# Prove it is a sub-ring

#### cristianoceli

##### New member
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
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?

• topsquark and cristianoceli

#### cristianoceli

##### New member
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
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

#### cristianoceli

##### New member
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
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
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 $\neq p$???

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
$1$ ,$-1$ 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 $R$?

• topsquark and Country Boy

#### cristianoceli

##### New member
$p$ does not divide $m$

#### cristianoceli

##### New member
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 $m$ ???

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
$R^{\times} = n \in \mathbb{Q}$ 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 $R^\times$.

• topsquark and cristianoceli

I understand.

Thank you