MHB Definition of a Right Artinian Ring - Cohn - page 66

[Math Amateur]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 2: Linear Algebras and Artinian Rings, on Page 66 we find a definition of right Artinian rings ...

The relevant text in Cohn's book is as follows:https://www.physicsforums.com/attachments/3338In the above text Cohn defines a right Artinian ring as follows:

" ... ... A ring R is said to be right Artinian if it is Artinian when regarded as a right module over itself ... ... "

I am slightly confused by this definition ... ... I thought a ring (right or left) was always a (right, left) module over itself ... so isn't an Artinian ring a right module over itself by definition ... ...can someone please draw out the reasons for and the implications of Cohn's definition for me ... ...Further, Cohn also writes regarding the above definition (see above text):

" ... ... This then means that R satisfies the minimum condition on right ideals ... ... "

Can someone please explain what Cohn means by "the minimum condition on right ideals"?

Help in this matter will be appreciated ... ...

Peter

 

[Euge]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 2: Linear Algebras and Artinian Rings, on Page 66 we find a definition of right Artinian rings ...

The relevant text in Cohn's book is as follows:https://www.physicsforums.com/attachments/3338In the above text Cohn defines a right Artinian ring as follows:

" ... ... A ring R is said to be right Artinian if it is Artinian when regarded as a right module over itself ... ... "

I am slightly confused by this definition ... ... I thought a ring (right or left) was always a (right, left) module over itself ... so isn't an Artinian ring a right module over itself by definition ... ...can someone please draw out the reasons for and the implications of Cohn's definition for me ... ...Further, Cohn also writes regarding the above definition (see above text):

" ... ... This then means that R satisfies the minimum condition on right ideals ... ... "

Can someone please explain what Cohn means by "the minimum condition on right ideals"?

Help in this matter will be appreciated ... ...

Peter

Hi Peter,

If $R$ is a ring and $A$ is a left $R$-module, then $A$ is Artinian if every descending chain

$$ X_0 \supseteq X_1 \supseteq X_2 \supseteq \cdots$$

of submodules of $A$ stabilizes (known as the descending chain condition, or, D.C.C.). Then a left (resp. right) Artinian ring, call it $S$, is an ${}_SS$ (resp. $S_S$) module. Please note the similarity of these definitions with those of Noetherian rings and modules. Now if $S$ satisfies the D.C.C. for both left and right ideals, then $S$ is Artinian (without reference to "left" or "right").

For a module ${}_RA$, $A$ is Artinian if and only if every nonempty set of submodules of $A$ has a minimal element. The corresponding equivalence for right Artinian rings is probably what Cohn meant by the "minimum condition on right ideals". Check back in your book though to see if he defined it somewhere. Then we can work from there.

 

[Math Amateur]

Hi Peter,

If $R$ is a ring and $A$ is a left $R$-module, then $A$ is Artinian if every descending chain

$$ X_0 \supseteq X_1 \supseteq X_2 \supseteq \cdots$$

of submodules of $A$ stabilizes (known as the descending chain condition, or, D.C.C.). Then a left (resp. right) Artinian ring, call it $S$, is an ${}_SS$ (resp. $S_S$) module. Please note the similarity of these definitions with those of Noetherian rings and modules. Now if $S$ satisfies the D.C.C. for both left and right ideals, then $S$ is Artinian (without reference to "left" or "right").

For a module ${}_RA$, $A$ is Artinian if and only if every nonempty set of submodules of $A$ has a minimal element. The corresponding equivalence for right Artinian rings is probably what Cohn meant by the "minimum condition on right ideals". Check back in your book though to see if he defined it somewhere. Then we can work from there.

Thanks so much for the help, Euge ...

I did not find an explicit reference to "the minimal condition" in Cohn , but I think "the minimal condition" is what you specify when you write:

" ... ... For a module ${}_RA$, $A$ is Artinian if and only if every nonempty set of submodules of $A$ has a minimal element. The corresponding equivalence for right Artinian rings is probably what Cohn meant by the "minimum condition on right ideals". ... ... "

I say this because when reading Chapter 5: Chain Conditions: Jordan-Holder Towers (that is composition series) in T.S. Blyth's book: Module Theory: An Approach to Linear Algebra we find a definition of "the maximum condition" on page 45. The relevant text from Blyth is as follows:View attachment 3342

On page 46, we find Blyth writing the following:

" ... ... We can of course define the dual concepts of descending chain condition and minimum condition in the obvious way. We say that M is artinian if it satisfies the descending chain condition on submodules ... ... "

Blyth then states theorem 5.2 which states that for every R-module M the statements:

(1) M is artinian

(2) M satisfies the minimum condition

However, ... ... Blyth does not deal with artinian rings ...

