MHB Are all subrings and units correctly identified in algebraic structures?

[evinda]

Hi! :)
I have a question..
Are all the following sentences right??
-$T_1$ subring of $R_1$,where $R_1=\{a+ b \sqrt{2}\}$ and $T_1=\{2a+b \sqrt{2}\}$
- $T_2$ subring of $R_2$,where $T_2=\begin{pmatrix}
1 & 0\\
0 & a
\end{pmatrix}: a\in \mathbb{R}$ and $R_2=M_2(\mathbb{R})$
-Does the set $R_6=\mathbb{Z} \times \mathbb{Q} \times \mathbb{Z}$ have infinitely many invertible elements? So,does it belong in $R^{*}$?

Or is the last wrong??

 

[I like Serena]

Hi! :)
I have a question..
Are all the following sentences right??
-$T_1$ subring of $R_1$,where $R_1=\{a+ b \sqrt{2}\}$ and $T_1=\{2a+b \sqrt{2}\}$
- $T_2$ subring of $R_2$,where $T_2=\begin{pmatrix}
1 & 0\\
0 & a
\end{pmatrix}: a\in \mathbb{R}$ and $R_2=M_2(\mathbb{R})$
-Does the set $R_6=\mathbb{Z} \times \mathbb{Q} \times \mathbb{Z}$ have infinitely many invertible elements? So,does it belong in $R^{*}$?

Or is the last wrong??

Hey! ;)

To check is a set $S$ is a subring, it suffices to verify for instance the following:
\begin{array}{}
1. &&&1 \in S &\text{identity element for multiplication is in } S\\
2. &\forall a,b \in S&: &a+b \in S & \text{closed for addition}\\
3. &\forall a \in S&: &-a \in S & \text{closed for additive inverse}\\
4. &\forall a,b \in S&: &ab \in S & \text{closed for multiplication} \\
\end{array}

You may have a similar list?

Perhaps you can check the sentences against that list?

 

[evinda]

Hey! ;)

To check is a set $S$ is a subring, it suffices to verify for instance the following:
\begin{array}{}
1. &&&1 \in S &\text{identity element for multiplication is in } S\\
2. &\forall a,b \in S&: &a+b \in S & \text{closed for addition}\\
3. &\forall a \in S&: &-a \in S & \text{closed for additive inverse}\\
4. &\forall a,b \in S&: &ab \in S & \text{closed for multiplication} \\
\end{array}

You may have a similar list?

Perhaps you can check the sentences against that list?

Yes,the first 2 sentences are right,but what's with the third one? (Thinking)

 

[Deveno]

Conditions 2 and 3 are often combined in the "one-step test":

2a. For all $a,b \in S$, we have $a - b \in S$.

It should be clear that 2a and (2 and 3 together) are equivalent. For example, if 2a holds, we get 3 by letting $a = 0$ (and we know that $0 \in S$ since if 2a holds, and $a \in S$, we have $a - a = 0 \in S$, and condition 1 tells us $S$ is non-emtpy, so we can always choose $a = 1$).

We then get from 3 and 2a, that for $a,b \in S$, that $a,-b \in S$ so that from 2a:

$a - (-b) = a+b \in S$.

On the other hand, if we have that 2 and 3 hold, then for $a,b \in S$, we know (via 3) that $a,-b \in S$, and thus (by 2) $a + (-b) = a - b \in S$.

 

[I like Serena]

Yes,the first 2 sentences are right,but what's with the third one? (Thinking)

I think the 2nd sentence is wrong!

What are your thoughts about the 3rd one?
Actually, the 3rd one is really 2 sentences... (Thinking)

 

[Deveno]

Ask yourself, is:

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} + \begin{bmatrix}1&0\\0&1 \end{bmatrix} \in T_2$?

 

[evinda]

I think the 2nd sentence is wrong!

What are your thoughts about the 3rd one?
Actually, the 3rd one is really 2 sentences... (Thinking)

Ask yourself, is:

$\begin{bmatrix}1&0\\0&1 \end{bmatrix} + \begin{bmatrix}1&0\\0&1 \end{bmatrix} \in T_2$?

With the third sentence,I meant this one: Does the set $R_6=\mathbb{Z} \times \mathbb{Q} \times \mathbb{Z}$ have infinitely many invertible elements?
Is this sentence right?

I think, $T_1$ is a subring of $R_1$ and $T_2$ is a subring of $R_2$. Or am I wrong?

