MHB Generation of Modules .... Dummit and Foote, Section 10.3 .... ....

[Math Amateur]

I am reading Dummit and Foote's book: "Abstract Algebra" (Third Edition) ...

I am currently studying Chapter 10: Introduction to Module Theory ... ...

I need some help with an aspect of Dummit and Foote's Section 10.3 Basic Generation of Modules, Direct Sums and Free Modules ... ...

The start of Section 10.3 reads as follows:View attachment 8224
View attachment 8225
In the above text from Dummit and Foote we read the following:

" ... ... for submodules $$N_1, \ ... \ ... \ , N_n$$ of $$M$$, $$N_1 + \ ... \ ... \ + N_n$$ is just the submodule generated by the set $$N_1 \cup \ ... \ ... \ \cup N_n$$ ... ... "I tried to prove the above ... but despite the relationship seeming intuitive, I found it difficult to formulate a rigorous proof ...
My attempt at a proof is as follows ...
Firstly we note that $$RN$$ where $$N = \cup_{ s = 1 }^n N_s $$ is as follows:$$RN = \{ r_1 n_1 + r_2 n_2 + \ ... \ ... \ + r_p n_p \mid r_s \in R, n_s \in N , p \in \mathbb{N} \}$$
To show $$\sum_{ s = 1 }^n N_s = RN$$ ... ... ... ... ... (1)
Firstly show that $$\sum_{ s = 1 }^n N_s \subseteq RN$$ ... ... ... ... ... (i)
Now ... $$x \in \sum_{ s = 1 }^n N_s \Longrightarrow x = x_1 + x_2 + \ ... \ ... \ + x_n$$ where $$x_s \in N_s$$But ... each $$x_s \in N_s \subseteq M \Longrightarrow x_s = r'_s m'_s$$$$\Longrightarrow x = r'_1 m'_1 + r'_2 m'_2 + \ ... \ ... \ + r'_n m'_n$$ ... where $$r'_s \in R$$ and $$m'_s \in N_s$$ ...$$\Longrightarrow x \in RN$$ ...Is that correct?
Now ... to show that $$RN \subseteq \sum_{ s = 1 }^n N_s$$$$x \in RN$$ $$\Longrightarrow x = r_1 n_1 + r_2 n_2 + \ ... \ ... \ + r_p n_p $$ where $$r_s \in R$$, $$n_s \in N = \cup_{ s = 1 }^n N_s \text{ and } p \in \mathbb{N} $$... BUT! ... the $$n_s$$ do not necessarily belong to $$N_s$$ ... ?How do we proceed ...?Can someone please help ...Help will be much appreciated ...

Peter

 

[steenis]

No, I think the above is not correct.

If $Y$ is generated by $X$, notation $Y =\langle X \rangle$ iff for every $y \in Y$: $y = \Sigma_{ j = 1 }^{ m } r_j x_j $, where $r_j \in R$, $x_j \in X$, i.e.,

$Y =\langle X \rangle = \{\Sigma_{ j = 1 }^{ m } r_j x_j \text{ } | \text{ } x_j \in X, r_j \in R \}$

Realize that

$\langle N_1 \cup \cdots \cup N_n \rangle = \{ \Sigma_{ j = 1 }^{ m } x_j \text{ } | \text{ } x_j \in N_1 \cup \cdots \cup N_n \}$, $N_i$ are submodules of $M$.

Now it is easy to prove that $N_1 + \cdots + N_n = \langle N_1 \cup \cdots \cup N_n \rangle$

 

[Math Amateur]

I am reading Dummit and Foote's book: "Abstract Algebra" (Third Edition) ...

I am currently studying Chapter 10: Introduction to Module Theory ... ...

I need some help with an aspect of Dummit and Foote's Section 10.3 Basic Generation of Modules, Direct Sums and Free Modules ... ...

