# Examples where $R/I\cong R$? [duplicate]

I had to prove on a test that if $R$ is a PID then every surjective endomorphism of $R$ is an injection. To do this, I supposed there was a surjective endomorphism $\varphi:R\to R$. Then $$R/\ker\varphi\cong R$$

and I had to use the fact that $R$ is a PID to show $\ker\varphi=\{0\}$. Prior to this test, I would have assumed this implied a zero kernel regardless if $R$ were a PID or not. Therefore

I'm wondering if anybody has an example of a ring $R$ and a nontrivial ideal $I$ such that $R/I\cong R$?

## 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(); } ); }); }); Nov 27 '15 at 6:45

Let $R_0$ be any ring and let $R=R_0[x_1,x_2,x_3,\dots]$ be a polynomial ring over $R_0$ in infinitely many variables. Then $R/I\cong R$, where $I=(x_1)$ (since $R/I=R_0[x_2,x_3,\dots]$, and you can just shift the variables over by one to get an isomorphism with $R$). There are many other similar examples. For instance, you could take a product $R=R_0^\mathbb{N}$ of infinitely many copies of $R_0$, and let $I$ be the ideal generated by $(1,0,0,0,\dots)\in R$.