Proof by Will Dyer

I is maximal ↔ R/I is a field

←

Assume R/I is a field. Let J be an ideal such that I ⊂ J. Because of that, ∃x ∈ J such that x ∉ I. As a result, I + x ≠ I + 0. Since R/I is a field, ∃(I + y) ∈ R/I such that (I + x)(I + y) = (I + 1), or (I + xy) = 1. This equation means xy ~ 1, or (xy – 1) ∈ I. Since I ⊂ J, xy -1 ∈ J. Then, consider 1 = xy – (xy – 1). x ∈ J, y ∈ R, and xy-1 ∈ J. From the definition of an ideal, we also know that xy ∈ J. Since all elements of the equation ∈ J, 1 ∈ J. Since 1 ∈ J, J = R, meaning I is maximal.

→

Assume I is maximal. Suppose (I + a) ∈ R/I such that (I + a) ≠ (I + 0). Consider J = I + Ra = {i + ra | i ∈ I and r ∈ R}. ∀i ∈ I, i = i + 0a, meaning i ∈ J, so I ⊆ J. Since I is maximal, J = I or J = R. However, a can be written as a = 0 + 1a, so a ∈ J, but a ∉ I. Therefore, I ⊂ J, so J = R. This leads to I + Ra = R, and 1 ∈ R, so 1 = i + ra for some i ∈ I and r ∈ R. 1 – ra = i ∈ I, so (I + 1) = (I + ra), or (I + r)(I + a) = (I + 1). (I + a) is the inverse of (I + r).