Proof. Items (a)–(d) follow easily from the definitions and the above discussions. To prove item (e), note that $\operatorname {Tor}^\Lambda (X_L,W_L)\mid \operatorname {Tor}^\Lambda (X,W)$ (even without asking that k is algebraically closed). The converse divisibility statement follows via a straightforward ‘spreading out and specialization at a k-point’ argument. This concludes the proof of the lemma.

Proof. Since $\operatorname {Tor}^\Lambda (X_L,W_L)$ divides $\operatorname {Tor}^\Lambda (X,W)$ , we can assume without loss of generality that $L=k$ . By work of Temkin [Reference Temkin47], we can pick an alteration $\tau :X'\to X$ whose degree is a power of the exponential characteristic. Moreover, we can choose $\tau $ to be of degree $1$ if X admits a resolution of singularities. We then let and $U'=X'\setminus W'$ . We further let $V\subset U$ be the locus over which $\tau $ is étale and define .

Let . By assumption, $m\cdot \delta _X\in \operatorname {CH}_0(X_{k(X)},\Lambda )$ vanishes when restricted to U. Hence,

$$ \begin{align*}m\cdot \tau^\ast \delta_U=0\in \operatorname{CH}_0(U^{\prime}_{k(U)},\Lambda). \end{align*} $$

The base change of this class to the field extension $k(U')$ of $k(U)$ still vanishes. If we spread this out and use the localization sequence, we find that

(3.1) $$ \begin{align} m\cdot \Gamma = [Z_1]+[Z_2]\in \operatorname{CH}_{\dim X'}(X'\times X',\Lambda), \end{align} $$

where $\Gamma \subset X'\times X'$ denotes the closure of the locus

$$ \begin{align*}(\tau|_{V'}\times \tau|_{V'})^{-1}(\Delta_V) = \{(x,y)\in V'\times V'\mid \tau(x)=\tau(y)\in V\}\subset X'\times X' , \end{align*} $$

and $Z_1$ , $Z_2$ denote some cycles with

$$ \begin{align*}\operatorname{supp} Z_1\subset W'\times X'\ \ \text{and}\ \ \operatorname{supp} Z_2\subset X'\times D \end{align*} $$

for some nowhere dense closed subset $D\subsetneq X'$ .

Let now $z\in \operatorname {CH}_0(U,\Lambda )$ . By Chow’s moving lemma, we can assume that $\operatorname {supp} z\subset V$ and $\operatorname {supp} z\cap \tau (D)=\emptyset $ . We aim to show $m\cdot z=0\in \operatorname {CH}_0(U,\Lambda )$ . To this end, let $p:X'\times X'\to X'$ and $q:X'\times X'\to X'$ denote the projections to the first and second factors, respectively. Since $X'$ is smooth and proper over k, we can define pullbacks along correspondences in $\operatorname {CH}^{\dim X}(X'\times X',\Lambda )$ .

By assumption, the support of z lies in the locus over which $\tau $ is étale: $\operatorname {supp} z\subset V$ . We can thus define the zero-cycle $\tau ^\ast z$ on $X'$ on the level of cycles via the preimages of the points in the support of z. We then apply the correspondence $\Gamma $ and get

$$ \begin{align*}m\cdot \Gamma^\ast(\tau^\ast z)=p_\ast (m\Gamma \cdot q^\ast \tau^\ast z)\in \operatorname{CH}_0(X',\Lambda). \end{align*} $$

Since the support of z is contained in the locus V over which $\tau $ is étale, a direct computation shows that the above zero-cycle has the property that

(3.2) $$ \begin{align} \tau_\ast (m\cdot \Gamma^\ast(\tau^\ast z))=(\deg \tau )^2 \cdot m\cdot z\in \operatorname{CH}_0(X,\Lambda). \end{align} $$

Conversely, by (3.1), we know that this cycle is rationally equivalent to

$$ \begin{align*}\tau_\ast( \Gamma^\ast(m\cdot \tau^\ast z))= \tau_\ast ( [Z_1]^\ast(\tau^\ast z)+[Z_2]^\ast(\tau^\ast z))\in \operatorname{CH}_0(X,\Lambda). \end{align*} $$

Since $ \operatorname {supp} Z_2\subset X'\times D$ and $\operatorname {supp} z\cap \tau (D)=\emptyset $ , we have $[Z_2]^\ast (\tau ^\ast z)=0$ and so

$$ \begin{align*}\Gamma ^\ast(m\cdot \tau^\ast z)=[Z_1]^\ast(\tau^\ast z) \in \operatorname{CH}_0(X',\Lambda). \end{align*} $$

Since $\operatorname {supp} Z_1\subset W'\times X'$ , the above class lies in the image of $\operatorname {CH}_0(W',\Lambda )\to \operatorname {CH}_0(X',\Lambda )$ . Hence,

$$ \begin{align*}\tau_\ast \Gamma^\ast(m\cdot \tau^\ast z)\in \operatorname{im}(\operatorname{CH}_0(W,\Lambda)\to \operatorname{CH}_0(X,\Lambda)). \end{align*} $$

By (3.2), it follows that

$$ \begin{align*}(\deg \tau )^2 \cdot m\cdot z \in \operatorname{im} (\operatorname{CH}_0(W,\Lambda)\longrightarrow \operatorname{CH}_0(X,\Lambda)). \end{align*} $$

By the localization sequence, this implies

$$ \begin{align*}\deg(\tau)^2 \cdot m\cdot z|_U=0\in \operatorname{CH}_0(U,\Lambda). \end{align*} $$

Since $\deg (\tau )^2$ is either one or a power of the exponential characteristic of k, it is invertible in $\Lambda $ by assumption. We deduce, as we want, that $z \in \operatorname {CH}_0(U,\Lambda )$ is m-torsion. This concludes the proof of the lemma.