The start of Section 10.3 reads as follows:

In the above text from Dummit and Foote we read the following:

" ... ... for submodules $$N_1, \ ... \ ... \ , N_n$$ of $$M$$, $$N_1 + \ ... \ ... \ + N_n$$ is just the submodule generated by the set $$N_1 \cup \ ... \ ... \ \cup N_n$$ ... ... "I tried to prove the above ... but despite the relationship seeming intuitive, I found it difficult to formulate a rigorous proof ...
My attempt at a proof is as follows ...
Firstly we note that $$RN$$ where $$N = \cup_{ s = 1 }^n N_s $$ is as follows:$$RN = \{ r_1 n_1 + r_2 n_2 + \ ... \ ... \ + r_p n_p \mid r_s \in R, n_s \in N , p \in \mathbb{N} \}$$
To show $$\sum_{ s = 1 }^n N_s = RN$$ ... ... ... ... ... (1)
Firstly show that $$\sum_{ s = 1 }^n N_s \subseteq RN$$ ... ... ... ... ... (i)
Now ... $$x \in \sum_{ s = 1 }^n N_s \Longrightarrow x = x_1 + x_2 + \ ... \ ... \ + x_n$$ where $$x_s \in N_s$$But ... each $$x_s \in N_s \subseteq M \Longrightarrow x_s = r'_s m'_s$$$$\Longrightarrow x = r'_1 m'_1 + r'_2 m'_2 + \ ... \ ... \ + r'_n m'_n$$ ... where $$r'_s \in R$$ and $$m'_s \in N_s$$ ...$$\Longrightarrow x \in RN$$ ...Is that correct?
Now ... to show that $$RN \subseteq \sum_{ s = 1 }^n N_s$$$$x \in RN$$ $$\Longrightarrow x = r_1 n_1 + r_2 n_2 + \ ... \ ... \ + r_p n_p $$ where $$r_s \in R$$, $$n_s \in N = \cup_{ s = 1 }^n N_s \text{ and } p \in \mathbb{N} $$... BUT! ... the $$n_s$$ do not necessarily belong to $$N_s$$ ... ?How do we proceed ...?Can someone please help ...Help will be much appreciated ...

Peter

========================================================================Since I could not see any specific errors, I have completed the proof ... please note, though, that I am not stubbornly insisting that my proof is correct ... indeed you are usually correct! ... but ... it is just that after examining my proof I cannot see any specific errors ... so I completed the proof ... but am most interested in its validity ... so would welcome specific errors or shortcomings to be pointed out ...
My proof reads as follows ...
Firstly we note that $$RN$$ where $$N = \cup_{ s = 1 }^n N_s $$ is as follows:$$RN = \{ r_1 n_1 + r_2 n_2 + \ ... \ ... \ + r_p n_p \mid r_s \in R, n_s \in N , p \in \mathbb{N} \}$$
To show $$\sum_{ s = 1 }^n N_s = RN$$ ... ... ... ... ... (1)
Firstly show that $$\sum_{ s = 1 }^n N_s \subseteq RN$$ ... ... ... ... ... (i)
Now ... $$x \in \sum_{ s = 1 }^n N_s \Longrightarrow x = x_1 + x_2 + \ ... \ ... \ + x_n$$ where $$x_s \in N_s$$But ... each $$x_s \in N_s \subseteq M \Longrightarrow x_s = r'_s m'_s$$$$\Longrightarrow x = r'_1 m'_1 + r'_2 m'_2 + \ ... \ ... \ + r'_n m'_n$$ ... where $$r'_s \in R$$ and $$m'_s \in N_s$$ ...$$\Longrightarrow x \in RN$$ ... ...
Now ... to show that $$RN \subseteq \sum_{ s = 1 }^n N_s$$ ... ... ... (ii)


$$x \in RN$$ $$\Longrightarrow x = r_1 n_1 + r_2 n_2 + \ ... \ ... \ + r_p n_p $$ where $$r_s \in R$$, $$n_s \in N = \cup_{ s = 1 }^n N_s \text{ and } p \in \mathbb{N} $$ ... ... ... ... (2)Now ... for each $$i$$ we have $$r_i n_i \in N_j$$ for some (one or more) $$j$$ since $$n_i \in N = \cup_ { s =1}^n N_s$$ and $$N_j$$ is a submodule ...

