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?
Best Answer
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 $j$, $I=(x_1)$.