Integral Domain-PID: Exploring the Proof Step-by-Step

[evinda]

Hello! (Wave)

The following holds for the ring $ \mathbb{Z}_p, p \in \mathbb{P}$:

The ring $ \mathbb{Z}_p $ is a principal ideal domain, especially an integral domain.

I try to understand the following proof:

Proof:

We will show that $ \mathbb{Z}_p $ is an integral domain.
The fact that $ \mathbb{Z}_p $ is a principal ideal domain can be concluded from the proposition "$\mathbb{Z}_p $ contains only the ideals $0$ and $p^n \mathbb{Z}_p $" for $n \in \mathbb{N}_0 $. It holds $ \bigcap_{n \in \mathbb{N}_0} p^n \mathbb{Z}_p=0$ and $ \mathbb{Z}_p / p^n \mathbb{Z}_p \cong \mathbb{Z} \ p^n \mathbb{Z}$.
Especially $ p \mathbb{Z}_p$ is the only maximal Ideal. "

Let $ x=(\overline{x_k}), y=(\overline{y_k}) \in \mathbb{Z}_p \setminus \{0\} $ and $ z=(\overline{z_k})=xy=(\overline{x_ky_k})$.

Since $ x,y \neq 0$ there are $ m,n \in \mathbb{N}_0 $ with $x_n \not\equiv 0 \mod{p^{n+1}}$ and $ y_m \not\equiv 0 \mod{p^{m+1}}$.
We set $ l:=n+m+1$. From $ x_l \equiv x_n \mod{p^{n+1}}$ and $ y_l \equiv y_m \mod{p^{m+1}}$ we deduce the existence of the partitions $ x_l=u \cdot p^{n'}$ and $ y_l=v \cdot p^{m'}$ with $gcd(u,p)=gcd(v,p)=1$, $ n' \leq n $ and $ m' \leq m$. Then since $ n'+m'<l $ we have that

$ z_l=uvp^{n'+m'} \not\equiv 0 \mod{p^{l+1}}$, and so $ z \neq 0$.Could you explain me step by step the above proof? (Thinking)

 

[caffeinemachine]

Hello! (Wave)

The following holds for the ring $ \mathbb{Z}_p, p \in \mathbb{P}$:

The ring $ \mathbb{Z}_p $ is a principal ideal domain, especially an integral domain.

I try to understand the following proof:

Proof:

We will show that $ \mathbb{Z}_p $ is an integral domain.
The fact that $ \mathbb{Z}_p $ is a principal ideal domain can be concluded from the proposition "$\mathbb{Z}_p $ contains only the ideals $0$ and $p^n \mathbb{Z}_p $" for $n \in \mathbb{N}_0 $. It holds $ \bigcap_{n \in \mathbb{N}_0} p^n \mathbb{Z}_p=0$ and $ \mathbb{Z}_p / p^n \mathbb{Z}_p \cong \mathbb{Z} \ p^n \mathbb{Z}$.
Especially $ p \mathbb{Z}_p$ is the only maximal Ideal. "

Let $ x=(\overline{x_k}), y=(\overline{y_k}) \in \mathbb{Z}_p \setminus \{0\} $ and $ z=(\overline{z_k})=xy=(\overline{x_ky_k})$.

Since $ x,y \neq 0$ there are $ m,n \in \mathbb{N}_0 $ with $x_n \not\equiv 0 \mod{p^{n+1}}$ and $ y_m \not\equiv 0 \mod{p^{m+1}}$.
We set $ l:=n+m+1$. From $ x_l \equiv x_n \mod{p^{n+1}}$ and $ y_l \equiv y_m \mod{p^{m+1}}$ we deduce the existence of the partitions $ x_l=u \cdot p^{n'}$ and $ y_l=v \cdot p^{m'}$ with $gcd(u,p)=gcd(v,p)=1$, $ n' \leq n $ and $ m' \leq m$. Then since $ n'+m'<l $ we have that

$ z_l=uvp^{n'+m'} \not\equiv 0 \mod{p^{l+1}}$, and so $ z \neq 0$.Could you explain me step by step the above proof? (Thinking)

Is $\mathbb Z_p$ the same as $\mathbb Z/p\mathbb Z$? In that case $\mathbb Z_p$ is a field and thus a PID. I ask this because I don't see why you'd consider $p^n\mathbb Z_p$ in your proof.

 

[evinda]

Is $\mathbb Z_p$ the same as $\mathbb Z/p\mathbb Z$? In that case $\mathbb Z_p$ is a field and thus a PID. I ask this because I don't see why you'd consider $p^n\mathbb Z_p$ in your proof.

$\mathbb{Z}_p$ is the set of integer p-adics.

We only use this sentence: "$\mathbb{Z}_p $ contains only the ideals $0$ and $p^n \mathbb{Z}_p $" for $n \in \mathbb{N}_0 $ because it is an other proposition of the theorem I am looking at.
So could we also prove it in an other way that $\mathbb{Z}_p$ is a PID? (Thinking)

 

[caffeinemachine]

$\mathbb{Z}_p$ is the set of integer p-adics.

We only use this sentence: "$\mathbb{Z}_p $ contains only the ideals $0$ and $p^n \mathbb{Z}_p $" for $n \in \mathbb{N}_0 $ because it is an other proposition of the theorem I am looking at.
So could we also prove it in an other way that $\mathbb{Z}_p$ is a PID? (Thinking)

Oh. Thanks for clarifying. Wish I could help. Don't know anything about $p$-adics.

 

[CellCoree]

Sure, I'd be happy to explain this proof step by step.

1. The first step in the proof is to show that $ \mathbb{Z}_p $ is an integral domain. This means that the ring has no zero divisors, i.e. if $ ab = 0 $, then either $ a = 0 $ or $ b = 0 $. In order to show this, we will first prove that $ \mathbb{Z}_p $ is a principal ideal domain (PID). This means that every ideal in the ring can be generated by a single element.

2. To prove that $ \mathbb{Z}_p $ is a PID, we will use the fact that the ring only contains the ideals $ 0 $ and $ p^n \mathbb{Z}_p $ for $ n \in \mathbb{N}_0 $. This means that any other ideal in the ring must be a multiple of $ p^n \mathbb{Z}_p $ for some $ n $. In other words, any element in the ideal can be written as $ ap^n $ for some $ a \in \mathbb{Z}_p $. This is the definition of a principal ideal.

3. Now, we need to show that the intersection of all the ideals $ p^n \mathbb{Z}_p $ is equal to $ 0 $. In other words, we need to show that the only element that is in all of these ideals is $ 0 $. This is equivalent to showing that the only element that is a multiple of $ p^n $ for all $ n $ is $ 0 $. This can be proven by using the fact that $ \mathbb{Z}_p / p^n \mathbb{Z}_p \cong \mathbb{Z} / p^n \mathbb{Z} $ and the fact that $ \mathbb{Z} / p^n \mathbb{Z} $ only contains the elements $ 0 $ and $ p^n $.

4. Next, we need to show that $ p \mathbb{Z}_p $ is the only maximal ideal in $ \mathbb{Z}_p $. This means that there are no other ideals in the ring that properly contain $ p \mathbb{Z}_p $. This can be proven by using the previous result, since $ p \mathbb{Z}_p $ is a multiple of