Peter

 

[Euge]

Thanks so much for the help, Euge ...

I did not find an explicit reference to "the minimal condition" in Cohn , but I think "the minimal condition" is what you specify when you write:

" ... ... For a module ${}_RA$, $A$ is Artinian if and only if every nonempty set of submodules of $A$ has a minimal element. The corresponding equivalence for right Artinian rings is probably what Cohn meant by the "minimum condition on right ideals". ... ... "

I say this because when reading Chapter 5: Chain Conditions: Jordan-Holder Towers (that is composition series) in T.S. Blyth's book: Module Theory: An Approach to Linear Algebra we find a definition of "the maximum condition" on page 45. The relevant text from Blyth is as follows:View attachment 3342

On page 46, we find Blyth writing the following:

" ... ... We can of course define the dual concepts of descending chain condition and minimum condition in the obvious way. We say that M is artinian if it satisfies the descending chain condition on submodules ... ... "

Blyth then states theorem 5.2 which states that for every R-module M the statements:

(1) M is artinian

(2) M satisfies the minimum condition

However, ... ... Blyth does not deal with artinian rings ...

Peter

Ok, that's what I expected. Let's consider two examples.

$\textbf{Exmaple 1}$. Let $\Bbb k$ be a field. Then $\Bbb k$ is Artinian, since a nonempty collection of submodules of $\Bbb k$ has minimal element $0$ or $\Bbb k$. However, integral domains are not always Artinian. The integers is one prime example; the descending chain $2\Bbb Z \supseteq 4\Bbb Z \supseteq 8\Bbb Z \supseteq \cdots$ does not stabilize.

$\textbf{Example 2}$. Let $\Bbb k$ be a field and $x$ an indeterminate. For any nonzero ideal $J$ of $\Bbb k[x]$, the quotient $\Bbb k[x]/J$ is Artinian. Indeed, since $\Bbb k$ is a field, $\Bbb k[x]$ is a PID; so let $d(x)$ be the monic polynomial generating $J$. If $X_0' \supseteq X_1' \supseteq X_2' \supseteq \cdots$ is a descending chain of ideals of $\Bbb k[x]/J$, it lifts to a descending chain $X_0 \supseteq X_1 \supseteq X_2 \supseteq \cdots$ for $\Bbb k[x]$ such that $X_i \supseteq J$ for all $i \ge 0$ (keep in mind each $X_i$ is the preimage of the corresponding $X_i'$ under the natural projection $\pi : \Bbb k[x] \to \Bbb k[x]/J$). For each $i$, let $d_i$ be the monic polynomial generating $X_i$. The descending chain of $X_i$ implies $d_0\, |\, d_1\, |\, d_2\, |\, \cdots$ and the containments $X_i \supseteq J$ for all $i$ imply $d_i\, |\, d$ for all $i$. So $\{\deg d_i\}_{i=1}^\infty$ is an increasing sequence of positive integers bounded above by $\deg d$. This means that the sequence is eventually constant, i.e., there exists an index $N$ such that $d_k = d_N$ for all $k \ge N$. Thus $X_k = X_N$ for all $k \ge N$ and consequently $X_k' = X_N'$ for all $k \ge N$. It follows that the DCC is satisfied for $\Bbb k[x]/J$, so it is Artinian.

Note that if we relax the condition that $\Bbb k$ is a field, then $\Bbb k[x]/J$ may not be Artinian. For instance, the factor ring $\Bbb Z[x]/(x)$, being isomorphic to the non-Artinian ring $\Bbb Z$, is non-Artinian.

Here are two exercises for you to try.

1. Show that if a left $R$-module $M$ is both Artinian and Noetherian, then it has finite length.

2. Show that if $A$ is left (unital) Artinian ring, then whenever $x,y \in A$, $xy = 1$ implies $yx = 1$.

 

[Math Amateur]

Ok, that's what I expected. Let's consider two examples.

$\textbf{Exmaple 1}$. Let $\Bbb k$ be a field. Then $\Bbb k$ is Artinian, since a nonempty collection of submodules of $\Bbb k$ has minimal element $0$ or $\Bbb k$. However, integral domains are not always Artinian. The integers is one prime example; the descending chain $2\Bbb Z \supseteq 4\Bbb Z \supseteq 8\Bbb Z \supseteq \cdots$ does not stabilize.

