MHB What are the ideals of the matrix ring?

[Math Amateur]

I am reading Paul E. Bland's book: Rings and Their Modules and am currently focused on Section 4.2 Noetherian and Artinian Modules ... ...

I need help with fully understanding Example 6 "Right Artinian but not Left Artinian" ... in Section 4.2 ... ...

Example 6 reads as follows:
View attachment 5971My problem is how to prove, explicitly and formally, that the matrix ring:$$\begin{pmatrix} \mathbb{Q} & \mathbb{R} \\ 0 & \mathbb{R} \end{pmatrix}$$Also, a related problem is that I am trying to determine the form of all the ideals of the above matrix ring ... but without success ... help with this issue would be appreciated as well ...
Reading around this problem leads me to believe that showing the above matrix ring to be right artinian but not left artinian would involve showing that all descending chains of right ideals terminate ... but that this does not hold for descending chains of left ideals ... ...
Note that the conditions regarding descending chains of right (left) ideals for right (left) Artinian rings are not given in the definition of right (left) Artinian rings by Bland, but I believe Bland's definition implies them ... is that correct? ... see Bland's definition of Artinian rings below ...
Bland's definition of Artinian rings and modules is as follows:https://www.physicsforums.com/attachments/5972Hope someone can help,

Peter

 

[Deveno]

Here is a start:

Note that as an abelian group, our ring is (isomorphic) to:

$\Bbb Q \oplus \{0\} \oplus \Bbb R \oplus \Bbb R$.

So, first, convince yourself that any right (or left) ideal must be of the form:

$H \oplus \{0\} \oplus K_1 \oplus K_2$, where $H$ is an additive subgroup of $\Bbb Q$, and $K_1,K_2$ are additive subgroups of $\Bbb R$.

Now, if we have a(n additive) subgroup $I$ and we call our ring $R$, for any $x \in I$ and $r \in R$, we must have $xr \in I$, for $I$ to be a right ideal.

Let's look at the multiplication, and how this plays out:

$\begin{pmatrix}x_1&x_2\\0&x_3\end{pmatrix}\begin{pmatrix}r_1&r_2\\0&r_3\end{pmatrix} = \begin{pmatrix}x_1r_1&x_1r_2+x_2r_3\\0&x_3r_3\end{pmatrix}$

Looking at the bottom right entry, it should be clear that we must have $K_2$ be an ideal of $\Bbb R$. Since $\Bbb R$ is a field, we have either two choices: $x_3 = 0$, or $x_3$ is a unit (of $\Bbb R$, that is: non-zero).

Let's suppose that $x_3$ is a unit. Now if $x_1$ is also a unit, then the matrix with the $x$'s is invertible (since it has non-zero determinant), and in fact is a unit in our ring since it has the inverse in our ring of:

$\begin{pmatrix}\dfrac{1}{x_1}&-\dfrac{x_2}{x_1x_3}\\0&\dfrac{1}{x_3}\end{pmatrix}$

So if $I$ has an element where $x_1,x_3$ are both units, then $I = R$.

Thus for any PROPER ideal, we must have $x_1 = 0$, or $x_3 = 0$. Investigate these.

 

[Math Amateur]

Here is a start:

Note that as an abelian group, our ring is (isomorphic) to:

$\Bbb Q \oplus \{0\} \oplus \Bbb R \oplus \Bbb R$.

So, first, convince yourself that any right (or left) ideal must be of the form:

$H \oplus \{0\} \oplus K_1 \oplus K_2$, where $H$ is an additive subgroup of $\Bbb Q$, and $K_1,K_2$ are additive subgroups of $\Bbb R$.

Now, if we have a(n additive) subgroup $I$ and we call our ring $R$, for any $x \in I$ and $r \in R$, we must have $xr \in I$, for $I$ to be a right ideal.