Proof. By inflation of local rings [Reference Bourbaki7, Chapter IX, Appendice §2, Corollaire du Théorème 1 and Exercice 4], there is an unramified extension of discrete valuation rings $R'/R$ such that the residue field of $R'$ is $\bar k$ . Up to a base change along $R'/R$ we can thus assume that k is algebraically closed. (This uses item (e) in Lemma 3.6.)

Let . Then there is a finite extension $K'/K$ such that the $\Lambda $ -torsion order of $X\times K'$ relative to $W_{X}\times K'$ is m. Let $R_{K'}\subset K'$ be the integral closure of R in $K'$ and let $R'\subset K'$ be the localization of $R_{K'}$ at a maximal ideal lying over the maximal ideal of R. Then $R'$ is a discrete valuation ring with $\operatorname {Frac} R'=K'$ and $R\subset R'$ . Up to a base change along $\operatorname {Spec} R'\to \operatorname {Spec} R$ , we can then assume that $K=K'$ . Hence, $m=\operatorname {Tor}^\Lambda ( X,W_{X})$ and it remains to show that

(3.3) $$ \begin{align} \operatorname{Tor}^\Lambda( Y,W_{Y})\mid m. \end{align} $$

Let $U=\mathcal X\setminus W_{\mathcal X}$ with generic fibre and special fibre . By assumptions, $U\to \operatorname {Spec} R$ is flat with nonempty geometrically integral fibres. Let A be the local ring of U at the generic point of the special fibre $U_0$ . Since $U_0$ is integral, A is a discrete valuation ring with residue field $k(U_0)$ and fraction field $K(U_\eta )$ . We then apply Fulton’s specialization map on Chow groups to the base change and obtain a group homomorphism

$$ \begin{align*}\operatorname{sp}: \operatorname{CH}_0(U_\eta\times K(U_\eta),\Lambda)\longrightarrow \operatorname{CH}_0(U_0\times k(U_0),\Lambda) \end{align*} $$

with $\operatorname {sp}(\delta _{U_\eta })=\delta _{U_0}$ , see Lemma 2.3. This implies (3.3), because $m=\operatorname {Tor}^\Lambda ( X,W_{X})$ is the order of $\delta _{U_\eta }$ , while $\operatorname {Tor}^\Lambda ( Y,W_{Y})$ is the order of $\delta _{U_0}$ .

Proof. Consider the part of the Jacobian given by the derivatives $\partial _t$ and $\partial _z$

$$ \begin{align*}\begin{pmatrix} 0 & x_0^{d-1} \\ x_0^2 & w \end{pmatrix}. \end{align*} $$

As the singular locus is given by the vanishing of all $2 \times 2$ minors of the Jacobian, we see that it is contained in $\{x_0 = 0\}$ as claimed.

Proof. The parameter t is nonzero on the generic fibre X of (5.3). We then work on the open subset $\{x_0z\neq 0\}$ and perform the substitution $w=-tx_0^2z^{-1}$ to obtain the equation

(5.5) $$ \begin{align} f + \sum\limits_{i=0}^l a_i x_0^i y_j^i + x_0^{d-1}z - x_0^{d-2}(\lambda y_j + x_0)tx_0^2 z^{-1} = 0. \end{align} $$

After the change of coordinates $y_j \mapsto y_jz / x_0 - \lambda ^{-1} x_0$ we arrive at the degree d hypersurface given by the vanishing of the polynomial

$$ \begin{align*}f + \sum\limits_{i = 0}^l a_i (y_j z - \lambda^{-1}x_0^2)^i + x_0^{d-1}z - t\lambda x_0^{d-1}y_j = 0. \end{align*} $$

Reordering the terms and renaming the coordinate z to $z_{s+1}$ , yields the claim that X is birational to a degree d hypersurface $X'$ of the form (5.4), where

(5.6) $$ \begin{align} a^{\prime}_i = z_{s+1}^i \left(\sum\limits_{m = i}^l \binom{m}{i} (-\lambda^{-1})^{m-i} x_0^{2 m - 2 i} a_{m}\right) - \delta_{i,1} t \lambda x_0^{d-1} + \delta_{i,0} x_0^{d-1}z_{s+1}. \end{align} $$

