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

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In summary, the group of units of a ring $R$ is the set of elements that are invertible with respect to multiplication.
  • #1
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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  • #2
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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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 $R$?
 
  • #6
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?
 
  • #7
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 $\neq p$?
 
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  • #8
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 $R$?
 
  • #9
$p$ 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 $m$ ?
 
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  • #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
 

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

What is a sub-ring?

A sub-ring is a subset of a ring that is itself a ring under the same operations as the original ring.

How do you prove that R is a sub-ring of $\mathbb{Q}$?

To prove that R is a sub-ring of $\mathbb{Q}$, we need to show that R satisfies all the properties of a ring, including closure under addition and multiplication, existence of additive and multiplicative identity, and existence of additive and multiplicative inverses.

What is a prime number?

A prime number is a positive integer that is only divisible by 1 and itself. In other words, it has no other factors besides 1 and itself.

Why is it important that p is a prime number in this context?

In this context, p being a prime number is important because it ensures that the sub-ring R is closed under multiplication. If p were not a prime number, there would exist non-trivial divisors of p in R, which would violate the closure property.

Can R be a sub-ring of $\mathbb{Q}$ if p is not a prime number?

No, R cannot be a sub-ring of $\mathbb{Q}$ if p is not a prime number. As mentioned before, this would violate the closure property of a ring, making R not a valid sub-ring of $\mathbb{Q}$.

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