MHB Can Submodules of a Noetherian Module Fail to Intersect with a Given Submodule?

[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 help to fully understand the proof of part of Proposition 4.2.5 ... ...

Proposition 4.2.5 reads as follows:
View attachment 8188
https://www.physicsforums.com/attachments/8189My questions are as follows:Question 1

In the above text from Bland we read the following:

" ... ... Conversely, suppose that $$N$$ and $$M/N$$ are noetherian. Let$$M_1 \subseteq M_2 \subseteq M_3 \subseteq \ ... \ ... $$be an ascending chain of submodules of $$M$$. Then $$M_1 \cap N \subseteq M_2 \cap N \subseteq M_3 \cap N \subseteq \ ... \ ...$$ ... ... "My question is ... what about the case where all the $$M_i$$ fail to intersect with $$N$$ ... is this possible ... if so how does the proof read then ...?
Question 2

In the above text from Bland we read the following:

" ... ... If $$i \ge n$$ and $$x \in M_i$$ then $$x + N \in (M_i + N)/N = (M_n + N)/N$$ ... ... "My question is ... why does $$x \in M_i \Longrightarrow x + N \in (M_i + N)/N$$ ... ... is it because ...

$$x \in M_i \Longrightarrow x + 0_N + N \in (M_i + N)/N$$ ...

... and $$x + 0_N + N = x + N$$ ... ... ?
Hope someone can help ...

Peter

 

[steenis]

Q1: $M_i$ and $N$ are both submodules of $M$, therefore $M_i \cap N$ can never be empty.

If $M_n \cap N = 0$ then of course $M_i \cap N = 0$ for $i \leq n$. ($0 = \{ 0 \}$ is the zero-module.)

The proof, however, remains the same.

Q2: if $x \in M_i$ then $x \in M_i + N$ and. because $N \leq M_i + N$, we have $x + N \in (M_i + N)/N$

 

[Math Amateur]

Q1: $M_i$ and $N$ are both submodules of $M$, therefore $M_i \cap N$ can never be empty.

If $M_n \cap N = 0$ then of course $M_i \cap N = 0$ for $i \leq n$. ($0 = \{ 0 \}$ is the zero-module.)

The proof, however, remains the same.

Q2: if $x \in M_i$ then $x \in M_i + N$ and. because $N \leq M_i + N$, we have $x + N \in (M_i + N)/N$

Thanks for thr help Steenis ...

Peter