This shows (2) and (3). Item (1) follows immediately from the above construction. Indeed, the coordinate transformation $y_j \mapsto y_jz/x_0 - \lambda ^{-1}x_0$ is invertible on the set $\{x_0z \neq 0\}$ with inverse given by ${y_j \mapsto x_0(y_j + \lambda ^{-1}x_0)/z}$ . Thus the birational map induces an isomorphism between the open subsets $\{x_0 \neq 0\} = \{x_0z \neq 0\} \subset X$ and $\{x_0 z_{s+1} \neq 0\} \subset X'$ . (Note that we renamed the z-coordinate to $z_{s+1}$ .) Next we prove (4). Since $a_0'$ contains the variable $z_{s+1}$ linearly by (5.6) and f does not contain $z_{s+1}$ , the condition (5.2) implies that $f + a_0'$ is irreducible, as claimed. The hypersurface $X'$ is geometrically integral by (4).

Proof. Since X is a complete intersection in ${\mathbb {P}}^{N+3}_K$ , it is equidimensional and Cohen-Macaulay. As X is Cohen-Macaulay, X has no embedded components. By Lemma 5.2, we know that the open subset $\{x_0 \neq 0\} \subset X$ is isomorphic to an open subset of the geometrically integral hypersurface $X' \subset {\mathbb {P}}^{N+2}_K$ in (5.4). Hence it suffices to show that the subset $\{x_0 = 0\} \subset X$ is not an irreducible component of X, which is clear because $f + a_0$ is an irreducible polynomial by (5.2).

Proof. The derivative of the defining equation of $Y_0$ with respect to z is given by

$$ \begin{align*}\partial_z \left(f + \sum\limits_{i = 0}^l a_ix_0^iy_j^i + x_0^{d-1}z\right) = x_0^{d-1}. \end{align*} $$

Hence, the singular locus of $Y_0$ is contained in $\{x_0 = 0\}$ .

The derivative of the defining equation of $Y_1$ with respect to w is given by

$$ \begin{align*}\partial_w \left(f + \sum\limits_{i = 0}^l a_ix_0^iy_j^i + x_0^{d-2}(\lambda y_j + x_0)w\right) = x_0^{d-2} (\lambda y_j + x_0). \end{align*} $$

Hence, the singular locus of $Y_1$ is contained in $\{x_0(\lambda y_j + x_0)=0\}$ as claimed. Finally, the last claim in the lemma is clear, as $Z \subset Y_1$ is a Cartier divisor in $Y_1$ (given by $\{w = 0\} \subset Y_1$ ).

Proof. Let $L/k$ be a field extension. Recall that

$$ \begin{align*} Y_0 \times_k L &= \left\{f + \sum\limits_{i=0}^l a_i x_0^i y_j^i + x_0^{d-1}z = 0\right\} \subset {\mathbb{P}}_L^{N+2}, \\ Y_1 \times_k L &= \left\{f + \sum\limits_{i=0}^l a_i x_0^i y_j^i + x_0^{d-2}(\lambda y_j + x_0)w = 0\right\} \subset {\mathbb{P}}_L^{N+2}. \end{align*} $$

We consider $Y_1 \times _k L$ . The projection away from $P = [0:\dots :0:1] \in {\mathbb {P}}_L^{N+2}$ induces a rational map

$$ \begin{align*}\varphi \colon {\mathbb{P}}_L^{N+2} \dashrightarrow {\mathbb{P}}_L^{N+1}. \end{align*} $$

Since w appears only linearly in the defining equation of $Y_1 \times _k L$ , the restriction of $\varphi $ to $Y_1 \times _k L$ yields a birational map

$$ \begin{align*}Y_1 \times_k L \dashrightarrow {\mathbb{P}}_L^{N+1}, \end{align*} $$

which induces an isomorphism between the complements of the closed subschemes $D_1 \times _k L \subset Y_1 \times _k L$ and , respectively. Since $H_1$ is a union of two hyperplanes in ${\mathbb {P}}_L^{N+1}$ , the pushforward along the natural inclusion

$$ \begin{align*}\operatorname{CH}_1(H_1) \longrightarrow \operatorname{CH}_1({\mathbb{P}}^{N+1}_L) \cong {\mathbb{Z}} \end{align*} $$

is surjective. Thus, by the localization exact sequence (see [Reference Fulton17, Proposition 1.8]), we find that

$$ \begin{align*}\operatorname{CH}_1(D_1 \times_k L) \longrightarrow \operatorname{CH}_1(Y_1 \times_k L) \end{align*} $$

is surjective, as $\operatorname {CH}_1(Y_1 \times _k L \setminus D_1 \times _k L) \cong \operatorname {CH}_1({\mathbb {P}}^{N+1}_L \setminus H_1) = 0$ by the above discussion. A similar argument shows that

$$ \begin{align*}\operatorname{CH}_1(D_0 \times_k L) \longrightarrow \operatorname{CH}_1(Y_0 \times_k L) \end{align*} $$

is surjective. This finishes the proof of the lemma.

Proof. Since $\Lambda $ is a ring of positive characteristic,

is a positive integer $m \in {\mathbb {Z}}_{\geq 1}$ .

