Help Prove Gaussian Integers with Ideal I is maximal!

Let R={a+bi: a, b elements of Reals} and let I be an Ideal of R generated by <1+i>. Prove that I is maximal and that R/I is isomorphic to Z2.

I feel I need to use the isomorphism theorems and the fact that if R/I is a field then I is maximal but I don't know where to start, any help would be greatly appreciated.

Define a map f: R --> Z2 by the equation f(a + bi) = a - b (mod 2). f is a homomorphism since

f((a + bi) + (c + di)) = f((a + c) + (b + d)i) = (a + c) - (b + d) = (a - b) + (c - d) = f(a + bi) + f(c + di)

and

f((a + bi)(c + di))

= f((ac - bd) + (ad + bc)i)

= (ac - bd) - (ad + bc)

= (ac - ad) - (bd + bc)

= a(c - d) - b(c + d)

= a(c - d) - b(c - d) (since 2d = 0 (mod 2))

= (a - b)(c - d) (mod 2)

= f(a + bi) f(c + di).

f is surjective, since for each a in Z2, f(a) = a. The kernel of f is Ker(f) = <1 + i>. For if a + bi is in Ker(f), then a - b = 0 (mod 2). Then a = b + 2k for some integer k. So

a + bi = (b + 2k) + bi = (b + bi) + 2k = b(1 + i) + (1 - i)(1 + i)k = [b + (1 - i)k](1 + i)

is in <1 + i>. Conversely, since f(1 + i) = 1 - 1 = 0 and f is a homomorphism, every element of <1 + i> equals 0.

We conclude by the first isomorphism theorem that R/I ≈ Z2. Then R/I is a field. Thus I is a maximal ideal.