Lemma 15.31.2. 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 15.30.2 and 15.30.3). 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.**
By Algebra, Lemma 10.70.6 it suffices to show that $R''$ is $a$-torsion free.

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

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

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

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 15.30.5 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 $R'' = R[y_2, \ldots , y_ r]/(a y_ i - a_ i)$ has no nonzero $a$-torsion elements as desired. $\square$

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## Comments (2)

Comment #3792 by Kestutis Cesnavicius on

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