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\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 15.30.1. Let $a, a_2, \ldots , a_ r$ be an $H_1$-regular sequence in a ring $R$ (for example a Koszul regular sequence or a regular sequence, see Lemmas 15.29.2 and 15.29.3). With $I = (a, a_2, \ldots , a_ r)$ the blowup algebra $R' = R[\frac{I}{a}]$ is isomorphic to $R'' = R[y_2, \ldots , y_ r]/(a y_ i - a_ i)$.

Proof. We claim $a, ay_2 - a_2, \ldots , ay_ n - a_ r$ is a $H_1$-regular sequence in $R[y_2, \ldots , y_ r]$. Namely, the map

\[ (a, ay_2 - a_2, \ldots , ay_ n - a_ r) : R[y_2, \ldots , y_ r]^{\oplus r} \longrightarrow R[y_2, \ldots , y_ r] \]

used to define the Koszul complex on $a, ay_2 - a_2, \ldots , ay_ n - a_ r$ is isomorphic to the map

\[ (a, a_2, \ldots , a_ r) : R[y_2, \ldots , y_ r]^{\oplus r} \longrightarrow R[y_2, \ldots , y_ r] \]

used to the define the Koszul complex on $a, a_2, \ldots , a_ r$ via the isomorphism

\[ R[y_2, \ldots , y_ r]^{\oplus r} \longrightarrow R[y_2, \ldots , y_ r]^{\oplus r} \]

sending $(b_1, \ldots , b_ r)$ to $(b_1 - b_2y_2 \ldots - b_ ry_ r, -b_2, \ldots , - b_ r)$. By Lemma 15.28.3 these Koszul complexes are isomorphic. By Lemma 15.29.5 applied to the flat ring map $R \to R[y_2, \ldots , y_ r]$ we conclude our claim is true. By Lemma 15.28.8 we see that the Koszul complex $K$ on $a, ay_2 - a_2, \ldots , ay_ n - a_ r$ is the cone on $a : L \to L$ where $L$ is the Koszul complex on $ay_2 - a_2, \ldots , ay_ n - a_ r$. Since $H_1(K) = 0$ by the claim, we conclude that $a : H_0(L) \to H_0(L)$ is injective, in other words that $R[y_2, \ldots , y_ r]/(a y_ i - a_ i)$ has no nonzero $a$-power torsion elements. This statement implies the lemma; see next paragraph.

There is a canonical map $R'' \to R'$ sending $y_ i$ to the class of $a_ i/a$. Since every element $x$ of $I$ can be written as $ra + \sum r_ i a_ i$ we see that $x/a = r + \sum r_ i a_ i/a$ is in the image of the map. Hence our map is surjective. On the other hand, $a$ is a nonzerodivisor on both rings and $R''_ a \to R'_ a$ is an isomorphism (as both rings evaluate to $R_ a$, see Algebra, Lemma 10.69.2 for the second one). Thus $R'' \to R'$ is injective as well and the proof is complete. $\square$

Comments (2)

Comment #3792 by Kestutis Cesnavicius on

It seems to me that the Noetherianness assumption is not needed in this lemma: one may reduce to the universal case , by using SGA 6, VII, 1.8 i). A direct proof of the general case is also given by Gabber--Ramero in their "Foundations for almost ring theory": Lemma 7.8.16 (ii) in the v13 of their arXiv manuscript

Comment #3914 by on

OK, I generalized this to -regular sequences and non-Noetherian rings... See changes here. Thanks for pointing this out.

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