$\textbf{Example 2}$. Let $\Bbb k$ be a field and $x$ an indeterminate. For any nonzero ideal $J$ of $\Bbb k[x]$, the quotient $\Bbb k[x]/J$ is Artinian. Indeed, since $\Bbb k$ is a field, $\Bbb k[x]$ is a PID; so let $d(x)$ be the monic polynomial generating $J$. If $X_0' \supseteq X_1' \supseteq X_2' \supseteq \cdots$ is a descending chain of ideals of $\Bbb k[x]/J$, it lifts to a descending chain $X_0 \supseteq X_1 \supseteq X_2 \supseteq \cdots$ for $\Bbb k[x]$ such that $X_i \supseteq J$ for all $i \ge 0$ (keep in mind each $X_i$ is the preimage of the corresponding $X_i'$ under the natural projection $\pi : \Bbb k[x] \to \Bbb k[x]/J$). For each $i$, let $d_i$ be the monic polynomial generating $X_i$. The descending chain of $X_i$ implies $d_0\, |\, d_1\, |\, d_2\, |\, \cdots$ and the containments $X_i \supseteq J$ for all $i$ imply $d_i\, |\, d$ for all $i$. So $\{\deg d_i\}_{i=1}^\infty$ is an increasing sequence of positive integers bounded above by $\deg d$. This means that the sequence is eventually constant, i.e., there exists an index $N$ such that $d_k = d_N$ for all $k \ge N$. Thus $X_k = X_N$ for all $k \ge N$ and consequently $X_k' = X_N'$ for all $k \ge N$. It follows that the DCC is satisfied for $\Bbb k[x]/J$, so it is Artinian.

Note that if we relax the condition that $\Bbb k$ is a field, then $\Bbb k[x]/J$ may not be Artinian. For instance, the factor ring $\Bbb Z[x]/(x)$, being isomorphic to the non-Artinian ring $\Bbb Z$, is non-Artinian.

Here are two exercises for you to try.

1. Show that if a left $R$-module $M$ is both Artinian and Noetherian, then it has finite length.

2. Show that if $A$ is left (unital) Artinian ring, then whenever $x,y \in A$, $xy = 1$ implies $yx = 1$.

Thank you for this post, Euge ... most helpful of you ... just working through it now ... Artinian rings and modules are quite fascinating!Peter

 

[Math Amateur]

Ok, that's what I expected. Let's consider two examples.

$\textbf{Exmaple 1}$. Let $\Bbb k$ be a field. Then $\Bbb k$ is Artinian, since a nonempty collection of submodules of $\Bbb k$ has minimal element $0$ or $\Bbb k$. However, integral domains are not always Artinian. The integers is one prime example; the descending chain $2\Bbb Z \supseteq 4\Bbb Z \supseteq 8\Bbb Z \supseteq \cdots$ does not stabilize.

$\textbf{Example 2}$. Let $\Bbb k$ be a field and $x$ an indeterminate. For any nonzero ideal $J$ of $\Bbb k[x]$, the quotient $\Bbb k[x]/J$ is Artinian. Indeed, since $\Bbb k$ is a field, $\Bbb k[x]$ is a PID; so let $d(x)$ be the monic polynomial generating $J$. If $X_0' \supseteq X_1' \supseteq X_2' \supseteq \cdots$ is a descending chain of ideals of $\Bbb k[x]/J$, it lifts to a descending chain $X_0 \supseteq X_1 \supseteq X_2 \supseteq \cdots$ for $\Bbb k[x]$ such that $X_i \supseteq J$ for all $i \ge 0$ (keep in mind each $X_i$ is the preimage of the corresponding $X_i'$ under the natural projection $\pi : \Bbb k[x] \to \Bbb k[x]/J$). For each $i$, let $d_i$ be the monic polynomial generating $X_i$. The descending chain of $X_i$ implies $d_0\, |\, d_1\, |\, d_2\, |\, \cdots$ and the containments $X_i \supseteq J$ for all $i$ imply $d_i\, |\, d$ for all $i$. So $\{\deg d_i\}_{i=1}^\infty$ is an increasing sequence of positive integers bounded above by $\deg d$. This means that the sequence is eventually constant, i.e., there exists an index $N$ such that $d_k = d_N$ for all $k \ge N$. Thus $X_k = X_N$ for all $k \ge N$ and consequently $X_k' = X_N'$ for all $k \ge N$. It follows that the DCC is satisfied for $\Bbb k[x]/J$, so it is Artinian.