Step 1. We will check that the assumptions in Theorem 4.3 are satisfied for the projective family $\mathcal {X} \to \operatorname {Spec} R$ from (5.3) with the closed subset

As in Theorem 4.3, let . We note that $W_X$ agrees with $W_{\mathcal {X}} \cap X$ . The generic fibre X of $\mathcal X \to \operatorname {Spec} R$ is geometrically integral by Corollary 5.4. The special fibre consists of two components $Y^\circ = Y_0^\circ \cup Y_1^\circ $ such that $Y_0^\circ $ , $Y_1^\circ $ , and their intersection are smooth and integral, see also Lemma 5.5. In particular, $Y^\circ $ is an snc scheme, see Definition 2.1. The singular locus of $\mathcal X$ is contained in $\{x_0 = 0\}$ by Lemma 5.1. It follows from this that $Y^\circ _i$ is a Cartier divisor on $\mathcal X^\circ $ for $i=0,1$ . The generic fibre of the R-scheme is equal to $X \setminus W_X$ and thus smooth by assumption. In particular, $\mathcal {X}^\circ \to \operatorname {Spec} R$ is strictly semi-stable, see Definition 2.2. It follows that the assumptions (1) and (2) in Theorem 4.3 are satisfied for $\mathcal X \to \operatorname {Spec} R$ and the closed subset $W_{\mathcal {X}} \subset \mathcal X$ . This concludes Step 1.

Recall that Z is integral and let $\delta _{Z^\circ }\in Z^\circ _{k(Z)}$ denote the diagonal point of $Z^\circ $ , which is dense open in Z.

Step 2. We will show that there is a one-cycle $\gamma \in \operatorname {CH}_1(Y_{1} \times _k k(Z),\Lambda )$ , supported on $\{x_0(\lambda y_j + x_0) =0 \} \subset Y_1\times _k k(Z)$ , such that

(5.10) $$ \begin{align} m \cdot \delta_{Z^\circ} = (\iota^\ast \gamma)|_{Z^\circ_{k(Z)}} \in \operatorname{CH}_0(Z^\circ_{k(Z)},\Lambda) , \end{align} $$

where $ \iota \colon Y_1^\circ \times _k k(Z) \hookrightarrow Y_1 \times _k k(Z)$ denotes the natural open embedding and $(\iota ^\ast \gamma )|_{Z^\circ _{k(Z)}}$ the pullback of the one-cycle to the Cartier divisor $Z^\circ _{k(Z)} \subset Y_1^\circ \times _k k(Z)$ along the natural regular embedding.

By Step 1, Theorem 4.3 implies that the cokernel of the map

$$ \begin{align*}\Psi^\Lambda_{Y_L^\circ} \colon \operatorname{CH}_1(Y_0^\circ \times_k L,\Lambda) \oplus \operatorname{CH}_1(Y_1^\circ \times_k L,\Lambda) \longrightarrow \operatorname{CH}_0(Z^\circ \times_k L,\Lambda) \end{align*} $$

from (4.1) (with $0 < 1$ ) is m-torsion for every field extension $L/k$ . In particular,

$$ \begin{align*}m \cdot \delta_{Z^\circ} \in \operatorname{im} \Psi^\Lambda_{Y^\circ_{k(Z)}}. \end{align*} $$

Note that $\operatorname {CH}_1(Y_0^\circ \times _k k(Z),\Lambda ) = 0$ by Lemma 5.6 and that the pull-back

$$ \begin{align*}\iota^\ast \colon \operatorname{CH}_1(Y_1 \times_k k(Z),\Lambda) \longrightarrow \operatorname{CH}_1(Y_1^\circ \times_k k(Z),\Lambda) \end{align*} $$

is surjective by [Reference Fulton17, Proposition 1.8]. Hence there exists a one-cycle $\gamma \in \operatorname {CH}_1(Y_{1} \times _k k(Z),\Lambda )$ such that $ m \cdot \delta _{Z^\circ } = (\iota ^\ast \gamma )|_{Z^\circ _{k(Z)}}$ . By Lemma 5.6 we can further assume that $\gamma $ is supported on $\{x_0(\lambda y_j + x_0) =0 \} \subset Y_1\times _k k(Z)$ . This concludes Step 2.

Step 3. We now specialize $\lambda \to 0$ and aim to compute the image of (5.10) under the corresponding specialization map on Chow groups (see Lemma 2.3); we will show that the specialization of the one-cycle $\gamma $ vanishes and so does the specialization of $m\cdot \delta _{Z^\circ }$ .

