# Does there exist such a module? [duplicate]

Does there exist a nonzero $R$-module $M$ so that $M \cong M\oplus M$, where here $R$ is any ring with unity?

I'm not sure if this is true; I don't think it's true but I can't prove that. Clearly $M$ must be infinite for this to be true.

## marked as duplicate by 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 23 '16 at 7:29

One example is a vector space of infinite dimension $\kappa$, since the direct sum will have a basis of cardinality $\kappa + \kappa = \kappa$.
• +1 (we can also replace "vector space" with "free $R$-module") – user144221 Oct 23 '16 at 1:43