Note that if we relax the condition that $\Bbb k$ is a field, then $\Bbb k[x]/J$ may not be Artinian. For instance, the factor ring $\Bbb Z[x]/(x)$, being isomorphic to the non-Artinian ring $\Bbb Z$, is non-Artinian.

Here are two exercises for you to try.

1. Show that if a left $R$-module $M$ is both Artinian and Noetherian, then it has finite length.

2. Show that if $A$ is left (unital) Artinian ring, then whenever $x,y \in A$, $xy = 1$ implies $yx = 1$.

I am still puzzled over the definition of an Artinian ring in terms of module ...

For example see the definition in Berrick and Keating's book: An Introduction to Rings and Modules With K-Theory in View.

The relevant text (page 146) is as follows:
https://www.physicsforums.com/attachments/3343
So, in the above text we read:

"A ring R is said to be right Artinian if it is Artinian when considered as a right module over itself. Put more directly, every descending chain of right ideals in R must terminate or be finite."Can someone please show me exactly how it is that a ring that is Artinian when considered as a right module over itself necessarily has the property that every descending chain of right ideals in R must terminate or be finite?

I would appreciate some help in this matter ...

... ... now back to the exercises set by Euge ... :)

Peter

 

[Euge]

Did you know that, viewing a ring R as a right module over itself, the R-submodules are the right ideals of R? Using this fact, the DCC for right Artinian modules translates to the DCC for right Artinian rings, which is displayed in your text.

 

[Math Amateur]

Ok, that's what I expected. Let's consider two examples.

$\textbf{Exmaple 1}$. Let $\Bbb k$ be a field. Then $\Bbb k$ is Artinian, since a nonempty collection of submodules of $\Bbb k$ has minimal element $0$ or $\Bbb k$. However, integral domains are not always Artinian. The integers is one prime example; the descending chain $2\Bbb Z \supseteq 4\Bbb Z \supseteq 8\Bbb Z \supseteq \cdots$ does not stabilize.

$\textbf{Example 2}$. Let $\Bbb k$ be a field and $x$ an indeterminate. For any nonzero ideal $J$ of $\Bbb k[x]$, the quotient $\Bbb k[x]/J$ is Artinian. Indeed, since $\Bbb k$ is a field, $\Bbb k[x]$ is a PID; so let $d(x)$ be the monic polynomial generating $J$. If $X_0' \supseteq X_1' \supseteq X_2' \supseteq \cdots$ is a descending chain of ideals of $\Bbb k[x]/J$, it lifts to a descending chain $X_0 \supseteq X_1 \supseteq X_2 \supseteq \cdots$ for $\Bbb k[x]$ such that $X_i \supseteq J$ for all $i \ge 0$ (keep in mind each $X_i$ is the preimage of the corresponding $X_i'$ under the natural projection $\pi : \Bbb k[x] \to \Bbb k[x]/J$). For each $i$, let $d_i$ be the monic polynomial generating $X_i$. The descending chain of $X_i$ implies $d_0\, |\, d_1\, |\, d_2\, |\, \cdots$ and the containments $X_i \supseteq J$ for all $i$ imply $d_i\, |\, d$ for all $i$. So $\{\deg d_i\}_{i=1}^\infty$ is an increasing sequence of positive integers bounded above by $\deg d$. This means that the sequence is eventually constant, i.e., there exists an index $N$ such that $d_k = d_N$ for all $k \ge N$. Thus $X_k = X_N$ for all $k \ge N$ and consequently $X_k' = X_N'$ for all $k \ge N$. It follows that the DCC is satisfied for $\Bbb k[x]/J$, so it is Artinian.

Note that if we relax the condition that $\Bbb k$ is a field, then $\Bbb k[x]/J$ may not be Artinian. For instance, the factor ring $\Bbb Z[x]/(x)$, being isomorphic to the non-Artinian ring $\Bbb Z$, is non-Artinian.

Here are two exercises for you to try.

1. Show that if a left $R$-module $M$ is both Artinian and Noetherian, then it has finite length.

2. Show that if $A$ is left (unital) Artinian ring, then whenever $x,y \in A$, $xy = 1$ implies $yx = 1$.

In the above post, Euge set the following exercise:

" ... ... Show that if a left $R$-module $M$ is both Artinian and Noetherian, then it has finite length. ... ..."