Consider the discrete valuation ring $B = k_0[\lambda ]_{(\lambda )}$ with residue field $k_0$ and fraction field $k_0(\lambda )$ . Recall that k is an algebraic closure of $k_0(\lambda )$ . Then consider the flat projective B-schemes

where f and $a_i$ are as in (5.1). Note that Z and $Y_1$ are the geometric generic fibres of $\mathcal {Z}$ and $\mathcal {Y}_1$ , respectively. Let $\mathcal {Z}^\circ \subset \mathcal {Z}$ and $\mathcal {Y}_{1}^\circ \subset \mathcal {Y}_1$ be the complement of the closure of $W_Z$ in $\mathcal {Z}$ and of $W_{Y_1}$ in $\mathcal {Y}_1$ , respectively. Note that $Z^\circ = Z \setminus W_Z$ and $Y_1^\circ = Y_1 \setminus W_{Y_1}$ are the geometric generic fibres of $\mathcal {Z}^\circ $ and $\mathcal {Y}_1^\circ $ , respectively. We denote the special fibres by $Z_0^\circ $ and $Y_{1,0}^\circ $ , respectively. By Lemma 2.3, there exist specialization maps induced by Fulton’s specialization map for the flat $\mathcal {O}_{\mathcal {Z},Z_0}$ -schemes $\mathcal {Z}^\circ \times _B \mathcal {O}_{\mathcal {Z},Z_0}$ , $\mathcal {Y}_1^\circ \times _B \mathcal {O}_{\mathcal {Z},Z_0}$ , and $\mathcal {Y}_1 \times _B \mathcal {O}_{\mathcal {Z},Z_0}$

$$ \begin{align*}\begin{aligned} \operatorname{sp}_{Z^\circ} &\colon \operatorname{CH}_0(Z^\circ \times_k k(Z),\Lambda) \longrightarrow \operatorname{CH}_0(Z_0^\circ \times_{k_0} k_0(Z_0),\Lambda), \\ \operatorname{sp}_{Y_1^\circ} &\colon \operatorname{CH}_1(Y_1^\circ \times_k k(Z),\Lambda) \longrightarrow \operatorname{CH}_1(Y_{1,0}^\circ \times_{k_0} k_0(Z_0),\Lambda), \\ \operatorname{sp}_{Y_1} &\colon \operatorname{CH}_1(Y_1 \times_k k(Z),\Lambda) \longrightarrow \operatorname{CH}_1(Y_{1,0} \times_{k_0} k_0(Z_0),\Lambda), \end{aligned} \end{align*} $$

where $Y_{1,0}$ denotes the special fibre of the B-scheme $\mathcal {Y}_1$ .

We now apply the specialization $\operatorname {sp}_{Z^\circ }$ to the zero-cycle (5.10). By Lemma 2.3, we get

(5.11) $$ \begin{align} m \cdot \delta_{Z_0^\circ} = \operatorname{sp}_{Z^\circ}(m \cdot \delta_{Z^\circ}) = \operatorname{sp}_{Z^\circ}((\iota^\ast \gamma)|_{Z^\circ_{k(Z)}}) = \operatorname{sp}_{Y_1^\circ}(\iota^\ast \gamma)|_{Z^\circ_{0,k_0(Z_0)}}, \end{align} $$

where $|_{Z^\circ _{k(Z)}}$ and $|_{Z^\circ _{0,k_0(Z_0)}}$ denote the pullback along the regular embedding of the Cartier divisor $Z^\circ _{k(Z)}$ and $Z^\circ _{0,k_0(Z_0)}$ in $Y_1^\circ \times _k k(Z)$ and $Y_{1,0}^\circ \times _{k_0} k_0(Z_0)$ , respectively. Recall that the one-cycle $\gamma $ is supported on $\{x_0(\lambda y_j + x_0) =0 \} \subset Y_1\times _k k(Z)$ . Thus the specialization $\operatorname {sp}_{Y_1}(\gamma ) \in \operatorname {CH}_1(Y_{1,0},\Lambda )$ is supported on $\{x_0^2 =0\} \subset Y_{1,0}$ . In particular

$$ \begin{align*}\operatorname{sp}_{Y_1^\circ}(\iota^\ast \gamma) = 0 \in \operatorname{CH}_1(Y_{1,0}^\circ), \end{align*} $$

as the subset $\{x_0 = 0\} \subset Y_{1,0}$ is contained in the specialization of $W_{Y_1}$ and $\operatorname {sp}$ commutes with pullbacks along open immersions. Hence, the right hand side of (5.11) vanishes, which concludes Step 3.

By Step 3,

$$ \begin{align*}\operatorname{Tor}^\Lambda(Z_0,W_{Z_0}) \mid m, \end{align*} $$

where $W_{Z_0} \subset Z_0$ is the specialization of $W_Z \subset Z$ . We note that $Z = Z_0 \times _{k_0} k$ and $W_Z = W_{Z_0} \times _{k_0} k$ , because the defining equations of Z and $W_Z$ are defined over $k_0$ . Hence, the proposition follows from Lemma 3.6 (e), as $m = \operatorname {Tor}^\Lambda (\bar {X},W_{\bar {X}})$ .

