# Finitely generated module with a submodule that is not finitely generated [duplicate]

Can someone give an example of a ring $R$, a left $R$-module $M$ and a submodule $N$ of $M$ such that $M$ is finitely generated, but $N$ is not finitely generated?

I tried a couple of examples of modules I know and got nothing...

## marked as duplicate by user26857, Jyrki Lahtonen abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 7 '16 at 20:31

Consider the simplest possible nontrivial (left) $R$-module: $R$ itself. It's certainly finitely generated, by $\{ 1 \}$. The submodules are exactly the (left) ideals of $R$. So you want a ring which has (left) ideals which are not finitely generated. For example, you could use a non-Noetherian commutative ring, such as $\mathbb{Z}[X_1, X_2, X_3, \ldots ]$.
• Even more is true: A ring $R$ is noetherian if and only if submodules of f.g. modules over $R$ are f.g. – Martin Brandenburg Mar 27 '12 at 10:46