MHB What is the Proof for the Nonempty Intersection of Submodules Being a Submodule?

[Math Amateur]

In Paul Bland's book: Rings and Their Modules, we read the following text at the start of Section 2.2 Free Modules:View attachment 3385In the above text we read the following:

" ... ... if $$N$$ is generated by $$X$$ and if $$\{ N_\alpha \}_\Delta$$ is the family of submodules of $$M$$ that contain $$X$$, ... ...

then

... ... $$N = \ \bigcap \nolimits_\Delta N_\alpha \ = \ \sum \nolimits_X xR$$ ... ... "In order to fully understand this statement I would like to prove it ... but I am unable to get started on a proof ...

Can someone please help ...

Peter***NOTE***

To ensure that MHB members reading this post can follow Bland's notation I am providing the relevant text from page 1 of his text, as follows:View attachment 3386

 

[Euge]

What Bland is saying is that $N$ is the smallest submodule of $M$ that contains $X$ (which is $\cap_{\Delta} N_\alpha$). Let $N' := \cap_\Delta N_\alpha$. Since $N$ is a submodule of $N$ that contains $X$ and $N'$ is the smallest such submodule, $N \supseteq N'$. On the other hand, since $X \subseteq N'$, $xR \in N'$ for all $x \in X$ and hence $\sum_X xR \subseteq N'$, i.e., $N \subseteq N'$. So $N = N'$.

 

[Math Amateur]

What Bland is saying is that $N$ is the smallest submodule of $M$ that contains $X$ (which is $\cap_{\Delta} N_\alpha$). Let $N' := \cap_\Delta N_\alpha$. Since $N$ is a submodule of $N$ that contains $X$ and $N'$ is the smallest such submodule, $N \supseteq N'$. On the other hand, since $X \subseteq N'$, $xR \in N'$ for all $x \in X$ and hence $\sum_X xR \subseteq N'$, i.e., $N \subseteq N'$. So $N = N'$.

Thanks Euge ... that post was VERY helpful ... BUT reflecting on it ...

You write:

" ... ... Let $N' := \cap_\Delta N_\alpha$. ... ..."

How do we know that N' is actually a submodule?

Peter

 

[mathbalarka]

How do we know that N' is actually a submodule?

Nonempty intersection of submodules is always a submodule. As $0 \in N_\alpha$, $0 \in \cap N_\alpha$. For $x, y \in \cap N_\alpha$, $x + ry \in N_\alpha$ (submodules) for $r \in R$, which implies $x + ry \in \cap N_\alpha$. As $\cap N_\alpha \subseteq N$, this completes the proof.

 

[Math Amateur]

Nonempty intersection of submodules is always a submodule. As $0 \in N_\alpha$, $0 \in \cap N_\alpha$. For $x, y \in \cap N_\alpha$, $x + ry \in N_\alpha$ (submodules) for $r \in R$, which implies $x + ry \in \cap N_\alpha$. As $\cap N_\alpha \subseteq N$, this completes the proof.

Thanks Mathbalarka ... Appreciate your help!

Peter