 

[Deveno]

I think $T_2$ is NOT a subring of $R_2$. Why do you suppose I think this?

For your 3rd question, let $n$ be any natural number, and consider the element:

$(1,\dfrac{1}{n},1) \in \Bbb Z \times \Bbb Q \times \Bbb Z$.

Is this a unit?

 

[I like Serena]

With the third sentence,I meant this one: Does the set $R_6=\mathbb{Z} \times \mathbb{Q} \times \mathbb{Z}$ have infinitely many invertible elements?
Is this sentence right?

It seems as if you only want a yes/no answer. (Sadface)

Let's investigate.
Which operation are we talking about?
Is it addition or multiplication?
Can you give an example of an invertible element?

I think, $T_1$ is a subring of $R_1$ and $T_2$ is a subring of $R_2$. Or am I wrong?

I'm afraid you're wrong. (Shake)
Can you explain why you think so?

 

[evinda]

Can you explain why you think so?

I thought that $T_1$ is a subring of $R_1$,because:

$ (2a+b \sqrt{2})-(2d+c \sqrt{2}) =(2a-2d)+(b-c) \sqrt{2} \in R_1 $

Also, $(2a+b \sqrt{2})(2d+c \sqrt{2})=4ad+2ac \sqrt{2}+2db \sqrt{2}+2cb=(4ad-2db)+(2ac-2db)\sqrt{2} \in R_1 $

Also,I thought that $T_2$ is a subring of $R_2$,because:

$\begin{pmatrix}
1 & 0\\
0 & a
\end{pmatrix} $- $\begin{pmatrix}
1 & 0\\
0 & b
\end{pmatrix} $ =$\begin{pmatrix}
0 & 0\\
0 & a-b \end{pmatrix}$ $ \in R_2$ and also:

$\begin{pmatrix}
1 & 0\\
0& a
\end{pmatrix} $ $\cdot $$\begin{pmatrix}
1 & 0\\
0 & b
\end{pmatrix}$=$\begin{pmatrix}
1 & 0\\
0& ab
\end{pmatrix} $ $\in R_2 $

Let's investigate.
Which operation are we talking about?
Is it addition or multiplication?
Can you give an example of an invertible element?

We are talking about multiplication..An invertble element $a \in R$ should satisfy the condition: $\exists a'$ such that $aa'=1_R$ and $a'a=1_R$

 

[evinda]

I think $T_2$ is NOT a subring of $R_2$. Why do you suppose I think this?

I thought that $T_2$ is a subring of $R_2$,because:

$\begin{pmatrix}
1 & 0\\
0 & a
\end{pmatrix} $- $\begin{pmatrix}
1 & 0\\
0 & b
\end{pmatrix} $ =$\begin{pmatrix}
0 & 0\\
0 & a-b \end{pmatrix}$ $ \in R_2$ and also:

$\begin{pmatrix}
1 & 0\\
0& a
\end{pmatrix} $ $\cdot $$\begin{pmatrix}
1 & 0\\
0 & b
\end{pmatrix}$=$\begin{pmatrix}
1 & 0\\
0& ab
\end{pmatrix} $ $\in R_2 $

For your 3rd question, let $n$ be any natural number, and consider the element:

$(1,\dfrac{1}{n},1) \in \Bbb Z \times \Bbb Q \times \Bbb Z$.

Is this a unit?

I think that it is..Am I wrong??

 

[I like Serena]

I thought that $T_1$ is a subring of $R_1$,because:

$ (2a+b \sqrt{2})-(2d+c \sqrt{2}) =(2a-2d)+(b-c) \sqrt{2} \in R_1 $

Also, $(2a+b \sqrt{2})(2d+c \sqrt{2})=4ad+2ac \sqrt{2}+2db \sqrt{2}+2cb=(4ad-2db)+(2ac-2db)\sqrt{2} \in R_1 $

The criterion is that the combinations are in $T_1$, not that they are in $R_1$.

As it is, the combinations are implicitly in $R_1$, since we already know that $R_1$ is a ring, but that is not what we are verifying - we want to know something about $T_1$!

Luckily, the elements you found are also in $T_1$, so $T_1$ is indeed a subring.