Let's look at the multiplication, and how this plays out:

$\begin{pmatrix}x_1&x_2\\0&x_3\end{pmatrix}\begin{pmatrix}r_1&r_2\\0&r_3\end{pmatrix} = \begin{pmatrix}x_1r_1&x_1r_2+x_2r_3\\0&x_3r_3\end{pmatrix}$

Looking at the bottom right entry, it should be clear that we must have $K_2$ be an ideal of $\Bbb R$. Since $\Bbb R$ is a field, we have either two choices: $x_3 = 0$, or $x_3$ is a unit (of $\Bbb R$, that is: non-zero).

Let's suppose that $x_3$ is a unit. Now if $x_1$ is also a unit, then the matrix with the $x$'s is invertible (since it has non-zero determinant), and in fact is a unit in our ring since it has the inverse in our ring of:

$\begin{pmatrix}\dfrac{1}{x_1}&-\dfrac{x_2}{x_1x_3}\\0&\dfrac{1}{x_3}\end{pmatrix}$

So if $I$ has an element where $x_1,x_3$ are both units, then $I = R$.

Thus for any PROPER ideal, we must have $x_1 = 0$, or $x_3 = 0$. Investigate these.

Thanks for the help, Deveno ...

Just reflecting on what you have written ...

Peter

 

[Math Amateur]

Here is a start:

Note that as an abelian group, our ring is (isomorphic) to:

$\Bbb Q \oplus \{0\} \oplus \Bbb R \oplus \Bbb R$.

So, first, convince yourself that any right (or left) ideal must be of the form:

$H \oplus \{0\} \oplus K_1 \oplus K_2$, where $H$ is an additive subgroup of $\Bbb Q$, and $K_1,K_2$ are additive subgroups of $\Bbb R$.

Now, if we have a(n additive) subgroup $I$ and we call our ring $R$, for any $x \in I$ and $r \in R$, we must have $xr \in I$, for $I$ to be a right ideal.

Let's look at the multiplication, and how this plays out:

$\begin{pmatrix}x_1&x_2\\0&x_3\end{pmatrix}\begin{pmatrix}r_1&r_2\\0&r_3\end{pmatrix} = \begin{pmatrix}x_1r_1&x_1r_2+x_2r_3\\0&x_3r_3\end{pmatrix}$

Looking at the bottom right entry, it should be clear that we must have $K_2$ be an ideal of $\Bbb R$. Since $\Bbb R$ is a field, we have either two choices: $x_3 = 0$, or $x_3$ is a unit (of $\Bbb R$, that is: non-zero).

Let's suppose that $x_3$ is a unit. Now if $x_1$ is also a unit, then the matrix with the $x$'s is invertible (since it has non-zero determinant), and in fact is a unit in our ring since it has the inverse in our ring of:

$\begin{pmatrix}\dfrac{1}{x_1}&-\dfrac{x_2}{x_1x_3}\\0&\dfrac{1}{x_3}\end{pmatrix}$

So if $I$ has an element where $x_1,x_3$ are both units, then $I = R$.

Thus for any PROPER ideal, we must have $x_1 = 0$, or $x_3 = 0$. Investigate these.

Thanks again for your help, Deveno ... but just need some further help in clarifying some things ... hope you can help ...Question 1

You write:

" ... ... Note that , our ring is (isomorphic) to:

$\Bbb Q \oplus \{0\} \oplus \Bbb R \oplus \Bbb R$. ... ... "I am currently struggling somewhat with this ... can you demonstrate why/how

$$\begin{pmatrix} \mathbb{Q} & \mathbb{R} \\ 0 & \mathbb{R} \end{pmatrix}$$ $$ \cong $$ $\Bbb Q \oplus \{0\} \oplus \Bbb R \oplus \Bbb R$Sorry if I am being a bit slow here ... ...
Question 2

I am also having problems seeing exactly why any right (or left) ideal must be of the form:

