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This is my own personal proof, it seems right, but I want to make sure I don't carry incorrect logic into my future work.


Let $R$ be a PID and $c\in R$

If $c$ is irreducible, then $\langle c\rangle$ is a maximal ideal of $R$.


Proof: If $c$ is irreducible, then $c=ab\implies$ $a$ or $b$ is a unit.

Now let $\langle c\rangle\trianglelefteq I$ where $I$ is some ideal of $R$. Now since we have $\langle c\rangle \trianglelefteq I$, firstly $\langle i\rangle =I$ since $R$ is a PID, and secondly $ri=c$ for some $r\in R$. Now since $c$ is irreducible, either $r$ or $i$ is a unit.

If $i$ is a unit, then $\langle i \rangle = R$ because $i^{-1}\in R \implies ii^{-1}=1\in \langle i\rangle\implies \forall r\in R, r\in \langle i\rangle$.

If $r$ is a unit, then $cr^{-1}=i\implies \langle c\rangle = I$

So in the former case, an ideal containing $\langle c\rangle $ is actually the ring itself, and in the latter case, $\langle c\rangle$ is the ideal containing it. This means that $\langle c\rangle$ is a maximal ideal. $\blacksquare$

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  • $\begingroup$ For $\langle c\rangle$ is maximal $\implies c$ is irreducible, it seems it follows easily: Assume $c$ is reducible, then $c=ab$ where $a$ and $b$ are not units. Then $\langle c \rangle \triangleleft \langle a\rangle$ since $b\ne 1$, and so this is a stricty containment. Also since $a$ is not a unit, it does not equal to the ring. Hence $\langle c \rangle $ is not maximal, contradiction, $c$ is irreducible. $\endgroup$ – Galois in the Field Sep 26 '15 at 11:36
  • $\begingroup$ Everything works. Good job. $\endgroup$ – Crostul Sep 26 '15 at 12:00
  • $\begingroup$ @Crostul Thanks sir. $\endgroup$ – Galois in the Field Sep 26 '15 at 12:01

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