I needed some help to get started on this exercise, so I consulted T.S. Blyth's book: Module Theory: An Approach to Linear Algebra.

In Blyth Chapter 5: Chain Conditions: Jordan-Holder Towers, we find Theorem 5.9 which more than covers the exercise set by Euge ...

[Note that Blyth in Theorem 5.9 uses the term "Jordan-Holder Towers" for what most authors refer to as "Composition Series" ]

Theorem 5.9 reads as follows:

View attachment 3344
View attachment 3345To see how Blyth answers Euge's exercise - that is, demonstrates that "if a left $R$-module $M$ is both Artinian and Noetherian, then it has finite length" ... ... we start reading Blyth's proof at the point where Blyth writes:

"Conversely, suppose that $$M$$ satisfies both chain conditions. Let $$C$$ be the collection ... ... "

I have some questions related to the proof that follows this text ... ...Question 1

At the start of the above proof we read:

" ... ... Let $$C$$ be the collection of all submodules of $$M$$ that have Jordan-Holder towers. Then $$ C \neq \phi $$ since every descending chain of non-zero submodules

$$M_0 \supset M_1 \supset M_2 \supset ... ... $$

terminates at $$M_p$$ say which must be simple and so has a Jordan-Holder tower of height $$1$$. ... ..."


So ... thinking about the above statement ... if we consider a descending chain of submodules:

$$M_0 \supset M_1 \supset M_2 \supset ... ... \supset M_i \supset M_{i+1} \supset ...
$$

... then to meet the descending chain condition it must terminate in some module $$M_p$$ ... ... which means that there is a natural number $$n$$ such that for $$k \gt n $$, $$M_k = M_p$$ ... ...

BUT why does Blyth conclude that $$M_p$$ is simple? ... ... and, further, why does this mean that $$M_p$$ has a Jordan-Holder tower of height $$1$$.



Question 2

In the above text from Blyth we read:

"We now note that $$C$$ has a maximal element, $$M^*$$ say; for otherwise the descending chain would be violated. ... ..."

Now the maximum condition is for Noetherian rings which satisfy the ascending chain condition (and not necessarily the descending chain condition) ... ... so shouldn't Blyth be dealing with a minimal element and saying:

"We now note that $$C$$ has a minimal element, $$M^*$$ say; for otherwise the descending chain would be violated. ... ..."

OR having $$M^*$$ as a maximal element and saying:

"We now note that $$C$$ has a maximal element, $$M^*$$ say; for otherwise the ascending chain would be violated. ... ..."Can someone please clarify this situation?***NOTE***

Blyth defines Noetherian and Artinian rings together with the maximum and minimum conditions on pages 45-46, so I am providing these pages below in order to ensure that MHB members reading this post have enough information and context to follow the post.View attachment 3346

View attachment 3347

 

[Euge]

In the above post, Euge set the following exercise:

" ... ... Show that if a left $R$-module $M$ is both Artinian and Noetherian, then it has finite length. ... ..."

I needed some help to get started on this exercise, so I consulted T.S. Blyth's book: Module Theory: An Approach to Linear Algebra.

In Blyth Chapter 5: Chain Conditions: Jordan-Holder Towers, we find Theorem 5.9 which more than covers the exercise set by Euge ...

[Note that Blyth in Theorem 5.9 uses the term "Jordan-Holder Towers" for what most authors refer to as "Composition Series" ]

Theorem 5.9 reads as follows:

View attachment 3344
View attachment 3345To see how Blyth answers Euge's exercise - that is, demonstrates that "if a left $R$-module $M$ is both Artinian and Noetherian, then it has finite length" ... ... we start reading Blyth's proof at the point where Blyth writes:

"Conversely, suppose that $$M$$ satisfies both chain conditions. Let $$C$$ be the collection ... ... "

I have some questions related to the proof that follows this text ... ...Question 1

At the start of the above proof we read:

" ... ... Let $$C$$ be the collection of all submodules of $$M$$ that have Jordan-Holder towers. Then $$ C \neq \phi $$ since every descending chain of non-zero submodules

$$M_0 \supset M_1 \supset M_2 \supset ... ... $$

terminates at $$M_p$$ say which must be simple and so has a Jordan-Holder tower of height $$1$$. ... ..."


