Let $X$ be a noetherian scheme and $\mathcal{I} \subset \mathcal{O}\_X$ a coherent sheaf of ideals. Suppose that $\mathcal{I}^d$ is locally-free for some power $d$. Then the blowing up $\mathrm{Bl}\_{\mathcal{I}^d} \to X$ is an isomorphism. However, $\mathrm{Bl}\_{\mathcal{I}} \cong \mathrm{Bl}\_{\mathcal{I}^d}$ as $X$-schemes so the map $\pi : \mathrm{Bl}\_{\mathcal{I}} \to X$ is also an isomorphism. By definition of blowing up this implies that the inverse image of $\mathcal{I}$ which is equal to $\mathcal{I}$ is is locally-free.


My question is about a direct algebraic proof of this fact translated into commutative algebra:


Let $A$ be a noetherian local ring and $I \subset A$ be an ideal such that $I^d = (f)$ for some non zero-divisor $f$. Then $I = (g)$ for some non zero-divisor $g$.


I can see how to prove this when $A$ is a UFD but I don't see how a direct proof would go of the general case. Is there an easy argument or is it necessary to argue with the Rees Algebra?



If power of an ideal is locally free then it is locally free
Here is a proof if $A$ is an integral domain.


Let $x\_1,\dots, x\_n$ generate $I$. Then $\prod\_{i=1}^n x\_i^{e\_i}$ generate $I^d =f$ for vectors $e\_i$ of nonnegative integers satisfying $\sum\_i e\_i=d$. These generators are all multiples of $f$ and can't all lie in the maximal ideal times $f$ so, since $A$ is local, one of them must be a unit times $f$. Fix such a generator $\prod\_{i=1}^n x\_i^{e\_i}$.


Without loss of generality, $e\_1>0$.


Then for all $j$ from $1$ to $n$, $(x\_j/x\_1)\prod\_{i=1}^n x\_i^{e\_i}$ is also in $I^d$, thus is a multiple of $f$, hence is a multiple of $\prod\_{i=1}^n x\_i^{e\_i}$. Dividing by $x\_1^{e\_1-1} \prod\_{i=2}^n x\_i^{e\_i}$, we see that $x\_j$ is a multiple of $x\_1$. Becuase this works for all $j$, $I=(x\_1)$.


