MHB Finitely Generated Modules and Maximal Submodules

[Math Amateur]

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

I am focused on Section 6.1 The Jacobson Radical ... ...

I need help with an aspect of Proposition 6.1.2 ... ...

Proposition 6.1.2 relies on Zorn's Lemma and the notion of inductive sets ... ... so I am providing a short note from Bland on Zorn's Lemma and inductive sets ... ... as follows:
View attachment 6304
NOTE: My apologies for the poor quality of the above image - due to some over-enthusiastic highlighting of Bland's text :(Now, Proposition 6.1.2 reads as follows:

View attachment 6305

Now ... in the above proof of Proposition 6.1.2, Bland writes the following:"... ... If $$\mathscr{C}$$ is a chain of submodules of $$\mathscr{S}$$, then $$x_1 \notin \bigcup_\mathscr{C}$$ , so $$\bigcup_\mathscr{C}$$ is a proper submodule of $$M$$ and contains $$N$$. Hence $$\mathscr{S}$$ is inductive ... ... My question is as follows: Why does Bland bother to show that $$\bigcup_\mathscr{C}$$ is a proper submodule of $$M$$ that contains $$N$$ ... presumably he is showing that any chain of submodules in $$\mathscr{S}$$ has an upper bound ... is that right?
... ... but why does he need to do this as the largest submodule in the chain would be an upper bound ... ... ?Hope someone can help ... ...

Peter
NOTE: My apologies for not being able to exactly reproduce Bland's embellished S in the above text ...

 

[caffeinemachine]

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

I am focused on Section 6.1 The Jacobson Radical ... ...

I need help with an aspect of Proposition 6.1.2 ... ...

Proposition 6.1.2 relies on Zorn's Lemma and the notion of inductive sets ... ... so I am providing a short note from Bland on Zorn's Lemma and inductive sets ... ... as follows:

NOTE: My apologies for the poor quality of the above image - due to some over-enthusiastic highlighting of Bland's text :(Now, Proposition 6.1.2 reads as follows:

Now ... in the above proof of Proposition 6.1.2, Bland writes the following:"... ... If $$\mathscr{C}$$ is a chain of submodules of $$\mathscr{S}$$, then $$x_1 \notin \bigcup_\mathscr{C}$$ , so $$\bigcup_\mathscr{C}$$ is a proper submodule of $$M$$ and contains $$N$$. Hence $$\mathscr{S}$$ is inductive ... ... My question is as follows: Why does Bland bother to show that $$\bigcup_\mathscr{C}$$ is a proper submodule of $$M$$ that contains $$N$$ ... presumably he is showing that any chain of submodules in $$\mathscr{S}$$ has an upper bound ... is that right?
... ... but why does he need to do this as the largest submodule in the chain would be an upper bound ... ... ?Hope someone can help ... ...

Peter
NOTE: My apologies for not being able to exactly reproduce Bland's embellished S in the above text ...

The poset here is $\mathcal S$, which has only proper submodules of $M$. To apply Zorn's lemma, one needs to exhibit an upper bound of a chain lying in the poset (of course). So to make sure that the union of all the elements in the chain is actually in $\mathcal S$, one needs to check that it is a proper submodule.

 

[Math Amateur]

The poset here is $\mathcal S$, which has only proper submodules of $M$. To apply Zorn's lemma, one needs to exhibit an upper bound of a chain lying in the poset (of course). So to make sure that the union of all the elements in the chain is actually in $\mathcal S$, one needs to check that it is a proper submodule.

Thanks caffeinemachine

My question was actually to do with why Bland needed $$\bigcup_\mathscr{C} N'$$ as an upper bound for the chain $$\mathscr{C}$$ when it seemed to me (at the time) to be possible to use the largest submodule of the chain as an upper bound ... BUT ... I now think that this does not account for the infinite case where there may be no largest element in the chain ...

My apologies for not making my question really clear ...

Peter