1) ). Prove that g is a homomorphism from F(R) onto RxR, and describe its kernel"

I'm trying to find the kernel. so the kernel is kerf={x in F(R); g(x)= e_RxR} ? Wouldn't it just be (0,0) ? If this is right how do i prove this? If this is wrong then how do i find the kernel?

Re: Quotient Rings and Homomorphic images in F(R)

it would help if we know exactly what you meant by F(R) and R. if R is a ring, then the identity of (RxR,+) is indeed (0,0), so the kernel of g is:

ker(g) = {f in F(R): f(0) = f(1) = 0}.

my guess is that by F(R) you mean the set of all functions from R to R with:

(f+h)(x) = f(x) + h(x) and

(fh)(x) = f(x)h(x).

note that the additive identity of F(R) is the function 0(x) = 0, for all x in R, and the multiplicative identity of F(R) is the (constant) function 1(x) = 1, for all x in R.

(note that ker(g) is non-trivial, the function f(x) = x(x-1) is in it).