Prove R is a Sub-ring of $\mathbb{Q}$ - Prime $p \in \mathbb{Z}$

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Discussion Overview

The discussion revolves around proving that the set \( R = \{m/n \in \mathbb{Q}: p \text{ does not divide } n \} \) is a sub-ring of the rational numbers \( \mathbb{Q} \), where \( p \) is a prime number. Participants explore the necessary conditions for \( R \) to qualify as a sub-ring and investigate the group of units \( R^{\times} \) within this context.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants reference the definition of a subring, noting that it must be closed under multiplication and subtraction, and contain the multiplicative identity.
  • Participants question whether the product of two elements in \( R \) remains in \( R \), specifically asking if \( \frac{m_1}{n_1} \times \frac{m_2}{n_2} \) is in \( R \) when \( p \) does not divide \( n_1 \) or \( n_2 \).
  • There is a discussion about identifying the units of \( R \), with some suggesting that \( 1 \), \( -1 \), and primes not equal to \( p \) are units.
  • Another participant proposes that the multiplicative inverse \( \frac{n}{m} \) must also be in \( R \) for \( \frac{m}{n} \) to be a unit, leading to further exploration of conditions under which this holds.
  • One participant concludes that the set of units \( R^{\times} \) includes elements where \( p \) does not divide both the numerator and denominator.

Areas of Agreement / Disagreement

Participants express uncertainty about the closure properties of \( R \) under multiplication and the complete characterization of its units. There is no consensus on the final characterization of \( R \) as a subring or the full set of units.

Contextual Notes

Participants have not fully resolved the implications of the closure properties or the conditions required for elements to be units in \( R \). There are also assumptions regarding the definitions of invertibility and the nature of the elements in \( R \) that remain unexamined.

fabiancillo
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Hello, I don't know to solve this exercise:

Let $p \in \mathbb{Z}$ be a prime number. Consider $R = \{m/n \in \mathbb{Q}: p$ does not divide $n \}$

How can I prove that $R $ is a sub-ring of $\mathbb{Q}$? (only the obvious parts) and find the group of units of $R, R^{\times}$I have no idea.How can I solve ?

Thanks
 
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Hi cristianoceli,

One of the equivalent definitions for a subring is:
"A subset S of R is a subring if and only if it is closed under multiplication and subtraction, and contains the multiplicative identity of R."


How far can we get with those conditions?
 
Klaas van Aarsen said:
Hi cristianoceli,

One of the equivalent definitions for a subring is:
"A subset S of R is a subring if and only if it is closed under multiplication and subtraction, and contains the multiplicative identity of R."


How far can we get with those conditions?

Can you explain? Sorry

In the exercise I do not know if this happens

contains the multiplicative identity of R.
 
Suppose we have the elements $\frac{m_1}{n_1}$ and $\frac{m_2}{n_2}$. It means that $p$ does not divide $n_1$, and $p$ does not divide $n_2$ either.
Is $\frac{m_1}{n_1}\times\frac{m_2}{n_2}$ always an element of $R$?
If it is, then $R$ is closed under multiplication.
 
Klaas van Aarsen said:
Suppose we have the elements $\frac{m_1}{n_1}$ and $\frac{m_2}{n_2}$. It means that $p$ does not divide $n_1$, and $p$ does not divide $n_2$ either.
Is $\frac{m_1}{n_1}\times\frac{m_2}{n_2}$ always an element of $R$?
If it is, then $R$ is closed under multiplication.
I understand but .

What are the units of $R$?
 
cristianoceli said:
What are the units of $R$?
The units of a ring $R$ are the elements that are invertible with respect to multiplication.
Which elements are invertible?
 
Klaas van Aarsen said:
The units of a ring $R$ are the elements that are invertible with respect to multiplication.
Which elements are invertible?
$1$ ,$-1$ and prime numbers $\neq p$?
 
Last edited:
cristianoceli said:
$1$ ,$-1$ and prime numbers $\neq p$?
Those are indeed units, but there are more.

The multiplicative inverse of $\frac{m}{n}$, if it exists in $R$, is $\frac{n}{m}$.
What do we need for $\frac{n}{m}$ to be an element of $R$?
 
$p$ does not divide $m$
 
  • #10
Klaas van Aarsen said:
Those are indeed units, but there are more.

The multiplicative inverse of $\frac{m}{n}$, if it exists in $R$, is $\frac{n}{m}$.
What do we need for $\frac{n}{m}$ to be an element of $R$?
$R^{\times} = m \in \mathbb{Q} $ such that $p$ does not divide $m$ ?
 
Last edited:
  • #11
cristianoceli said:
$R^{\times} = n \in \mathbb{Q} $ such that $p$ does not divide $m$ ?
That should be: the set of units of $R$ is $\{m/n\in\mathbb Q : p\text{ does not divide }m \land p\text{ does not divide }n\}$.

This set happens to be a group with respect to multiplication named $R^\times$.
 
  • #12
I understand.

Thank you
 

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