How to prove that a maximal ideal of a ring is unique?

840 Views Asked by At

So I have a commutative ring with unity, and I also have an ideal I of the ring such that every element not in the ideal is a unit of R. I have already proven that I is maximal by showing that R/I is a field, but I'm really not sure how to sure that it is unique. Let's say I have any maximal ideal of R (possibly I itself). Now I want to show that it is equal to I. How would I do that? I'm quite stuck, so any hints would be appreciated.

Thanks in advance.

2

There are 2 best solutions below

2
On BEST ANSWER

Let $\mathfrak{m} \subseteq R$ be a maximal ideal. Can $\mathfrak{m}$ have any units?

Next step: Is $\mathfrak{m}$ always contained in some ideal that you already know? How do we now use maximality?

Rings with a unique maximal ideal are called local, by the way.

0
On

I'm really not sure how to sure that it is unique.

Any other ideal which isn't I would contain a unit by your hypothesis and would therefore equal R.