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Let $P$ be a principal proper ideal in an integral domain.

Is it $P^2 \subset P$ in general? If yes, how to prove it?

For example, if you look at the ideal $(3)=3\mathbb{Z}$ in $\mathbb{Z}$, it is quite simple to show that $3$ is not in $(3)^2$, but how to prove a similar property in a more general context?
Now suppose that we have a Dedekind domain.

If $P$ is a principal prime ideal, does $P^2 \neq P$ (and so $P^2 \subset P$) follow from the fact that there is a sort of unique factorization of proper ideals in terms of prime ideals?

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By $\subset$ you mean $\subsetneq$ (gotten from \subsetneq). It is usually a good idea to be emphatic about this in notation. – Mariano Suárez-Alvarez Feb 15 '12 at 6:15
Hint: contains = divides for principal ideals and ideals in Dedekind domains. – Math Gems Feb 15 '12 at 6:27
up vote 7 down vote accepted

If $P=(x)$, then $P^2=(x^2)$. If $P\subseteq P^2$, there is an $a$ such that $x=ax^2$, so $x(1-ax)=0$. If $x\neq0$, then $1-ax=0$ and $x$ is a unit.

Therefore $P=P^2$ implies either $P=0$ or $P=A$, the whole ring.

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Nice one. And if we replace "principal" with "prime", does the property still hold? – Oo3 Feb 15 '12 at 18:44

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