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

[fabiancillo]

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

 

[I like Serena]

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?

 

[fabiancillo]

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.

 

[I like Serena]

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.

 

[fabiancillo]

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 $R$?

 

[I like Serena]

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?

 

[fabiancillo]

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 $\neq p$?

 

[I like Serena]

$1$ ,$-1$ 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 $R$?

 

[fabiancillo]

$p$ does not divide $m$

 

[fabiancillo]

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 $m$ ?

 

[I like Serena]

$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 $R^\times$.

 

[fabiancillo]

I understand.

Thank you