# I Artinian Modules - Bland - Proposition 4.24

1. Oct 29, 2016

### 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 4.2.4 ... ...

I need help to fully understand Part of the proof proving that $(2) \Longrightarrow (3)$ ...

In that part of the proof Bland seems to be assuming that

$\bigcap_F M_\alpha = N$

if and only if

$\bigcap_F (M_\alpha / N ) = 0$

In other words, if $F = \{ 1, 2, 3 \}$ then

$M_1 \cap M_2 \cap M_3$

if and only if

$M_1 / N \cap M_2 / N \cap M_3 / N$

But why exactly is this the case ... ...

... ... how do we formally and rigorously demonstrate that this is true ...

Hope someone can help ...

Peter

====================================================

Proposition 4.2.4 refers to the (possibly not well known) concept of cogeneration so I am providing Section 4.1 Generating as Cogenerating Classes ... ... as follows ...

#### Attached Files:

File size:
147.4 KB
Views:
104
File size:
84.4 KB
Views:
95
File size:
87.7 KB
Views:
94
File size:
80.6 KB
Views:
90
• ###### Bland - 3 - Section 4.1 - PART 3.png
File size:
131.6 KB
Views:
103
2. Oct 29, 2016

### andrewkirk

Start by observing that
$$(\bigcap_{\alpha\in F} M_\alpha)/N=N/N$$
which is the zero element of the module $M/N$.

So we have to show that
$$\bigcap_{\alpha\in F} (M_\alpha/N)=(\bigcap_{\alpha\in F} M_\alpha)/N\ \ \ \ \ \ \ (1)$$

First observe that if $m\in \bigcap_{\alpha\in F} M_\alpha$ then $\forall\alpha\in F:\ m\in M_\alpha$
so that $\forall\alpha\in F:\ m+N \in M_\alpha/N$ whence
$m+N \in \bigcap_{\alpha\in F}M_\alpha/N$.

Hence we have
$$(\bigcap_{\alpha\in F} M_\alpha)/N\subseteq \bigcap_{\alpha\in F} (M_\alpha/N)$$

For the other direction of inclusion, consider an arbitrary element $m+N$ of the LHS of (1). That is, $m+N\in \bigcap_{\alpha\in F} (M_\alpha/N)$. Then for any $\alpha\in F$ we have $m+N\in M_\alpha/N$, so there exists $m'\in M_\alpha$ and $n\in N$ such that $m=m'+n$, which is in $M_\alpha$ since $n\in N\subseteq M_\alpha$. Hence $m\in \bigcap_{\alpha\in F} M_\alpha$ and therefore $m+N$ is an element of the RHS of (1).

This is not as elegant as I would like because it is late at night and I am tired. But hopefully it makes sense.

3. Oct 29, 2016

### Math Amateur

Thanks for the help Andrew ... much appreciated...

Will be working through your post very shortly...

Thanks again ...

Peter

4. Oct 29, 2016

### Math Amateur

Thanks Andrew ... found that very clear and convincing ...

Peter