$H \oplus \{0\} \oplus K_1 \oplus K_2$, where $H$ is an additive subgroup of $\Bbb Q$, and $K_1,K_2$ are additive subgroups of $\Bbb R$.Can you please help in this matter as well ...Thanks again for your help,

Peter

 

[Deveno]

Thanks again for your help, Deveno ... but just need some further help in clarifying some things ... hope you can help ...Question 1

You write:

" ... ... Note that , our ring is (isomorphic) to:

$\Bbb Q \oplus \{0\} \oplus \Bbb R \oplus \Bbb R$. ... ... "I am currently struggling somewhat with this ... can you demonstrate why/how

$$\begin{pmatrix} \mathbb{Q} & \mathbb{R} \\ 0 & \mathbb{R} \end{pmatrix}$$ $$ \cong $$ $\Bbb Q \oplus \{0\} \oplus \Bbb R \oplus \Bbb R$Sorry if I am being a bit slow here ... ...
Question 2

I am also having problems seeing exactly why any right (or left) ideal must be of the form:

$H \oplus \{0\} \oplus K_1 \oplus K_2$, where $H$ is an additive subgroup of $\Bbb Q$, and $K_1,K_2$ are additive subgroups of $\Bbb R$.Can you please help in this matter as well ...Thanks again for your help,

Peter

We have the abelian group isomorphism:

$\begin{pmatrix}a&b\\0&c\end{pmatrix} \mapsto (a,0,b,c)$

Now let's say $A$ is an abelian subgroup of $(R,+)$, where we consider $R$ as $\Bbb Q \oplus \{0\} \oplus \Bbb R \oplus \Bbb R$.

Let $H' = \{(a,0,0,0) \in A\}$.

By closure, we have $(a,0,0,0) + (a',0,0,0) = (a+a',0,0,0) \in A$, and this is clearly an element of $H'$. So we have closure in $H'$.

Since $A$ is a subgroup, for any $(a,0,0,0) \in A$, we have $-(a,0,0,0) = (-a,0,0,0) \in A$, and this is also in $H'$. Thus $H'$ contains all additive inverses, and is a subgroup of $A$.

The isomorphism above (along with the canonical projection $(a,0,0,0) \mapsto a$) induces an abelian group monomorphism $H' \to \Bbb Q$. In this way, we see $H = \text{im}(H')$ must be a subgroup of $\Bbb Q$.

We can carry out similar constructions for $K_1$ and $K_2$, so we see that:

$A = H' + K_1' + K_2' \cong H \oplus\{0\} \oplus K_1 \oplus K_2$

because $(a,0,b,c) = (a,0,0,0) + (0,0,b,0) + (0,0,0,c)$.

 

[Math Amateur]

We have the abelian group isomorphism:

$\begin{pmatrix}a&b\\0&c\end{pmatrix} \mapsto (a,0,b,c)$

Now let's say $A$ is an abelian subgroup of $(R,+)$, where we consider $R$ as $\Bbb Q \oplus \{0\} \oplus \Bbb R \oplus \Bbb R$.

Let $H' = \{(a,0,0,0) \in A\}$.

By closure, we have $(a,0,0,0) + (a',0,0,0) = (a+a',0,0,0) \in A$, and this is clearly an element of $H'$. So we have closure in $H'$.

Since $A$ is a subgroup, for any $(a,0,0,0) \in A$, we have $-(a,0,0,0) = (-a,0,0,0) \in A$, and this is also in $H'$. Thus $H'$ contains all additive inverses, and is a subgroup of $A$.

The isomorphism above (along with the canonical projection $(a,0,0,0) \mapsto a$) induces an abelian group monomorphism $H' \to \Bbb Q$. In this way, we see $H = \text{im}(H')$ must be a subgroup of $\Bbb Q$.

We can carry out similar constructions for $K_1$ and $K_2$, so we see that:

$A = H' + K_1' + K_2' \cong H \oplus\{0\} \oplus K_1 \oplus K_2$

because $(a,0,b,c) = (a,0,0,0) + (0,0,b,0) + (0,0,0,c)$.