So ... thinking about the above statement ... if we consider a descending chain of submodules:

$$M_0 \supset M_1 \supset M_2 \supset ... ... \supset M_i \supset M_{i+1} \supset ...
$$

... then to meet the descending chain condition it must terminate in some module $$M_p$$ ... ... which means that there is a natural number $$n$$ such that for $$k \gt n $$, $$M_k = M_p$$ ... ...

BUT why does Blyth conclude that $$M_p$$ is simple? ... ... and, further, why does this mean that $$M_p$$ has a Jordan-Holder tower of height $$1$$.



Question 2

In the above text from Blyth we read:

"We now note that $$C$$ has a maximal element, $$M^*$$ say; for otherwise the descending chain would be violated. ... ..."

Now the maximum condition is for Noetherian rings which satisfy the ascending chain condition (and not necessarily the descending chain condition) ... ... so shouldn't Blyth be dealing with a minimal element and saying:

"We now note that $$C$$ has a minimal element, $$M^*$$ say; for otherwise the descending chain would be violated. ... ..."

OR having $$M^*$$ as a maximal element and saying:

"We now note that $$C$$ has a maximal element, $$M^*$$ say; for otherwise the ascending chain would be violated. ... ..."Can someone please clarify this situation?***NOTE***

Blyth defines Noetherian and Artinian rings together with the maximum and minimum conditions on pages 45-46, so I am providing these pages below in order to ensure that MHB members reading this post have enough information and context to follow the post.View attachment 3346

View attachment 3347

It looks like you're getting more confused, so let me give you a hint for the first exercise. Since $M$ is Artinian, the collection of nonzero submodules of $M$ has a minimal element, $N_1$. Consider the collection of submodules of $M$ which strictly contain $N_1$. Since $M$ is Artinian, it has a minimal element, $N_2$. Continue the process to construct an ascending chain $N_1 \subset N_2 \subset N_3 \subset \cdots$, and use the Noetherian property of $M$ to deduce that this chain stabilizes. Think of how you can use part of this chain to construct a composition series for $M$ of finite length.

 

[Math Amateur]

It looks like you're getting more confused, so let me give you a hint for the first exercise. Since $M$ is Artinian, the collection of nonzero submodules of $M$ has a minimal element, $N_1$. Consider the collection of submodules of $M$ which strictly contain $N_1$. Since $M$ is Artinian, it has a minimal element, $N_2$. Continue the process to construct an ascending chain $N_1 \subset N_2 \subset N_3 \subset \cdots$, and use the Noetherian property of $M$ to deduce that this chain stabilizes. Think of how you can use part of this chain to construct a composition series for $M$ of finite length.

Thanks Euge ... I will work on that ...

Are you also able to clarify (briefly anyway) the two sentences from Blyth ... I think it would increase my understanding of Artinian and Noetherian rings ... but anyway will work on your suggestion ... ...

Thanks again,

Peter***EDIT***

Referring to my post above, I must say that I have a reasonable (but slightly vague) idea why we can say that:

" ... ... every descending chain of non-zero submodules

$$M_0 \supset M_1 \supset M_2 \supset ... ... $$

terminates at $$M_p$$ say which must be simple and so has a Jordan-Holder tower of height $$1$$. ... ..."

BUT ... ... I am completely puzzled by the statement:

"We now note that $$C$$ has a maximal element, $$M^*$$ say; for otherwise the descending chain would be violated. ... ..."

 

[Euge]

Thanks Euge ... I will work on that ...

Are you also able to clarify (briefly anyway) the two sentences from Blyth ... I think it would increase my understanding of Artinian and Noetherian rings ... but anyway will work on your suggestion ... ...

Thanks again,

Peter***EDIT***

Referring to my post above, I must say that I have a reasonable (but slightly vague) idea why we can say that:

" ... ... every descending chain of non-zero submodules

$$M_0 \supset M_1 \supset M_2 \supset ... ... $$

terminates at $$M_p$$ say which must be simple and so has a Jordan-Holder tower of height $$1$$. ... ..."

BUT ... ... I am completely puzzled by the statement:

"We now note that $$C$$ has a maximal element, $$M^*$$ say; for otherwise the descending chain would be violated. ... ..."

For your first question, $M_p$ is simple because it doesn't properly contain any smaller nonzero submodules (remember the chain of $M$'s stabilizes at $M_p$). Since $M_p$ is simple, the chain $M_p \supset (0)$ is a composition series for $M_p$. This means that $M_p$ has height one.

