# Given a short exact sequence, is the following true? [duplicate]

Say $0 \to M_1 \to M_2 \to M_3 \to 0$ is a short exact sequence and each $M_i$ is an $R$-module. If $M_2$ is finitely generated, then so also is $M_1$?

## marked as duplicate by Dietrich Burde, user26857 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(); } ); }); }); Oct 10 '16 at 9:07

• No – unless $M_2$ is a noetherian $R$-module, e.g. if $R$ is a noetherian ring. – Bernard Oct 10 '16 at 8:44
Your question can be rephrased to the question: Is a submodule of a finitely generated module again finitely generated, because for $M_1\subset M_2$ the sequence $$0\rightarrow M_1 \rightarrow M_2 \rightarrow M_1/ M_2\rightarrow 0$$ is always exact.