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 $R'' = R[y_2, \ldots , y_ r]/(a y_ i - a_ i)$.

## 15.30 More on Koszul regular sequences

We continue the discussion from Section 15.29.

**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

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 15.29.5 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. $\square$

Lemma 15.30.2. Let $A \to B$ be a ring map. Let $f_1, \ldots , f_ r$ be a sequence in $B$ such that $B/(f_1, \ldots , f_ r)$ is $A$-flat. Let $A \to A'$ be a ring map. Then the canonical map

is an isomorphism. Here $B' = B \otimes _ A A'$ and $f_ i' \in B'$ is the image of $f_ i$.

**Proof.**
The sequence

is a complex of $A$-modules with $B/J$ flat over $A$ and cohomology group $H_1 = H_1(K_\bullet (B, f_1, \ldots , f_ r))$ in the spot $B^{\oplus r}$. If we tensor this with $A'$ we obtain a complex

which is exact at $B'$ and $B'/J'$. In order to compute its cohomology group $H'_1 = H_1(K_\bullet (B', f'_1, \ldots , f'_ r))$ at $(B')^{\oplus r}$ we split the first sequence above into the exact sequences $0 \to J \to B \to B/J \to 0$, $0 \to K \to B^{\oplus r} \to J \to 0$ and $\wedge ^2(B^{\oplus r}) \to K \to H_1 \to 0$. Tensoring over $A$ with $A'$ we obtain the exact sequences

where the first one is exact as $B/J$ is flat over $A$, see Algebra, Lemma 10.38.12. We conclude that $J' = J \otimes _ A A'$ and $\mathop{\mathrm{Ker}}((B')^{\oplus r} \to B') = K \otimes _ A A'$. Thus

as $-\otimes _ A A'$ is right exact. This proves the lemma. $\square$

Lemma 15.30.3. Let $A \to B$ and $A \to A'$ be ring maps. Set $B' = B \otimes _ A A'$. Let $f_1, \ldots , f_ r \in B$. Assume $B/(f_1, \ldots , f_ r)B$ is flat over $A$

If $f_1, \ldots , f_ r$ is a quasi-regular sequence, then the image in $B'$ is a quasi-regular sequence.

If $f_1, \ldots , f_ r$ is a $H_1$-regular sequence, then the image in $B'$ is a $H_1$-regular sequence.

**Proof.**
Assume $f_1, \ldots , f_ r$ is quasi-regular. Set $J = (f_1, \ldots , f_ r)$. By assumption $J^ n/J^{n + 1}$ is isomorphic to a direct sum of copies of $B/J$ hence flat over $A$. By induction and Algebra, Lemma 10.38.13 we conclude that $B/J^ n$ is flat over $A$. The ideal $(J')^ n$ is equal to $J^ n \otimes _ A A'$, see Algebra, Lemma 10.38.12. Hence $(J')^ n/(J')^{n + 1} = J^ n/J^{n + 1} \otimes _ A A'$ which clearly implies that $f_1, \ldots , f_ r$ is a quasi-regular sequence in $B'$.

Assume $f_1, \ldots , f_ r$ is $H_1$-regular. By Lemma 15.30.2 the vanishing of the Koszul homology group $H_1(K_\bullet (B, f_1, \ldots , f_ r))$ implies the vanishing of $H_1(K_\bullet (B', f'_1, \ldots , f'_ r))$ and we win. $\square$

Lemma 15.30.4. Let $A' \to B'$ be a ring map. Let $I \subset A'$ be an ideal. Set $A = A'/I$ and $B = B'/IB'$. Let $f'_1, \ldots , f'_ r \in B'$. Assume

$A' \to B'$ is flat and of finite presentation,

$I$ is locally nilpotent,

the images $f_1, \ldots , f_ r \in B$ form a quasi-regular sequence,

$B/(f_1, \ldots , f_ r)$ is flat over $A$.

Then $B'/(f'_1, \ldots , f'_ r)$ is flat over $A'$.

**Proof.**
Set $C' = B'/(f'_1, \ldots , f'_ r)$. We have to show $A' \to C'$ is flat. Let $\mathfrak r' \subset C'$ be a prime ideal lying over $\mathfrak p' \subset A'$. We let $\mathfrak q' \subset B'$ be the inverse image of $\mathfrak r'$. By Algebra, Lemma 10.38.19 it suffices to show that $A'_{\mathfrak p'} \to C'_{\mathfrak q'}$ is flat. Algebra, Lemma 10.127.6 tells us it suffices to show that $f'_1, \ldots , f'_ r$ map to a regular sequence in

with obvious notation. What we know is that $f_1, \ldots , f_ r$ is a quasi-regular sequence in $B$ and that $B/(f_1, \ldots , f_ r)$ is flat over $A$. By Lemma 15.30.3 the images $\overline{f}_1, \ldots , \overline{f}_ r$ of $f'_1, \ldots , f'_ r$ in $B \otimes _ A \kappa (\mathfrak p)$ form a quasi-regular sequence. Since $(B \otimes _ A \kappa (\mathfrak p))_\mathfrak q$ is a Noetherian local ring, we conclude by Lemma 15.29.7. $\square$

Lemma 15.30.5. Let $A' \to B'$ be a ring map. Let $I \subset A'$ be an ideal. Set $A = A/I$ and $B = B'/IB'$. Let $f'_1, \ldots , f'_ r \in B'$. Assume

$A' \to B'$ is flat and of finite presentation (for example smooth),

$I$ is locally nilpotent,

the images $f_1, \ldots , f_ r \in B$ form a quasi-regular sequence,

$B/(f_1, \ldots , f_ r)$ is smooth over $A$.

Then $B'/(f'_1, \ldots , f'_ r)$ is smooth over $A'$.

**Proof.**
Set $C' = B'/(f'_1, \ldots , f'_ r)$ and $C = B/(f_1, \ldots , f_ r)$. Then $A' \to C'$ is of finite presentation. By Lemma 15.30.4 we see that $A' \to C'$ is flat. The fibre rings of $A' \to C'$ are equal to the fibre rings of $A \to C$ and hence smooth by assumption (4). It follows that $A' \to C'$ is smooth by Algebra, Lemma 10.135.16.
$\square$

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