# Thread: [SOLVED] characterizing all ideals of a certain ring... :-\

1. ## [SOLVED] characterizing all ideals of a certain ring... :-\

Hello, and thank you very much for reading.

Hate these kinds of questions:
Let p be a prime number.
Let R= Z(p) be the ring defined as followed:
Z(p) = {x/y : gcd(y,p)=1} (notice that it's not the ring {0,1,...,p-1}!)
I need to characterize all the ideals in this ring, and all of it's quotient rings...

I already proved Z(p) is a ring (I needed to do so before this question).
I also noticed that an element x/y is invertible if and only if x is not in pZ (meaning, if and only if gcd(x,p)=1).
I know that if an Ideal cosist an invertible element then it is all of R, so I'm seeking for ideals that consist of elements x/y such that gcd(x,p)=1. However, I cannot see how to find how many ideals of this type there are, and moreover - how to show that there are no other types of ideals... :-\
I'll think of quotient rings after I find the ideals...

I'm too stuck on this one...

Thank you in advance.
Yours truely
Tomer.

2. Originally Posted by aurora
Hello, and thank you very much for reading.

Hate these kinds of questions:
Let p be a prime number.
Let R= Z(p) be the ring defined as followed:
Z(p) = {x/y : gcd(y,p)=1} (notice that it's not the ring {0,1,...,p-1}!)
I need to characterize all the ideals in this ring, and all of it's quotient rings...

I already proved Z(p) is a ring (I needed to do so before this question).
I also noticed that an element x/y is invertible if and only if x is not in pZ (meaning, if and only if gcd(x,p)=1).
I know that if an Ideal cosist an invertible element then it is all of R, so I'm seeking for ideals that consist of elements x/y such that gcd(x,p)=1. However, I cannot see how to find how many ideals of this type there are, and moreover - how to show that there are no other types of ideals... :-\
I'll think of quotient rings after I find the ideals...

I'm too stuck on this one...

Thank you in advance.
Yours truely
Tomer.
suppose $I \neq (0)$ is an ideal of $R.$ note that if $\frac{x}{y} \in I,$ for some $y$ coprime with $p,$ then for any $z \in \mathbb{Z}$ coprime with $p$ we have $\frac{x}{z}=\frac{y}{z} \cdot \frac{x}{y} \in I.$

let $n=\min \{x \in \mathbb{N}: \ \ \frac{x}{y} \in I, \ \text{for some} \ y \ \text{with} \ \gcd(p,y)=1 \}.$ (note that if $\frac{x}{y} \in I,$ then $\frac{-x}{y} \in I$ too. so the set we defined is not empty.)

so $\frac{n}{y} \in I,$ for some $y$ coprime with $p.$ thus by the above remark $\frac{n}{z} \in I$ for any $z$ coprime with $p.$ now let $\frac{x}{z} \in I.$ we have $x=sn+r,$ where

$0 \leq r < n.$ therefore: $\frac{r}{z}=\frac{x}{z}-\frac{sn}{z} \in I,$ because both $\frac{x}{z}$ and $\frac{n}{z}$ are in $I.$ but this will contradict the minimality of $n$ unless $r=0.$ so $\frac{x}{z}=\frac{sn}{z}.$

this proves that $I=\{\frac{sn}{z}: \ \ s,z \in \mathbb{Z}, \ \gcd(p,z)=1 \}=nR.$ obviously $I=R$ if and only if $\gcd(n,p)=1. \ \Box$

Note: so every ideal of $R$ is cyclic. this problem is a good introduction to "localization", which is a powerful tool in the theory of commutative rings.

3. Thank you!!
Your detailed proof, and the fact that you mentioned the ideal is cyclic, motivated me to attack the problem from this direction and I thin I mannaged to prove something pretty elegent -
That every ideal I that's not trivial (meaning, not {0} or Z(p) ) is of the form (p^k) for any natural k, where (p^k) = p^k * Z(p)!
So all the ideals are of the form (p^k)!

Thank you thank you thank you.

4. Originally Posted by aurora
Thank you!!
Your detailed proof, and the fact that you mentioned the ideal is cyclic, motivated me to attack the problem from this direction and I thin I mannaged to prove something pretty elegent -
That every ideal I that's not trivial (meaning, not {0} or Z(p) ) is of the form (p^k) for any natural k, where (p^k) = p^k * Z(p)!
So all the ideals are of the form (p^k)!

Thank you thank you thank you.
correct! so $\mathfrak{m}=pR$ is the unique maximal ideal of $R$ and $\{{\mathfrak{m}}^k: \ k \geq 0 \}$ is the set of all nonzero ideals of $R.$ this is not just an accident! the reason is that "a local PID is a discrete valuation ring."

(also a discrete valuation ring is a local PID which is not a field!) so if you replace $\mathbb{Z}$ in your problem with any principal ideal domain, you'll get the same result.

5. I'll remember that