Proof. Consider the pair $(Z_\rho ,W_{Z_\rho })$ and let $(Z_0,W_{Z_0})$ be the pair obtained by specializing $\rho \to 0$ , that is,

$$ \begin{align*}\begin{aligned} Z_0 &= \{l' \cdot F = 0\} \subset {\mathbb{P}}^{n+r+1}_k, \\ W_{Z_0} &= \{y_{r+1}l'l" =0 \} \cup Z_{0}^{\operatorname{sing}} \subset Z_0. \end{aligned} \end{align*} $$

Note that the scheme $Z_0$ is reducible, but the open subscheme is integral as the polynomial F is irreducible. Hence, the torsion order $\operatorname {Tor}^{{\mathbb {Z}}/m}(Z_0,W_{Z_0})$ is defined, see Remark 3.5. We observe that $U_0 = Z \setminus W_Z$ , where the pair $(Z,W_Z)$ is as in Theorem 6.1 with $l = l' \cdot l" \in k[x_0,\dots ,x_n]$ . Thus, we get $\operatorname {Tor}^{{\mathbb {Z}}/m}(Z_0,W_{Z_0}) = m$ by Theorem 6.1. Hence, m divides $\operatorname {Tor}^{{\mathbb {Z}}/m}(\bar {Z}_\rho ,W_{\bar {Z}_\rho })$ by Lemma 3.8. Conversely, any ${\mathbb {Z}}/m$ -torsion order can be at most m. Hence, $\operatorname {Tor}^{{\mathbb {Z}}/m}(\bar {Z}_\rho ,W_{\bar {Z_{\rho }}}) =m$ , as claimed.

Proof of Theorem 6.1

This follows from arguments similar to those in the proof of [Reference Schreieder45, Theorem 6.1 and 7.1]; we give some details for the reader’s convenience. Let $P=\{x_0=\dots =x_n=0 \}\subset {\mathbb {P}}^{n+r+1}_k$ and consider the blow-up , which can be described via the vanishing locus of

(6.4) $$ \begin{align} g(x_0,\dots,x_n) x_0^{m+n-\deg(g)} y_0^m + \sum\limits_{j = 1}^{r} x_0^{n-\deg c_j} c_j(x_1,\dots,x_n) y_j^m + (-1)^n x_1 x_2\dots x_n y_{r+1}^m, \end{align} $$

inside the projective bundle $ {\mathbb {P}}_{{\mathbb {P}}^n}(\mathcal O(-1)\oplus \mathcal O^{\oplus (r+1)})$ over ${\mathbb {P}}^n$ , where $y_0$ denotes a local coordinate that trivializes $\mathcal O(-1)$ and $y_1,\dots ,y_{r+1}$ trivialize $\mathcal O^{\oplus (r+1)}$ . The projection to the x-coordinates induces a morphism $f:Y\to {\mathbb {P}}^n_k$ . We furthermore pick an alteration $\tau ':Y'\to Y$ of order coprime to m, which can be done by [Reference Temkin47], see also [Reference Illusie and Temkin22]. The corresponding alteration of Z is denoted by $\tau :Y'\to Z$ .

As detailed in [Reference Schreieder45, §7], an application of [Reference Schreieder45, Theorem 5.3] shows that the pullback of the class $\alpha =(x_1/x_0,\dots ,x_n/x_0)\in H^n(k({\mathbb {P}}^n),\mu _m^{\otimes n})$ yields an unramified class

of order m such that for any subvariety $E\subset Y'$ which does not dominate ${\mathbb {P}}^n$ via $ f\circ \tau '$ , the class $(\tau ')^\ast \gamma $ vanishes in $H^n(k(E),\mu _m^{\otimes n})$ . Moreover, the class $f^\ast \alpha $ vanishes at the generic point of the exceptional divisor $D_0$ of $Y\to Z$ , which is cut out by $y_0=0$ , and at the generic point of the strict transform $D_{r+1}$ of the divisor $\{y_{r+1} = 0\} \subset Z$ under the blow-up morphism $Y \to Z$ . Indeed, the divisors $D_0$ and $D_{r+1}$ in Y map via f onto ${\mathbb {P}}^n$ and the generic fibres $D_{0,\eta }$ and $D_{r+1,\eta }$ of $\left .f\right |{}_{D_i}$ are the hypersurfaces in (different) ${\mathbb {P}}^r_{k({\mathbb {P}}^n)}$ given by

$$ \begin{align*}\begin{aligned} &\sum\limits_{j = 1}^{r} c_j(x_1,\dots,x_n) y_j^m + (-1)^n x_1 x_2 \dots x_n y_{r+1}^m =0 \quad \text{and} \\ &\pi \cdot \left(1+\sum\limits_{i=1}^n x_i^{\left\lceil \frac{n+1}{m}\right\rceil}\right)^m y_0^m - (-1)^n x_1 x_2 \cdots x_n y_0^m + \sum\limits_{j = 1}^{r} c_j(x_1,\dots,x_n) y_j^m =0, \end{aligned} \end{align*} $$

respectively, see (6.4) and (6.1). Thus $D_{0,\eta }$ and $D_{r+1,\eta }$ are each isomorphic as $k({\mathbb {P}}^n)$ -varieties to a subvariety of the hypersurface given by the vanishing of the n-th Fermat-Pfister form of degree m

$$ \begin{align*}\sum\limits_{\varepsilon \in \{0,1\}^n} (-x_1)^{\varepsilon_1} \dots (-x_n)^{\varepsilon_n} y^m_{j(\varepsilon)} \end{align*} $$

with $j(\varepsilon ) = \sum \limits _{i=0}^n \varepsilon _i 2^{i-1}$ as defined in [Reference Schreieder45, Equation (7)]. The vanishing of $f^\ast \alpha $ along the generic points of $D_0$ and $D_{r+1}$ follows from [Reference Schreieder45, Corollary 4.3]. Since the generic fibre of f is smooth (m is invertible in k), we conclude via [Reference Bloch and Ogus6] that

