# Prove it is a sub-ring

• MHB
• fabiancillo
In summary, the group of units of a ring $R$ is the set of elements that are invertible with respect to multiplication.f

#### 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

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?

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.

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 $R$?

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?

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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$1$ ,$-1$ 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 $m$

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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$R^{\times} = n \in \mathbb{Q}$ 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 $R^\times$.

I understand.

Thank you