Thanks so much Deveno ... REALLY helpful ...

Peter

 

[Math Amateur]

We have the abelian group isomorphism:

$\begin{pmatrix}a&b\\0&c\end{pmatrix} \mapsto (a,0,b,c)$

Now let's say $A$ is an abelian subgroup of $(R,+)$, where we consider $R$ as $\Bbb Q \oplus \{0\} \oplus \Bbb R \oplus \Bbb R$.

Let $H' = \{(a,0,0,0) \in A\}$.

By closure, we have $(a,0,0,0) + (a',0,0,0) = (a+a',0,0,0) \in A$, and this is clearly an element of $H'$. So we have closure in $H'$.

Since $A$ is a subgroup, for any $(a,0,0,0) \in A$, we have $-(a,0,0,0) = (-a,0,0,0) \in A$, and this is also in $H'$. Thus $H'$ contains all additive inverses, and is a subgroup of $A$.

The isomorphism above (along with the canonical projection $(a,0,0,0) \mapsto a$) induces an abelian group monomorphism $H' \to \Bbb Q$. In this way, we see $H = \text{im}(H')$ must be a subgroup of $\Bbb Q$.

We can carry out similar constructions for $K_1$ and $K_2$, so we see that:

$A = H' + K_1' + K_2' \cong H \oplus\{0\} \oplus K_1 \oplus K_2$

because $(a,0,b,c) = (a,0,0,0) + (0,0,b,0) + (0,0,0,c)$.

Hi Deveno,

I am still reflecting on the nature of the abelian group isomorphism (and further ring isomorphism)

$$S =$$ $$\begin{pmatrix} \mathbb{Q} & \mathbb{R} \\ 0 & \mathbb{R} \end{pmatrix}$$$$ \ \cong \ $$ $\Bbb Q \oplus \{0\} \oplus \Bbb R \oplus \Bbb R$Now, if we want to talk about ideals, surely we need to extend the abelian group isomorphism to a ring isomorphism ... ...So ... thinking ... if we are given two elements of the ring $$S$$ , say ...

$$
\begin{pmatrix} s_1 & s_2 \\ 0 & s_3 \end{pmatrix}$$ and $$\begin{pmatrix} r_1 & r_2 \\ 0 & r_3 \end{pmatrix}$$... ... then we have ...$$\begin{pmatrix} s_1 & s_2 \\ 0 & s_3 \end{pmatrix} \begin{pmatrix} r_1 & r_2 \\ 0 & r_3 \end{pmatrix}
$$

$$= \begin{pmatrix} s_1r_1 & s_1 r_2 + s_2 r_3 \\ 0 & s_3 r_3 \end{pmatrix}
$$Now ... if we want a ring isomorphism $S \cong \Bbb Q \oplus \{0\} \oplus \Bbb R \oplus \Bbb R$where we must define multiplication in $\Bbb Q \oplus \{0\} \oplus \Bbb R \oplus \Bbb R$ as follows:Given two elements $$(s_1, 0, s_2, s_3) , (r_1, 0, r_2, r_3) \in$$ $\Bbb Q \oplus \{0\} \oplus \Bbb R \oplus \Bbb R$ we have:$$(s_1, 0, s_2, s_3) \bullet (r_1, 0, r_2, r_3) = (s_1r_1, 0, s_1 r_2 + s_2 r_3, s_3 r_3)$$Is this correct?

Does such a multiplication (different from the usual componentwise multiplication in direct sums and products) for the ring S affect the fact that any right (or left) ideal must be of the form:

$H \oplus \{0\} \oplus K_1 \oplus K_2$, where $H$ is an additive subgroup of $\Bbb Q$, and $K_1,K_2$ are additive subgroups of $\Bbb R$.Can you comment and help ...

I am still puzzled as to how to proceed to show that S is right artinian but not left artinian ...

Can you please help further ...

Peter