Understanding Bland's Proposition 4.2.10 in Rings and Modules

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 4.2: Noetherian and Artinian Modules and need some further help to fully understand the proof of part of Proposition 4.2.10 ... ...

Proposition 4.2.10 reads as follows:View attachment 8215In the above proof by Bland we read the following:

" ... ... Continuing in this way we obtain an ascending chain \(\displaystyle Y \subseteq Y \oplus Y' \subseteq Y \oplus Y' \oplus Y'' \subseteq\) ... ... "Can someone please explain in some detail exactly how/why \(\displaystyle Y \subseteq Y \oplus Y'\) and \(\displaystyle Y \oplus Y' \subseteq Y \oplus Y' \oplus Y''\) etc ...
Help will be appreciated ...

Peter

=========================================================================Definition 4.2.9 is relevant to the above post so I am providing the text of Definition 4.2.9 ... as follows ... View attachment 8216=========================================================================

 

[steenis]

It is supposed that $M$ is decomposable but cannot be written as finite direct sum

$$M = M_1 \oplus M_2 \oplus \cdots \oplus M_n$$

of indecomposable submodules $M_i \leq M$

$M$ is decomposable so $M$ can be broken into two submodules, say $M = X \oplus Y$.
This is a finite direct sum (by hypothesis $M$ cannot be a finite sum of indecomposable submodules), so one of the submodules is not indecomposable, say $X$ is not indecomposable. Thus $X$ can be broken into two submodules say $X = X’ \oplus Y’$. Thus $M = X’ \oplus Y’ \oplus Y$.

Again this is a finite direct sum, so one of the submodules is not indecomposable, say $X’$ is not indecomposable. Thus $X’$ can be broken into two submodules say $X’ = X’’ \oplus Y’’$. Thus $M = X’’ \oplus Y’’ \oplus Y’ \oplus Y$.

Again this is a finite direct sum, so one of the submodules is not indecomposable, say $X’’$ is not indecomposable. Thus $X’’$ can be broken into two submodules say $X’’ = X’’’ \oplus Y’’’$. Thus $M = X’’’ \oplus Y’’’ \oplus Y’’ \oplus Y’ \oplus Y$.
And so on $\cdots$
The sequence starts with breaking $M$ into two submodules,

Each time the “$X$”-module is supposed to be not indecomposable:

$M = X \oplus Y$, then $X = X’ \oplus Y’$, so

$M = X’ \oplus Y’ \oplus Y$, then $X’ = X’’ \oplus Y’’$, so

$M = X’’ \oplus Y’’ \oplus Y’ \oplus Y$, then $X’’ = X’’’ \oplus Y’’’$, so

$M = X’’’ \oplus Y’’’ \oplus Y’’ \oplus Y’ \oplus Y$ and so on $\cdots$
Of course $Y \subset Y \oplus Y’ \subset M$

and $Y \oplus Y’ \subset Y \oplus Y’ \oplus Y’’ \subset M$

and $Y \oplus Y’ \oplus Y’’ \subset Y \oplus Y’ \oplus Y’’ \oplus Y’’’ \subset M$

and so on, thus

$ Y \subset Y \oplus Y’ \subset Y \oplus Y’ \oplus Y’’ \subset Y \oplus Y’ \oplus Y’’ \oplus Y’’’ \cdots $

 

[Math Amateur]

It is supposed that $M$ is decomposable but cannot be written as finite direct sum

$$M = M_1 \oplus M_2 \oplus \cdots \oplus M_n$$

of indecomposable submodules $M_i \leq M$

$M$ is decomposable so $M$ can be broken into two submodules, say $M = X \oplus Y$.
This is a finite direct sum (by hypothesis $M$ cannot be a finite sum of indecomposable submodules), so one of the submodules is not indecomposable, say $X$ is not indecomposable. Thus $X$ can be broken into two submodules say $X = X’ \oplus Y’$. Thus $M = X’ \oplus Y’ \oplus Y$.

