Lemma 31.32.6. Let $b : X' \to X$ be the blowing up of the scheme $X$ along a closed subscheme $Z$. Let $U = \mathop{\mathrm{Spec}}(A)$ be an affine open of $X$ and let $I \subset A$ be the ideal corresponding to $Z \cap U$. Let $a \in I$ and let $x' \in X'$ be a point mapping to a point of $U$. Then $x'$ is a point of the affine open $U' = \mathop{\mathrm{Spec}}(A[\frac{I}{a}])$ if and only if the image of $a$ in $\mathcal{O}_{X', x'}$ cuts out the exceptional divisor.

**Proof.**
Since the exceptional divisor over $U'$ is cut out by the image of $a$ in $A' = A[\frac{I}{a}]$ one direction is clear. Conversely, assume that the image of $a$ in $\mathcal{O}_{X', x'}$ cuts out $E$. Since every element of $I$ maps to an element of the ideal defining $E$ over $b^{-1}(U)$ we see that elements of $I$ become divisible by $a$ in $\mathcal{O}_{X', x'}$. Thus for $f \in I^ n$ we can write $f = \psi (f) a^ n$ for some $\psi (f) \in \mathcal{O}_{X', x'}$. Observe that since $a$ maps to a nonzerodivisor of $\mathcal{O}_{X', x'}$ the element $\psi (f)$ is uniquely characterized by this. Then we define

Here we use the description of blowup algebras given following Algebra, Definition 31.32.1. The uniqueness mentioned above shows that this is an $A$-algebra homomorphism. This gives a morphism $\mathop{\mathrm{Spec}}(\mathcal{O}_{X', x"}) \to \mathop{\mathrm{Spec}}(A') = U'$. By the universal property of blowing up (Lemma 31.32.5) this is a morphism over $X'$, which of course implies that $x' \in U'$. $\square$

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