(6.5) $$ \begin{align} ((\tau')^\ast \gamma)|_E=0\in H^n(k(E),\mu_m^{\otimes n}) \end{align} $$

for any subvariety $E\subset Y'$ with $\tau (E)\subset W_Z = Z^{\operatorname {sing}}\cup \{ y_{r+1} l=0\}$ , where $l\in k[x_0,\dots ,x_n]$ is any nontrivial homogeneous polynomial in $x_0,\dots ,x_n$ as in the statement of the theorem.

A computation with the Merkurjev pairing similar to [Reference Schreieder44, §3] or [Reference Schreieder and Farkas46, Theorem 8.6] then shows $\operatorname {Tor}^{{\mathbb {Z}}/m}(Z,W_Z)=m$ ; we sketch the argument for convenience. For a contradiction assume that there is a positive integer $m'< m$ with

$$ \begin{align*}m'\cdot \delta_{Z}=z+m\cdot \zeta \in \operatorname{CH}_0(Z_{k(Z)}) \end{align*} $$

for some zero-cycle z whose support $\operatorname {supp} z$ lies in $(W_Z)_{k(Z)}$ and some zero-cycle $\zeta \in \operatorname {CH}_0(Z_{k(Z)})$ that reflects the fact that we work with ${\mathbb {Z}}/m$ -torsion orders. We restrict the above identity to the regular locus of $Z_{k(Z)}$ and pull this back to $\tau ^{-1}(Z^{\operatorname {sm}}_{k(Z)})$ . The localization sequence then yields

$$ \begin{align*}m'\cdot \delta_\tau=z'+m\cdot \zeta'\in \operatorname{CH}_0(Y^{\prime}_{k(Z)}) , \end{align*} $$

where $\delta _\tau =\tau ^\ast \delta _Z$ is the point induced by the graph of $\tau :Y'\to Z$ , $z'$ is a zero-cycle with $\operatorname {supp} z'\subset \tau ^{-1}(W_Z)_{k(Z)}$ and $\zeta '\in \operatorname {CH}_0(Y^{\prime }_{k(Z)}) $ . Note that this used $Z^{\operatorname {sing}}\subset W_Z$ . We pair the above zero-cycle via the Merkurjev pairing (see [Reference Merkurjev32, §2.4] or [Reference Schreieder and Farkas46, §5]) with the unramified class $\gamma $ from above. This yields

$$ \begin{align*}\langle m'\cdot \delta_\tau, \tau^\ast \gamma \rangle =0 , \end{align*} $$

because $\gamma $ is m-torsion, hence pairs to zero with $m\cdot \zeta '$ , and it restricts to zero on generic points of subvarieties of $\tau ^{-1}(W_Z )$ by (6.5), hence pairs to zero with $z'$ . Conversely, the definition of the Merkurjev pairing directly implies that

$$ \begin{align*} 0&=\langle m'\cdot \delta_\tau, \tau^\ast \gamma \rangle= m' \cdot \tau_\ast \tau^\ast \gamma=m'\cdot \deg(\tau)\cdot \gamma \in H^n(k(Z),\mu_m^{\otimes n}), \end{align*} $$

as $\delta _\tau \in Y^{\prime }_{k(Z)}$ is the point associated to the graph $\Gamma _\tau \subset Y'\times Z$ of $\tau $ . This contradicts the fact that $ \deg (\tau )$ is coprime to m, that $1\leq m'<m$ , and that $\gamma $ has order m. This concludes the proof of the theorem.

Proof of Theorem 7.1

Note that the torsion order of any variety is divisible by the torsion order of the base-change to any field extension. Moreover, the definition of very general (see Definition 2.4) is stable under extension of the base field. Up to replacing k by a field extension, we can therefore assume that k is algebraically closed and uncountable.

We fix positive integers $n \geq 2$ and $r \leq 2^{n} -2$ . For a non-negative integer s as in the theorem, we define inductively an integral degree d-hypersurface $Z = Z_s$ of dimension $N = n + r + s$ .