Again this is a finite direct sum, so one of the submodules is not indecomposable, say $X’$ is not indecomposable. Thus $X’$ can be broken into two submodules say $X’ = X’’ \oplus Y’’$. Thus $M = X’’ \oplus Y’’ \oplus Y’ \oplus Y$.

Again this is a finite direct sum, so one of the submodules is not indecomposable, say $X’’$ is not indecomposable. Thus $X’’$ can be broken into two submodules say $X’’ = X’’’ \oplus Y’’’$. Thus $M = X’’’ \oplus Y’’’ \oplus Y’’ \oplus Y’ \oplus Y$.
And so on $\cdots$
The sequence starts with breaking $M$ into two submodules,

Each time the “$X$”-module is supposed to be not indecomposable:

$M = X \oplus Y$, then $X = X’ \oplus Y’$, so

$M = X’ \oplus Y’ \oplus Y$, then $X’ = X’’ \oplus Y’’$, so

$M = X’’ \oplus Y’’ \oplus Y’ \oplus Y$, then $X’’ = X’’’ \oplus Y’’’$, so

$M = X’’’ \oplus Y’’’ \oplus Y’’ \oplus Y’ \oplus Y$ and so on $\cdots$
Of course $Y \subset Y \oplus Y’ \subset M$

and $Y \oplus Y’ \subset Y \oplus Y’ \oplus Y’’ \subset M$

and $Y \oplus Y’ \oplus Y’’ \subset Y \oplus Y’ \oplus Y’’ \oplus Y’’’ \subset M$

and so on, thus

$ Y \subset Y \oplus Y’ \subset Y \oplus Y’ \oplus Y’’ \subset Y \oplus Y’ \oplus Y’’ \oplus Y’’’ \cdots $

Thanks for the help, Steenis ...

Will be working through your post in detail shortly ...

Peter

 

[Math Amateur]

It is supposed that $M$ is decomposable but cannot be written as finite direct sum

$$M = M_1 \oplus M_2 \oplus \cdots \oplus M_n$$

of indecomposable submodules $M_i \leq M$

$M$ is decomposable so $M$ can be broken into two submodules, say $M = X \oplus Y$.
This is a finite direct sum (by hypothesis $M$ cannot be a finite sum of indecomposable submodules), so one of the submodules is not indecomposable, say $X$ is not indecomposable. Thus $X$ can be broken into two submodules say $X = X’ \oplus Y’$. Thus $M = X’ \oplus Y’ \oplus Y$.

Again this is a finite direct sum, so one of the submodules is not indecomposable, say $X’$ is not indecomposable. Thus $X’$ can be broken into two submodules say $X’ = X’’ \oplus Y’’$. Thus $M = X’’ \oplus Y’’ \oplus Y’ \oplus Y$.

Again this is a finite direct sum, so one of the submodules is not indecomposable, say $X’’$ is not indecomposable. Thus $X’’$ can be broken into two submodules say $X’’ = X’’’ \oplus Y’’’$. Thus $M = X’’’ \oplus Y’’’ \oplus Y’’ \oplus Y’ \oplus Y$.
And so on $\cdots$
The sequence starts with breaking $M$ into two submodules,

Each time the “$X$”-module is supposed to be not indecomposable:

$M = X \oplus Y$, then $X = X’ \oplus Y’$, so

$M = X’ \oplus Y’ \oplus Y$, then $X’ = X’’ \oplus Y’’$, so

$M = X’’ \oplus Y’’ \oplus Y’ \oplus Y$, then $X’’ = X’’’ \oplus Y’’’$, so

$M = X’’’ \oplus Y’’’ \oplus Y’’ \oplus Y’ \oplus Y$ and so on $\cdots$
Of course $Y \subset Y \oplus Y’ \subset M$

and $Y \oplus Y’ \subset Y \oplus Y’ \oplus Y’’ \subset M$

and $Y \oplus Y’ \oplus Y’’ \subset Y \oplus Y’ \oplus Y’’ \oplus Y’’’ \subset M$

and so on, thus

$ Y \subset Y \oplus Y’ \subset Y \oplus Y’ \oplus Y’’ \subset Y \oplus Y’ \oplus Y’’ \oplus Y’’’ \cdots $

