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I'm currently reading F Kasch "Modules and Rings" and in the problem section corresponding to the basic modules definitions I found the following:

"Exhibit a module $M$ without a finite set of generators in which every proper submodule is contained in a maximal submodule"

My attepmt:

Since we want a (left) $R$-module that is not finitely generated, we can take an infinite dimensional vector space $V$ over a field $K$, say of countable dimension.
Consider a basis for $V$, say $\beta=\{x_{i}|i\in\mathbb{N}\}$ and for some $j\in\mathbb{N}$ let $x_{j}\in\beta$, then $\left(\beta-\{x_{j}\}\right]=gen(\beta-\{x_{j}\})$ is a maximal subspace of $V$.

My problem is that I don't really know if every proper submodule (subspace in this case) is contained in the submodule I gave. So far the only not finitely generated modules I know are infinite dimensional vector spaces and $\mathbb{Q}$ as a $\mathbb{Z}$-module.

Any help provided will be appreciated.

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Your example works.

If $W$ is a proper subspace of $V$, then $V=W\oplus W'$ for a nonzero subspace $W'$. Since $W'$ is then at least $1$-dimensional, you can project $V$ onto $K$ by isolating one dimension in $W'$ and projecting onto that coordinate. Explicitly, I mean that if $0\neq x\in W'$, just extend $\{x\}$ to a basis of $V$ and map $v\mapsto v\text{'s $x$ coefficient}$

The kernel of that projection has codimension $1$ (so it is certainly maximal) and it contains $W$.

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  • $\begingroup$ Working it out: Let $0 \neq x\in W'$ and consider a basis $\beta$ for $V$ s.t. $x\in \beta$.. Consider the left modules morphism $\gamma:_{K}V\rightarrow _{K}K$ s.t. $\gamma\left(v\right)=k'$. Where $k'$ is the $x$'s coefficient in $v$'s essentially unique representation on $\beta$. Then: $ker(\gamma)=gen(V-\{x\})$ So: $codim(ker(\gamma))=dim(_{K}K/Im(\gamma))=1$ So: $ker(\gamma))$ is the desired submodule. Am I getting it wrong? And if it is the case that I'm getting it right, how is $codim(ker(\gamma))=1$? Thank you very much, I really appreciate your help. $\endgroup$ – eNR Sep 1 '17 at 17:35
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    $\begingroup$ @eNR the first part about the map is right. But for the rest, $codim(\ker(\gamma)):=\dim(V/\ker(\gamma))=\dim(K)=1$. $\endgroup$ – rschwieb Sep 1 '17 at 23:09
  • $\begingroup$ You are tremendously kind. I think I wasn't getting the obvious part. $\endgroup$ – eNR Sep 2 '17 at 0:33

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