Properties of a localization of ℤ

[kremulum]

Hello forum,

I am new to group and ring theory, so exercices like this are making me problems:

Let R={ab∈ℚ∣b is odd}.

(1) Prove that R is isomorphic to ℤP, where ℤP is the localization at P, for a prime ideal P of R.

(ii) Find U(R). Prove that I:=R∖U(R) is a maximal ideal. Determine which field R/I is.

(iii) Find all the irreducibles of R and prove that they are primes.

(iv) Prove that R is an unique factorisation domain. Find all the ideals of R. Is R a principal ideal domain? And an Euclidean domain?

Any help would be appreciated!

 

[jvicens]

Hello,

Welcome to the forum! I can understand that group and ring theory can be challenging, especially when dealing with exercises like the one you mentioned. Let's break it down and see if we can work through it together.

For part (1), we need to prove that R is isomorphic to ℤP, where ℤP is the localization at P, for a prime ideal P of R. To do this, we can define a map from R to ℤP and show that it is a bijective homomorphism. Let's call this map φ.

First, we need to define the prime ideal P of R. Since R is a subset of ℚ, we can take P to be the set of all elements in R that can be written as a fraction a/b, where b is an odd integer and a is any integer. This is a prime ideal because if we multiply two elements in P, their denominators will still be odd, and if we add two elements in P, their denominators will still be odd.

Now, let's define our map φ: R → ℤP. We can do this by sending an element ab∈R to the element a/b in ℤP. This map is well-defined because if we have two different fractions a/b and c/d in R that map to the same element in ℤP, their difference (ad-bc)/bd is also in R and maps to 0 in ℤP. This means that ad-bc=0, which implies that ad=bc. Since b and d are odd, this can only happen if a=c.

Next, we need to show that φ is a homomorphism. We can do this by showing that it preserves addition and multiplication. Let's take two elements ab and cd in R and see what happens when we add them. φ(ab+cd) = φ((ad+bc)/bd) = (ad+bc)/(bd) = (a/b) + (c/d) = φ(ab) + φ(cd). Similarly, we can show that φ(abcd) = φ(ab)φ(cd).

Finally, we need to show that φ is a bijection. This means that it is both injective and surjective. To show injectivity, let's take two elements ab and cd in R and assume that φ(ab) = φ(cd). This means that (a/b) = (c/d) in