Question

Let R an integral domain and P⊂R a prime ideal.

(a.) Show that M={ p/d ∣p∈P,d∈D}
is a maximal ideal in Rₚ
​ (b.) Show that M is the only maximal ideal in Rₚ
(A ring with only 1 maximal ideal is called a local ring.)

Answer 1

M is the set p∈P, d∈D in the localization of R at a prime ideal P, denoted Rₚ. It can be shown to be a maximal ideal by demonstrating it's ideal and no larger ideal exists within Rₚ. M is also the unique maximal ideal, which makes Rₚ a local ring.

Step-by-step explanation:

To show that M is a maximal ideal in Rₚ, where R is an integral domain and P is a prime ideal in R. Let's first define M as M= p/d , where D is the set of non-zero elements in R that are not in P, and Rₚ is the localization of R at P, meaning it consists of fractions with numerators in R and denominators not in P.

For (a), we must show two things. First, that M is an ideal, and second, that it is maximal. Since P is a prime ideal, and due to the properties of localization, the set M will indeed be an ideal. To prove it's maximal, we assume there exists an ideal N such that M ⊆ N ⊆ Rₚ, and show that either N = M or N = Rₚ. We achieve this through contradiction, assuming there's an element in N not in M, and using properties of prime ideals and the localization to arrive at a contradiction.

For (b), to demonstrate that M is the only maximal ideal in Rₚ, we observe that any maximal ideal M' in Rₚ would necessarily contain fractions whose numerators are in P. If M' contained a fraction with a numerator not in P, it would be invertible, making M' the whole ring. Therefore, due to the definition of localization and the properties of prime ideals, M will be the unique maximal ideal thus making Rₚ a local ring.