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Proof: By Induction

(related to Proposition: Principal Ideals being Maximal Ideals)

By hypothesis, $(R, + ,\cdot)$ is an integral domain, $a\in R,$ and $a\neq 0.$

"$\Rightarrow$"

• Assume, $(a)\lhd R$ is a principal ideal, which is also maximal among all principal ideals in $R.$
• Furthermore, assume $a=rs$ for some $r,s\in R.$
• Since $R$ is an integral domain, and $a\neq 0$ by hypothesis, we have $r\neq 0$ and $s\neq 0.$
• In particular, by divisibility of principal ideals, since $r\mid a$ is a divisor of $a,$ we have $(r)\mid ( a)$, in other words $(a)\subseteq ( r).$
• But since $(a)$ is maximal, there is no ideal $(r)$ being a proper superset of $(a).$
• Therefore, either $(a)=(r),$ or $(r)=R.$
• In the case $(r)=R,$ by principal ideal generated by a unit, we have that $r\in R^\ast$ is a unit.
• In the case $(a)=(r),$ by equality of principal ideals, we have that $a\sim b$ are associates, i.e. $s$ is a unit.
• It follows that $a$ is an irreducible element in $R,$ since we have shown that $a\neq 0,$ and from $a=rs$ it followed that $r$ is a unit or $s$ is a unit.

"$\Leftarrow$"

• Conversely, assume, $a$ is irreducible in $R.$
• Furthermore, assume $( r )\lhd R$ is a principal ideal with $(a)\subset ( r )\subset R.$
• This means $(r)\mid ( a ),$ and it follows that $r\mid a$, i.e. $a=rs$ for some $s\in R.$
• But since $a$ is irreducible, we have that $r$ is a unit or $s$ is a unit.
• Since $( r)\subset R,$ we have that $( r)\neq R,$ and by principal ideal generated by a unit $r$ is not a unit.
• Therefore $s$ is a unit, but in this case, $a\sim r$ and by equality of principal ideals, $(a)=( r)$ which contradicts $(a)\subset ( r).$
• Therefore, the assumption is incorrect, i.e. there is no $( r )\lhd R$ with $(a)\subset ( r )\subset R.$
• In other words, $(a)$ is a maximal ideal in $R$ among all principal ideals in $R.$

Finally, we can conclude that if $R$ is a principal ideal ring, then every principal ideal $(a)$ generated by an irreducible element $a\in R$ is a maximal ideal in $R.$

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References

Bibliography

1. Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013