To answer your second question, if $C$ has no maximal element, then you can construct a descending chain of submodules that does not stabilize. This violates the descending chain condition.

 

[Deveno]

Can someone please show me exactly how it is that a ring that is Artinian when considered as a right module over itself necessarily has the property that every descending chain of right ideals in R must terminate or be finite?

I would appreciate some help in this matter ...

... ... now back to the exercises set by Euge ... :)

Peter

Did you know that, viewing a ring R as a right module over itself, the R-submodules are the right ideals of R? Using this fact, the DCC for right Artinian modules translates to the DCC for right Artinian rings, which is displayed in your text.

Just for future reference, this is one of the reasons for working with modules, rather than rings.

Abelian groups are "nice", any subgroup is normal (every inner automorphism is trivial), and any quotient group is the original group modulo some subgroup.

In other words: kernel = subgroup.

Rings are "not so nice", not every quotient ring is the original ring modulo a sub-ring. In rings we have:

Kernel = ideal $\neq$ subring.

Fact: For a ring $R$, a left $R$-submodule of $R$ is the same thing as a left ideal $I$ of $R$.

For suppose $I$ is a left ideal of $R$. We surely have closure under addition, since $(I,+)$ is subgroup of $(R,+)$.

Note that for $a \in R$, and $x \in I$, we simply take $a\cdot x = ax$, where the RHS is just the ring multiplication. Since $I$ is a left ideal, it is closed under the $R$-action so defined.

On the other hand, suppose $J$ is a (left) submodule of $R$. Since $J$ is closed under scalar multiplication, for any $a \in R$, and $x \in J$ we have:

$a\cdot x = ax \in J$ (by closure), so $J$ is a left ideal.

Often, one limits one's attention to COMMUTATIVE rings, where left/right doesn't make any difference.

The general idea, here, is to understand ideals, it makes more sense to look at modules, where ideals become the "sub-things" they ought to be.

 

[Math Amateur]

Just for future reference, this is one of the reasons for working with modules, rather than rings.

Abelian groups are "nice", any subgroup is normal (every inner automorphism is trivial), and any quotient group is the original group modulo some subgroup.

In other words: kernel = subgroup.

Rings are "not so nice", not every quotient ring is the original ring modulo a sub-ring. In rings we have:

Kernel = ideal $\neq$ subring.

Fact: For a ring $R$, a left $R$-submodule of $R$ is the same thing as a left ideal $I$ of $R$.

For suppose $I$ is a left ideal of $R$. We surely have closure under addition, since $(I,+)$ is subgroup of $(R,+)$.

Note that for $a \in R$, and $x \in I$, we simply take $a\cdot x = ax$, where the RHS is just the ring multiplication. Since $I$ is a left ideal, it is closed under the $R$-action so defined.

On the other hand, suppose $J$ is a (left) submodule of $R$. Since $J$ is closed under scalar multiplication, for any $a \in R$, and $x \in J$ we have:

$a\cdot x = ax \in J$ (by closure), so $J$ is a left ideal.

Often, one limits one's attention to COMMUTATIVE rings, where left/right doesn't make any difference.

The general idea, here, is to understand ideals, it makes more sense to look at modules, where ideals become the "sub-things" they ought to be.

Hi Deveno,

Thanks for an extremely helpful and timely post ... ...

Peter

 

[Math Amateur]

It looks like you're getting more confused, so let me give you a hint for the first exercise. Since $M$ is Artinian, the collection of nonzero submodules of $M$ has a minimal element, $N_1$. Consider the collection of submodules of $M$ which strictly contain $N_1$. Since $M$ is Artinian, it has a minimal element, $N_2$. Continue the process to construct an ascending chain $N_1 \subset N_2 \subset N_3 \subset \cdots$, and use the Noetherian property of $M$ to deduce that this chain stabilizes. Think of how you can use part of this chain to construct a composition series for $M$ of finite length.

Hi Euge ... thanks for your help so far but I need further help ...

I follow your (very clear!) argument so far ...

So we have a chain:

$N_1 \subset N_2 \subset N_3 \subset \cdots$ that stabilizes ... that is there is a natural number $$n$$ such that for all $$k \geq n$$ we have $$N_k = N_n$$ and so we have the stabilized chain

$N_1 \subset N_2 \subset N_3 \subset \cdots \subset N_k$ ... ... ... (1)

Now for (1) to be a composition series we need

(a) $$N_k = M$$

(b) each 'factor' $$N_i/N_{i-1}$$ is simple for $$i \leq k$$

BUT ... I cannot see how to do this ... can you help?

