# If $M$ isn't Noetherian, $M$ has a submodule maximal with respect to being not finitely generated.

Let $M$ be a module. Show that if $M$ is not Noetherian then $M$ has a submodule $N$ such that $N$ is not finitely generated but $A$ is finitely generated whenever $N < A ≤ M$.

The solution starts as follows (I'm following a given solution):
Let $Y$ be the set of non-finitely-generated submodules of $M$. If $(N_α)$ is a chain in $Y$ then $Q= \bigcup_αN_α$ is in $Y$, else $Q$ is f.g., which implies $Q=N_α$ for some $α$, whence $N_α$ is f.g., contradiction.

I don't understand why if $Q$ is f.g., then $Q=N_α$ for some $α$? Am I missing some fact?

Thanks!

• each generator of Q will be in some N_α. consider chain – user147308 May 28 '14 at 9:09
• @user26857 if $M$ is not finitely generated, take $N=M$ (there is no such $A$ as in the second part of the claim so this works). – Simone May 28 '14 at 12:13
• @Simone indeed: the formulation is logically correct, but ever so slightly tricky. – Andreas Caranti Feb 7 '16 at 11:06

By definition of union, an element $x\in M$ belongs to $\bigcup_\alpha N_\alpha$ if and only if it belongs to some $N_\alpha$.

If $Q=\bigcup_\alpha N_\alpha$ is f.g., then choose a finite family $\{x_1,\dots,x_n\}$ of generators of $Q$. For any $i\in\{1,\dots,n\}$, there exists $N_i\in\{N_\alpha\}_\alpha$ such that $x_i\in N_i$. Since your modules are linearly ordered, you can find a permutation $\sigma:\{1,\dots,n\}\to\{1,\dots, n\}$ such that $$N_{\sigma(1)}\leq\ldots\leq N_{\sigma(n)}.$$ This shows that $x_1,\dots,x_n\in N_{\sigma(n)}$, so that $Q\subseteq N_{\sigma(n)}$.

On the other hand, it is clear that $N_{\sigma(n)}\subseteq Q$, hence $Q=N_{\sigma(n)}\in\{N_\alpha\}_\alpha$.

• @user26857 well, in the question they are taking a chain... which means that the modules are already ordered. Why don't you want to assume their labels are ordered? I see no restriction in that... – Simone Feb 6 '16 at 23:58

A chain $(N_\alpha)_{\alpha\in A}$ of submodules always has the property that, given a finite subset $\{x_1,x_2,\dots,x_n\}$ of $Q$, then there exists $\alpha$ such that $\{x_1,x_2,\dots,x_n\}\subseteq N_\alpha$.

This is clear for $n=1$; suppose it is for subsets having $n$ elements. Suppose $\{x_1,x_2,\dots,x_n,x_{n+1}\}\subseteq Q$ and take $\alpha$ and $\beta$ such that $\{x_1,x_2,\dots,x_n\}\subseteq N_\alpha$ and $x_{n+1}\in N_\beta$.

Since we're given a chain, either $N_\alpha\subseteq N_\beta$ or $N_\beta\subseteq N_\alpha$. In the first case $\{x_1,x_2,\dots,x_n,x_{n+1}\}\subseteq N_\beta$, in the second case $\{x_1,x_2,\dots,x_n,x_{n+1}\}\subseteq N_\alpha$.

This in particular proves that $Q$ is a submodule, because we just need to consider two element subsets of $Q$.

You want to apply Zorn's lemma, so you want to show that any chain of nonfinitely generated submodules has an upper bound; the union is a good candidate, of course.

So, let $(N_\alpha)_{\alpha\in A}$ be a chain in the family $\mathcal{Y}$ of nonfinitely generated submodules of $M$ and set $Q=\bigcup_{\alpha\in A}N_\alpha$.

If $Q$ were finitely generated by $\{x_1,x_2,\dots,x_n\}$ and $\{x_1,x_2,\dots,x_n\}\subseteq N_\alpha$, then $Q\subseteq N_\alpha$, so $Q=N_\alpha$.

This is a contradiction to the hypothesis that $N_\alpha$ is not finitely generated.