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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 Nov 27 '15 at 6:45

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  • 1
    $\begingroup$ math.stackexchange.com/questions/570177/… $\endgroup$ – Prahlad Vaidyanathan Nov 27 '15 at 5:42
  • $\begingroup$ @PrahladVaidyanathan thanks; your answer in that thread is nice btw. This will probably get some sort of duplicate marking but I'll leave it up just in case anybody else has something to add. $\endgroup$ – Alex Mathers Nov 27 '15 at 5:45
  • $\begingroup$ Btw, the claim in the title of the original question holds true and it is more general than your test problem. $\endgroup$ – user26857 Nov 27 '15 at 6:54
  • $\begingroup$ @user26857 thanks. $\endgroup$ – Alex Mathers Nov 27 '15 at 7:00

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 $(1,0,0,0,\dots)\in R$.

  • $\begingroup$ Thanks. This is essentially the same idea @PrahladVaidyanathan linked me to in his comment. $\endgroup$ – Alex Mathers Nov 27 '15 at 6:44

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