If $$r_i n_i$$ belongs to several $$N_j$$ then allocate or assign $$r_i n_i$$ to the lowest index that is not already allocated ...

If $$r_m n_m$$ and $$r_k n_k$$ both belong to $$N_j$$ then create a new element $$r_q n_q = r_m n_m + r_k n_k \in N_j$$ since $$N_j$$ is a submodule ...

Proceed as above until all elements in (2) are allocated ... and re-label $$r_i n_i$$ as $$x_j$$ ...

Then ... possibly after some re-numbering ... we can write the following ...

$$x = x_1 + x_2 + \ ... \ ... \ + x_r + 0 + 0 + \ ... \ ... \ + 0$$ ... ($$n$$ elements ... ) ... ... where $$x_i \in N_i$$ and the number of zeros is $$z$$ where $$1 \lt z \lt n-1$$

Therefore $$x \in \sum_{ s = 1 }^n N_s$$ ...
Is that correct ... ?

If it is not correct can you please point out specific errors ...

If it is correct ... do you have a less messy proof ...

Peter

 

[steenis]

You have to prove that: $N_1 + \cdots + N_n = \langle N_1 \cup \cdots \cup N_n \rangle$

You defined $N = N_1 \cup \cdots \cup N_n$

You defined $RN$ in such a way that it coincides with $\langle N_1 \cup \cdots \cup N_n \rangle$

So you have to prove that: $N_1 + \cdots + N_n = RN$

If $x \in \sum_{ s = 1 }^{n} N_s$ then $x = x_1 + x_2 + \ ... \ ... \ + x_n$ for $x_i \in N_i$

Thus $x = 1x_1 + 1x_2 + \ ... \ ... \ + 1x_n$, with $x_i \in N$, so $x$ belongs to $RN$ by definition.

So the first part of your proof is ok.

To be honest, I cannot follow the second part. I think it can be done much easier.

Take $x \in RN$ then $x = \sum_{j = 1 }^{p} r_j x_j$ for some $p \in \mathbb{N}$, where $r_j \in R$, $x_j \in N$.

Since $N = N_1 \cup \cdots \cup N_n$. for each $j = 1, \cdots, p$ there is a $i \in \{1, \cdots, n\}$ such that $x_j \in N_i$, so also $r_j x_j \in N_i$

Therefore
$x = \sum_{j = 1 }^{p} r_j x_j \in N_1 + \cdots + N_n$

I think this is the second part but a little bit easier.

 

[Math Amateur]

You have to prove that: $N_1 + \cdots + N_n = \langle N_1 \cup \cdots \cup N_n \rangle$

You defined $N = N_1 \cup \cdots \cup N_n$

You defined $RN$ in such a way that it coincides with $\langle N_1 \cup \cdots \cup N_n \rangle$

So you have to prove that: $N_1 + \cdots + N_n = RN$

If $x \in \sum_{ s = 1 }^{n} N_s$ then $x = x_1 + x_2 + \ ... \ ... \ + x_n$ for $x_i \in N_i$

Thus $x = 1x_1 + 1x_2 + \ ... \ ... \ + 1x_n$, with $x_i \in N$, so $x$ belongs to $RN$ by definition.

So the first part of your proof is ok.

To be honest, I cannot follow the second part. I think it can be done much easier.

Take $x \in RN$ then $x = \sum_{j = 1 }^{p} r_j x_j$ for some $p \in \mathbb{N}$, where $r_j \in R$, $x_j \in N$.

Since $N = N_1 \cup \cdots \cup N_n$. for each $j = 1, \cdots, p$ there is a $i \in \{1, \cdots, n\}$ such that $x_j \in N_i$, so also $r_j x_j \in N_i$

Therefore
$x = \sum_{j = 1 }^{p} r_j x_j \in N_1 + \cdots + N_n$

I think this is the second part but a little bit easier.

Thanks Steenis ...

That proof seems really clear ...

Will work through it again shortly...Peter