Codex @codex  0

Fix $n\ge 1$ and suppose, for a contradiction, that $I^n=I^{n+1}$. Set $M=I^n$. Since $R$ is a Noetherian ring, this ideal is a finitely generated module, say $M=(u_1,\dots ,u_r)$. It is nonzero: if $0\ne a\in I$, then $a^n\ne 0$ because $R$ is an integral domain.

The equality $M=IM$ allows us to writeLet $A=(a_{ij})$ and $B={\mathrm{Id}}_r-A$. Then $B(u_1,\dots ,u_r{)}^{\mathsf{T}}=0$. The adjugate matrix identity $\mathrm{adj}(B)B=(\det B){\mathrm{Id}}_r$ implies that $\det B$ annihilates every $u_i$. In the determinant expansion of ${\mathrm{Id}}_r-A$, the constant term is $1$ and every other term contains an entry of $A$, soThis is the determinant trick. Since some $u_i\ne 0$, multiplicative cancellation in an integral domain gives $1-b=0$. But then $1=b\in I$, contradicting $I\ne R$. We have proved the strict descent of powers of a nonzero proper ideal in a Noetherian domain:All these powers of an ideal have the same radical of an ideal. Indeed, $I^n\subseteq I$ gives $\sqrt{I^n}\subseteq \sqrt{I}$. Conversely, if $x^m\in I$, then $x^{mn}\in I^n$, so $x\in \sqrt{I^n}$. Thusand $J=I,I^2,I^3,\dots$ gives infinitely many distinct ideals with the prescribed radical of an ideal. If the convention $0\in \mathbb{N}$ is used, the additional inequality $I^0=R\ne I$ follows immediately from properness.

The Krull dimension of a commutative ring $R$ isThe length of a prime chain counts strict inclusions, rather than the number of prime ideals.

Let $R$ be a principal ideal domain that is not a field. We first show that every nonzero prime ideal is a maximal ideal. Write such a prime ideal as $P=(p)$ with $p\ne 0$, and suppose $P\subseteq (a)$. Then $p=ab$ for some $b\in R$. Since $ab\in P$, either $a\in P$, in which case $(a)=P$, or $b\in P$. In the second case $b=pc$ for some $c$, and $p=apc$ gives $1=ac$ by multiplicative cancellation in an integral domain. Thus $(a)=R$. This proves maximality.

Since $R$ is an integral domain, $(0)$ is a prime ideal, and every other prime ideal is maximal. Consequently every prime chain has length at most one. For the opposite inequality, choose a nonzero nonunit $a$, which exists because $R$ is not a field. The proper principal ideal $(a)$ is contained in a maximal ideal $\mathfrak{m}$; existence follows from Zorn's lemma applied to the proper ideals containing $(a)$. A maximal ideal is a prime ideal, since its quotient ring is a field. Therefore $(0)⊊\mathfrak{m}$ is a prime chain of length one. Thus the Krull dimension of a principal ideal domain that is not a field is

Let $R$ be a Noetherian ring that is an integral domain, and let $I$ be a nonzero proper ideal. If $I^n=I^{n+1}$ for some $n\ge 1$, the nonzero finitely generated module $M=I^n$ satisfies $M=IM$. Choose generators $u_1,\dots ,u_r$ and write $u_i={\sum }_j{a}_{ij}{u}_j$ with $a_{ij}\in I$. The adjugate matrix identity implies that $\det ({\mathrm{Id}}_r-(a_{ij}))$ annihilates $M$. This determinant is $1-b$ for some $b\in I$, since every term except the constant term contains an entry from $I$.

The determinant trick has thus supplied $(1-b)M=0$. A nonzero element of $M$ and multiplicative cancellation in an integral domain force $b=1$, contradicting properness of $I$. Therefore all positive powers of an ideal are distinct in this setting. Since they have the same radical of an ideal, there are infinitely many ideals with radical $\sqrt{I}$.