Proof.
As blowing up commutes with restrictions to open subschemes (Lemma 31.32.3) the first statement just means that $X' = X$ if $Z = \emptyset $. In this case we are blowing up in the ideal sheaf $\mathcal{I} = \mathcal{O}_ X$ and the result follows from Constructions, Example 27.8.14.
The second statement is local on $X$, hence we may assume $X$ affine. Say $X = \mathop{\mathrm{Spec}}(A)$ and $Z = \mathop{\mathrm{Spec}}(A/I)$. By Lemma 31.32.2 we see that $X'$ is covered by the spectra of the affine blowup algebras $A' = A[\frac{I}{a}]$. Then $IA' = aA'$ and $a$ maps to a nonzerodivisor in $A'$ according to Algebra, Lemma 10.70.2. This proves the lemma as the inverse image of $Z$ in $\mathop{\mathrm{Spec}}(A')$ corresponds to $\mathop{\mathrm{Spec}}(A'/IA') \subset \mathop{\mathrm{Spec}}(A')$.
Consider the canonical map $\psi _{univ, 1} : b^*\mathcal{I} \to \mathcal{O}_{X'}(1)$, see discussion following Constructions, Definition 27.16.7. We claim that this factors through an isomorphism $\mathcal{I}_ E \to \mathcal{O}_{X'}(1)$ (which proves the final assertion). Namely, on the affine open corresponding to the blowup algebra $A' = A[\frac{I}{a}]$ mentioned above $\psi _{univ, 1}$ corresponds to the $A'$-module map
\[ I \otimes _ A A' \longrightarrow \left(\Big(\bigoplus \nolimits _{d \geq 0} I^ d\Big)_{a^{(1)}}\right)_1 \]
where $a^{(1)}$ is as in Algebra, Definition 10.70.1. We omit the verification that this is the map $I \otimes _ A A' \to IA' = aA'$.
$\square$
Comments (0)
There are also: