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Let $X$ be a scheme, let $i : Z \to X$ be a closed subscheme, let $Y := \mathrm{Bl}_{Z}(X)$ be the blowup of $X$ at $Z$ with projection $\pi : Y \to X$. Suppose $U \supseteq X \setminus Z$ is an open subscheme of $X$ for which $\pi^{-1}(U) \to U$ is an isomorphism. Does there exist a closed subscheme $Z' \to X$ with support contained in $X \setminus U$ and for which the blowup $Y' := \mathrm{Bl}_{Z'}(X)$ is $X$-isomorphic to $Y$?

Motivation: If we blow up a subscheme of a normal scheme $X$, the maximal open subscheme $U$ for which $\pi^{-1}(U) \to U$ is an isomorphism contains all the codimension 1 points of $X$, so I was naively wondering whether all blow ups of a normal scheme are obtained by blowing up a closed subscheme of codimension at least 2.

Thoughts: If $A$ is a ring and $I$ is an ideal of $A$ and $a \in A$ is a nonzerodivisor, then the Rees algebras $\bigoplus_{n \ge 0} I^{n}$ and $\bigoplus_{n \ge 0} (aI)^{n}$ are isomorphic so if the center of the blowup is of the form $\operatorname{Spec} A/(aI)$, I can replace it with $\operatorname{Spec} A/I$, but I don't know how to recognize an ideal as being of the form $aI$.

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Note that if $Y = Bl_I(X)$ then $I$, up to a twist and raising to a power, is the pushforward of a relative ample line bundle for $Y \to X$, so if the relative Picard number is 1, there are not so many choices for $I$.

Now, as an example, consider $$ X = C(\mathbb{P}^1 \times \mathbb{P}^1) $$ be the cone over a smooth 2-dimensional quadric. Let $$ Y = Tot_{\mathbb{P}^1}(\mathcal{O} \oplus \mathcal{O}(-1) \oplus \mathcal{O}(-1)) $$ be one of its two small resolutions. Note that $Y$ is the blowup of a Weil divisor on $X$ --- the strict transform of one of the rulings of $\mathbb{P}^1 \times \mathbb{P}^1$. In particular, its ideal is NOT supported at the vertex of the cone (while the vertex is the complement of $U$ in this case). On the other hand, the relative Picard number is 1, hence any ideal $I$ such that $Y \cong BL_I(X)$ is equal to the ideal of the above Weil divisor up to a twist and a power. In particular, any such ideal is not supported at the vertex.

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