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Prove it is a sub-ring

cristianoceli

New member
Jun 16, 2016
20
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
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,467
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?
 

cristianoceli

New member
Jun 16, 2016
20
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.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,467
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.
 

cristianoceli

New member
Jun 16, 2016
20
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 $R$?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,467
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?
 

cristianoceli

New member
Jun 16, 2016
20
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:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,467
$1$ ,$-1$ 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 $R$?
 

cristianoceli

New member
Jun 16, 2016
20
$p$ does not divide $m$
 

cristianoceli

New member
Jun 16, 2016
20
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:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,467
$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 $R^\times$.
 

cristianoceli

New member
Jun 16, 2016
20
I understand.

Thank you