In: Advanced Math
Let α be a root of x^2 +1 ∈ Z3[x]. Show that Z3(α) = {a+bα : a, b ∈ Z3} is isomorphic to Z3[x]/(x^2 + 1).
Define, f : Z_3[x]/(x²+1) ----> Z_3() be defined by,
f(a+bx) = a+b
f((a+bx)(c+dx)) = f{ac+(ad+bc)x+bdx²} = f{(ac-bd) + (ad+bc)x} = ac-bd + (ad+bc)
Also, f(a+bx)•f(c+dx) = (a+b)(c+d) = (ac-bd) + (ad-bc) (Since, ² = -1)
So, f((a+bx)(c+dx)) = f(a+bx)•f(c+dx)
So, f is a homomorphism.
Injective :
We have, f(a+bx) = f(c+dx)
So, a+b = c+d
So, a = c & b = d
So, a+bx = c+dx
So, f is one-to-one.
Surjective :
Let, a+b belong to Z_3()
So, there exists (a+bx) in Z_3[x]/(x²+1) such that, f(a+bx) = a+b
So, f is surjective.
Hence, f is bijective.
So, f is a bijective homomorphism.
Hence, f is an isomorphism.
So, Z_3() Z_3[x]/(x²+1) = {a+bx : a, b in Z}