Peter***EDITS***


*EDIT 1*

To show (1) to be a composition series do we also need to show that $$N_1 = \{ 0 \}$$?



*EDIT 2*

Presumably to show each factor $$N_i/N_{i-1}$$ is simple we assume that there exists $$L$$ such that $$ \{ 0 \} \subset L \subset N_i/N_{i-1}$$ and then derive a contradiction

OR

assume that there exists $$L$$ such that $${0} \subseteq L \subseteq N_i/N_{i-1}$$ and then show that $$L$$ must equal $$\{ 0 \}$$ or $$N_i/N_{i-1}$$

Is that correct?

 

[Euge]

Hi Euge ... thanks for your help so far but I need further help ...

I follow your (very clear!) argument so far ...

So we have a chain:

$N_1 \subset N_2 \subset N_3 \subset \cdots$ that stabilizes ... that is there is a natural number $$n$$ such that for all $$k \geq n$$ we have $$N_k = N_n$$ and so we have the stabilized chain

$N_1 \subset N_2 \subset N_3 \subset \cdots \subset N_k$ ... ... ... (1)

Now for (1) to be a composition series we need

(a) $$N_k = M$$

(b) each 'factor' $$N_i/N_{i-1}$$ is simple for $$i \leq k$$

BUT ... I cannot see how to do this ... can you help?

Peter***EDITS***


*EDIT 1*

To show (1) to be a composition series do we also need to show that $$N_1 = \{ 0 \}$$?



*EDIT 2*

Presumably to show each factor $$N_i/N_{i-1}$$ is simple we assume that there exists $$L$$ such that $$ \{ 0 \} \subset L \subset N_i/N_{i-1}$$ and then derive a contradiction

OR

assume that there exists $$L$$ such that $${0} \subseteq L \subseteq N_i/N_{i-1}$$ and then show that $$L$$ must equal $$\{ 0 \}$$ or $$N_i/N_{i-1}$$

Is that correct?

You can't show $N_1 = \{0\}$ because $N_1$ is nonzero by construction. I suggest not using the chain $N_1 \subset N_2 \subset \cdots \subset N_k$ from the start. First argue that $N_r = M$ for some $r$. This can done indirectly. If $N_i \neq M$ for all $i$, then by construction of the $N_i$, the $N$-chain would be infinite, violating the ACC for Artinian modules. Therefore, we have a finite chain

$$(*) \quad \{0\} = N_0 \subset N_1 \subset N_2 \subset \cdots \subset N_r = M$$

By minimality of the $N_i$, $i = 1, 2, \dots, r$, $N_{i-1}$ is a maximal submodule of $N_i$ for $i = 2,\ldots r$. Therefore, $N_i/N_{i-1}$ is simple for $i = 2,\ldots, r$. Since $N_1$ is minimal in the collection of all nonzero submodules of $M$, $N_1$ contains no other nonzero submodule. Hence $N_1$ is simple and consequently, $(*)$ is a composition series of $M$ of length $r$.

 

[Math Amateur]

You can't show $N_1 = \{0\}$ because $N_1$ is nonzero by construction. I suggest not using the chain $N_1 \subset N_2 \subset \cdots \subset N_k$ from the start. First argue that $N_r = M$ for some $r$. This can done indirectly. If $N_i \neq M$ for all $i$, then by construction of the $N_i$, the $N$-chain would be infinite, violating the ACC for Artinian modules. Therefore, we have a finite chain

$$(*) \quad \{0\} = N_0 \subset N_1 \subset N_2 \subset \cdots \subset N_r = M$$

By minimality of the $N_i$, $i = 1, 2, \dots, r$, $N_{i-1}$ is a maximal submodule of $N_i$ for $i = 2,\ldots r$. Therefore, $N_i/N_{i-1}$ is simple for $i = 2,\ldots, r$. Since $N_1$ is minimal in the collection of all nonzero submodules of $M$, $N_1$ contains no other nonzero submodule. Hence $N_1$ is simple and consequently, $(*)$ is a composition series of $M$ of length $r$.

Thanks for all your help in this matter, Euge ...

By the way, you write:

"You can't show $N_1 = \{0\}$ because $N_1$ is nonzero by construction."

Indeed, how careless/silly of me ... thanks for the correction ...

Peter