Also,I thought that $T_2$ is a subring of $R_2$,because:

$\begin{pmatrix}
1 & 0\\
0 & a
\end{pmatrix} $- $\begin{pmatrix}
1 & 0\\
0 & b
\end{pmatrix} $ =$\begin{pmatrix}
0 & 0\\
0 & a-b \end{pmatrix}$ $ \in R_2$ and also:

Here it is different.
That matrix is not an element of $T_2$.
Therefore $T_2$ is not a subring.

We are talking about multiplication..An invertible element $a \in R$ should satisfy the condition: $\exists a'$ such that $aa'=1_R$ and $a'a=1_R$

Good! :)

So for instance (1, 2, 1) has the inverse (1, 1/2, 1).

Which inverse does (2, 1, 1) have?
And (1,3,1)?

 

[evinda]

The criterion is that the combinations are in $T_1$, not that they are in $R_1$.

As it is, the combinations are implicitly in $R_1$, since we already know that $R_1$ is a ring, but that is not what we are verifying - we want to know something about $T_1$!

Luckily, the elements you found are also in $T_1$, so $T_1$ is indeed a subring.
Here it is different.
That matrix is not an element of $T_2$.
Therefore $T_2$ is not a subring.

I understand...

So for instance (1, 2, 1) has the inverse (1, 1/2, 1).

Which inverse does (2, 1, 1) have?
And (1,3,1)?

$(2,1,1)$ has not an inverse,because $\frac{1}{2} \notin \mathbb{Z}$,right? The inverse of $(1,3,1)$ is $(1,\frac{1}{3},1)$,or not?

 

[I like Serena]

$(2,1,1)$ has not an inverse,because $\frac{1}{2} \notin \mathbb{Z}$,right? The inverse of $(1,3,1)$ is $(1,\frac{1}{3},1)$,or not?

Correct! :)

How do elements look that are invertible and what is their inverse?

 

[Deveno]

$(2,1,1)$ has not an inverse,because $\frac{1}{2} \notin \mathbb{Z}$,right? The inverse of $(1,3,1)$ is $(1,\frac{1}{3},1)$,or not?

If I recall correctly, your original question was whether or not $R_6$ has infinitely many units. The elements that aren't units are of no interest to us...

 

[evinda]

Correct! :)

How do elements look that are invertible and what is their inverse?

They look like that: $(\pm 1, q, \pm 1) , q \in \mathbb{Q}$,right?But are these elements infinitely many?

 

[I like Serena]

They look like that: $(\pm 1, q, \pm 1) , q \in \mathbb{Q}$,right?But are these elements infinitely many?

Almost... (Thinking)

Let me pick an element $q$ from $\mathbb Q$.
I pick... $q=0$. (Evilgrin)
Is (1, 0, 1) invertible?How many elements does $\mathbb Q$ have?
Is each element invertible?

 

[evinda]

Almost... (Thinking)

Let me pick an element $q$ from $\mathbb Q$.
I pick... $q=0$. (Evilgrin)
Is (1, 0, 1) invertible?How many elements does $\mathbb Q$ have?
Is each element invertible?

No,$(1,0,1)$ is not invertible! (Shake)

$\mathbb Q$ has infinite many elements,or not?
I think that each element $q \in \mathbb{Q}$ except from $q=0$ is invertible..
So,the set has infinite many invertible elements..or not?? (Thinking)(Thinking)

 

[I like Serena]

No,$(1,0,1)$ is not invertible! (Shake)

Right! (Happy)

$\mathbb Q$ has infinite many elements,or not?
I think that each element $q \in \mathbb{Q}$ except from $q=0$ is invertible..
So,the set has infinite many invertible elements..or not?? (Thinking)(Thinking)

Yes.

 

[evinda]

Right! (Happy)
Yes.

Nice,thank you! :)

 

[Deveno]

I think $T_2$ is NOT a subring of $R_2$. Why do you suppose I think this?

For your 3rd question, let $n$ be any natural number, and consider the element:

$(1,\dfrac{1}{n},1) \in \Bbb Z \times \Bbb Q \times \Bbb Z$.

Is this a unit?

I think that it is..Am I wrong??

First, let's establish that (1,1,1) is the identity:

$(a,b,c)(1,1,1) = (a1,b1,c1) = (a,b,c)$, so that works.

Now:

$(1,\dfrac{1}{n},1)(1,n,1) = (1,\dfrac{n}{n},1) = (1,1,1)$

so we clearly have a unit.

Since there are an infinite number of distinct $n$ we can choose (one for every natural number) the cardinailty of the set of units is at least as big as the cardinality of the set of natural numbers, that is to say: infinite.