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Let $aR$ be a nonzero ideal in a PID $R$. Show that $R/aR$ is a ring with only finitely many ideals.

Honestly, I do not know how to start. Appreciate any tips.

Robert Cardona
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user222262
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1 Answers1

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Hint $\ $ By the correspondence theorem, ideals in $R/aR$ are in bijection with ideals $\,bR\,$ in $R$ that contain $\,aR.\ $ But $\,bR\supseteq aR\,$ iff $\, b\mid a,\,$ since contains = divides for principal ideals. Hence the problem reduces to the finitness of the number of divisors (up to associateness), which is immediate if you know that PIDs are UFDs.

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Bill Dubuque
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  • R/aR = set of cosets of aR, right? and bR is an ideal in R, how to see bR relates to R/aR? – user222262 Mar 09 '15 at 18:43
  • @user222262 The relation is given by the correspondence theorem - see the answers given in the linked question. – Bill Dubuque Mar 09 '15 at 18:47
  • That shows a = bm and it is finitely factors. In other words, it relates to aR only, right? I can't see the link between the number of elements of aR and the number of ideals of R/aR. – user222262 Mar 09 '15 at 19:09
  • @user222262 There are only finitely many ideals in $,R/aR,$ because they are in bijection with the finite set of ideals $,bR\supseteq aR,$ in $,R,,$ i.e. the factors $,b,$ of $,a,$ in the UFD $,R,$ (up to associateness, i.e. ignoring unit factors) – Bill Dubuque Mar 09 '15 at 19:17
  • Thanks for being patient to me. I think I got it. :) – user222262 Mar 09 '15 at 19:21
  • @user222262 Great! – Bill Dubuque Mar 09 '15 at 19:22