Thanks steenis ... I think I have understood all that you wrote ...
An explicit proof of \(\displaystyle Y \subseteq Y \oplus Y'\) follows ...
\(\displaystyle Y = \{ y \ \mid y \ \in Y \}\) \(\displaystyle Y \oplus Y' = \{ y + y' \ \mid \ y \in Y , y' \in Y' \} \)
Show \(\displaystyle Y \subseteq Y \oplus Y' \)
\(\displaystyle y_1 \in Y \)

\(\displaystyle \Longrightarrow y_1 + 0_{ Y' } \in Y \oplus Y' \)

\(\displaystyle \Longrightarrow y_1 \in Y \oplus Y'\)

\(\displaystyle \Longrightarrow Y \subseteq Y \oplus Y'\)------------------------------------------------------------------------------------------------------------------

*** What confused me was thinking that since we are dealing with a finite direct sum we have

\(\displaystyle Y \oplus Y' \cong Y \times Y'\)

and viewed in this way \(\displaystyle Y \oplus Y' = \{ (y, y') \ \mid \ y \in Y , y' \in Y' \}\)

... and then it is difficult to see how \(\displaystyle Y \subseteq Y \oplus Y'\)

... but I guess the answer is to realize that \(\displaystyle Y \times Y' \cong Y + Y'\) ... (Dummit and Foote, page 353)

-------------------------------------------------------------------------------------------------------------------
Now (I think) that Theorem 4.2.10 assumes that \(\displaystyle Y \oplus Y'\) is a submodule where \(\displaystyle Y, Y'\) are both submodules of M ...So ... we need to demonstrate that \(\displaystyle Y \oplus Y'\) is a submodule where \(\displaystyle Y, Y'\) are both submodules of \(\displaystyle M\) ...So ... let \(\displaystyle x, y \in Y \oplus Y'\) and \(\displaystyle a \in R\) ...

Then ... \(\displaystyle x = x_1 + x_2\) where \(\displaystyle x_1 \in Y\) and \(\displaystyle x_2 \in Y'\) ... ...

... and ... \(\displaystyle y = y_1 + y_2\) where \(\displaystyle y_1 \in Y\) and \(\displaystyle y_2 \in Y'\) ... ...

... then \(\displaystyle x + y = ( x_1 + x_2 ) + ( y_1 + y_2 )\)

\(\displaystyle \Longrightarrow x + y = ( x_1 + y_1 ) + ( x_2 + y_2 )\)

But \(\displaystyle ( x_1 + y_1 ) + ( x_2 + y_2 ) \in Y \oplus Y'\) ... since both \(\displaystyle Y\) and \(\displaystyle Y'\) are submodules ...

\(\displaystyle \Longrightarrow x + y \in Y \oplus Y'\)

... and ...

\(\displaystyle xa = ( x_1 + x_2 ) a = x_1 a + x_2 a \in Y \oplus Y'\) ... since both \(\displaystyle Y\) and \(\displaystyle Y'\) are submodules ... Since \(\displaystyle x+ y\) and \(\displaystyle xa\) \(\displaystyle \in Y \oplus Y'\) we have that \(\displaystyle Y \oplus Y'\) is a submodule of \(\displaystyle M\) ...
Is that correct ...?

Peter

 

[steenis]

The first part is ok.

What you have proven in the second part is ok, but has already been proven in proposition 1.4.4.

We are dealing with internal direct sums here, so we can use the "sum"-notation.

If $Y$ and $Z$ are modules, then we can say that $Y$ is a submodule of the external direct sum $Y \times Z$ if we consider $Y$ as $Y \times 0$, in fact $Y \cong Y \times 0$ and $Y \times 0$ is a submodule of $Y \times Z$.