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I came across this solution that seeks to prove that any submodule of a cyclic module is cyclic.

Proof: Let $M$ be a cyclic module, so that $\phi:R \rightarrow M$ is a surjection under $\phi(r)=r \cdot m$ for some fixed $m$ . Therefore $M$ is isomorphic to $R/ \ker \phi$, i.e. a quotient of $R$ by some ideal. Further, any submodule of $M$ is isomorphic to an ideal of $R/ \ker \phi$. Since $R/ \ker \phi$ is a PID also, any submodule of it is generated by one element. Hence any submodule $N \subset M$ is cyclic.

My question is: How to prove that $R/\ker \phi$ is also a PID?

user26857
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user10024395
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2 Answers2

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As pointed out in the comments, $R/I$ may not be a domain, however:

Let $I$ be any ideal of the principal ideal domain $R$. Let $\mathscr J$ be an ideal of $R/I$, let $\pi:R \to R/I$ be the canonical projection, viz. $\pi(r) = r + I$ for $r \in R$, and let $J = \pi^{-1} (\mathscr J)$. Then I claim that $J$ is an ideal of $R$ and $I \subset J$. The second assertion is pretty obvious: any $s \in I$ maps to $\bar 0 = 0 + I = I$ in $R/I$, and since $\bar 0 \in \mathscr J$, the set $\pi^{-1}(\bar 0) \subset J$ contains $s$; thus $I \subset J$. As for the first, it is easy to see that $J$ is an ideal of $R$, straight from the definitons: if $a, b \in J$, then $\pi(a - b) = \pi(a) - \pi(b) \in \mathscr J$ since $\pi(a), \pi(b) \in \mathscr J$; thus $a - b \in \pi^{-1}(\mathscr J) = J$; if $r \in R$, then $\pi(ra) = \pi(r) \pi(a) \in \mathscr J$, an ideal of $R/I$; this implies of course that $ra \in J$. Bearing these facts in mind we note that $J = (j)$ for some $j \in R$ since $R$ is a PID. Then any $l \in J = \pi^{-1}(\mathscr J)$ is of the form $pj$ for some $p \in R$. Thus $\pi(l) = \pi(pj) = \pi(p) \pi(j) = \bar p \bar j$; this shows that $\mathscr J = (\bar j) = (j + I)$; it follows that $R/I$ is a principal ideal ring, though not in general a domain. QED.

Robert Lewis
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    Something missing? Fiat lux? – Jyrki Lahtonen Mar 27 '15 at 07:42
  • @Jyrki Lahtonen: indeed it is missing. Your co-moderator Pedro didn't like it, and some other anonymous mod bitched me out rudely about it, and since for personal psychological reasons (i.e., I need a venue for mathematical expression much as a musician needs a place to perform; some form of addiction, I guess) I can't afford to risk having my account messed with, I have chosen the regretable course of bridling my spontenaity and creativity to a certain extent. – Robert Lewis Mar 27 '15 at 08:06
  • @Jyrki Lahtonen: I don't mind telling you that this choice has slowed me down, mathematically, to no small degree. If you as a moderator can do anything to ease what I see as an unfortunate narrowness of interpretation of "the rules", I for one would greatly appreciate it. – Robert Lewis Mar 27 '15 at 08:11
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    I will bring it up. It is kinda your trademark. I think we should be allowed to have a personal style. Some others prefer messing with the fonts/colors. – Jyrki Lahtonen Mar 27 '15 at 08:19
  • @Jyrki Lahtonen: thank you so much! I truly appreciate your kindness and concern! *Fiat Lux!!!* – Robert Lewis Mar 27 '15 at 08:24
  • @Jyrki Lahtonen: my "dialog" with Pedro can be found here:http://math.stackexchange.com/questions/1090362/if-every-vector-is-an-eigenvector-the-operator-must-be-a-scalar-multiple-of-the/1090463#1090463 – Robert Lewis Mar 27 '15 at 08:35
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Hint: let $I$ be an ideal in $R/\ker(\phi)$. What can you say about its preimage $J=\pi^{-1}(I)$ under the canonical projection $\pi:R\to R/\ker(\phi)$?

It's an ideal, hence $J=aR$ for some $a\in R$ since $R$ is a PID, and $\pi(a)$ is a generator of $I$.

Olivier Bégassat
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