Ring theory proof?

Question

I'm completely stumped on this one:

Let R be a commutative ring with unity and let I be a proper ideal with the property that every element of R that is not in I is a unit of R. Prove that I is the unique maximal ideal of R.

Answer 1

To show I is maximal let J be an ideal properly containing I. Say x is in J but not in I. Then x has an inverse y and 1 = xy is in J. Therefore J = R and I is maximal.

By a similar argument I cannot contain a unit. Therefore I consists exactly of the non-units and so is unique.

Answer 2

First we show that I consists of exactly the non-units of R:
1 is not an element of I. (If it were, then for any a in R, a1 = a would have to be in I, so I wouldn't be a proper ideal.)
No unit is an element of I. (If u is a unit in I, then u^-1u = 1 is in I, but we already showed 1 isn't in I.)
Since everything outside of I is a unit, and nothing inside of I is a unit, I consists of exactly the non-units of R.

Next we show that R is a local ring (that is, it has a unique maximal ideal):
0 is an element of I, because I is a subgroup under addition, and we already showed that a1 isn't an element of I, so 0 is not equal to 1.
Since I consists of all the non-units, and it's closed under addition, the sum of any two non-units is in I and therefore is a non-unit.
Those two statements are sufficient to show that R is a local ring.

In a local ring R, the Jacobson radical J(R) is the unique maximal ideal, and it consists of exactly the non-units. But that's what I is, so I = J(R), so I is the unique maximal ideal of R.