Step 1. Suppose there exists an integral degree d-hypersurface $Z_s \subset {\mathbb {P}}_k^{n+r+s+1}$ given by the vanishing of a homogeneous polynomial of the form

(7.1) $$ \begin{align} f_0^{(s)} + a_0^{(s)} + a_{r+1}^{(s)} y_{r+1}^m + \sum\limits_{j=1}^{r} \sum\limits_{i=1}^m a_{i,j}^{(s)} \cdot y_j^i \in k[x_0,\dots,x_n,y_1,\dots,y_{r+1},z_1,\dots,z_s] \end{align} $$

for some homogeneous polynomials

$$ \begin{align*}f_0^{(s)}, a_0^{(s)}, a_{r+1}^{(s)}, a_{i,j}^{(s)} \in k[x_0,\dots,x_n,z_1,\dots,z_s] \end{align*} $$

such that

1. (1) $f_0^{(s)} + a_0^{(s)} \in k[x_0,\dots ,x_n,z_1,\dots ,z_s]$ is an irreducible polynomial of degree d;

2. (2) for each j, if $x_0^{e m} \mid a_{m,j}^{(s)}$ for a non-negative integer e, then $x_0^{e i} \mid a_{i,j}^{(s)}$ for all $1 \leq i \leq m$ . We denote the maximal such e by $e_j^{(s)}$ ;

3. (3) $\operatorname {Tor}^{{\mathbb {Z}}/m}(Z_s,W_{s}) = m$ , where for some homogeneous polynomial $h^{(s)} \in k[x_0,\dots ,x_n,z_1,\dots ,z_s]$ .

Assume that there exists some $1 \leq j_0 \leq r$ such that $e_{j_0}^{(s)} \geq 1$ . Then we construct an integral degree d-hypersurface $Z_{s+1}$ of the same form (7.1) satisfying conditions (1), (2), and (3) as follows.

Consider $k' = \overline {k(\lambda )}$ an algebraic closure of the purely transcendental field extension $k(\lambda )$ of k. Rewrite the equation (7.1) as

$$ \begin{align*}\left(f_0^{(s)} + a_{r+1}^{(s)} y_{r+1}^m + \sum\limits_{j\neq j_0} \sum\limits_{i=1}^m a_{i,j}^{(s)} \cdot y_j^i\right) + a_0^{(s)} + \sum\limits_{i=1}^m a_{i,j_0}^{(s)} y_{j_0}^i. \end{align*} $$

We are now in the situation of Section 5. Note that condition (1) implies the assumption (5.2). Let $Z_{s+1}$ be the integral degree d-hypersurface $X' \times _{K} \bar {K} \subset {\mathbb {P}}^{n+r+s+2}_{\bar {K}}$ , where $X'$ is as in Lemma 5.2 and $K = k'(t)$ . We choose an isomorphism $k \cong \bar {K}$ , which exists because both fields are algebraically closed, have the same characteristic, and have the same uncountable transcendence degree over their prime fields. Thus we can view $Z_{s+1}$ as a variety over k. We aim to check that $Z_{s+1}$ satisfies the assumptions above. Recall from Lemma 5.2 that $Z_{s+1}\subset {\mathbb {P}}^{n+r+s+2}_k$ is cut out by the homogeneous polynomial

(7.2) $$ \begin{align} \left(f_0^{(s)} + a_{r+1}^{(s)} y_{r+1}^m + \sum\limits_{j\neq j_0} \sum\limits_{i=1}^m a^{(s)}_{i,j} \cdot y_j^i\right) + a^{(s+1)}_0 + \sum\limits_{i=1}^m a^{(s+1)}_{i,j_0} y_{j_0}^i, \end{align} $$

where $a^{(s+1)}_0,a^{(s+1)}_{i,j_0}\in k[x_0,\dots ,x_n,z_1,\dots ,z_{s+1}]$ are defined as in (5.6). Note that this uses also Lemma 5.2 (3). Hence, the defining equation of $Z_{s+1}$ has the form (7.1) with

Since $f_{0}^{(s+1)}$ and $a_0^{(s+1)}$ do not contain any $y_i$ , condition (1) follows from Lemma 5.2 (4). As $a_{i,j}^{(s+1)} = a_{i,j}^{(s)}$ for $j \neq j_0$ , condition (2) is clearly satisfied for $j \neq j_0$ and we note that $e_j^{(s+1)} = e_j^{(s)}$ for $j \neq j_0$ . For $j = j_0$ , condition (2) follows from Lemma 5.2 (2). Proposition 5.7 shows that $\operatorname {Tor}^{{\mathbb {Z}}/m}(Z_{s+1},W_{s+1}) = m$ , where $W_{s+1} = \{x_0 a_{r+1}^{(s)} y_{r+1} h^{(s)} z_{s+1} = 0\} \subset Z_{s+1}$ . Note that the derivative of (7.2) with respect to $y_{r+1}$ is equal to $m a_{r+1}^{(s)} y_{r+1}^{m-1}$ ; thus the singular locus of $Z_{s+1}$ is contained in $\{a_{r+1}^{(s)} y_{r+1}=0\} \subset Z_{s+1}$ . By definition, $a_{r+1}^{(s+1)} = a_{r+1}^{(s)}$ and so condition (3) is satisfied for $h^{(s+1)} = h^{(s)} z_{s+1}$ .

Step 2. Consider the hypersurface with defining equation $\rho h + x_0^{d-m-n} F$ as in Corollary 6.2, where $h \in k[x_0,\dots ,x_n]$ is an irreducible polynomial of degree d, for example,

$$ \begin{align*}h = \begin{cases} x_0^d + \sum\limits_{i = 1}^n x_{i-1} x_i^{d-1} & \text{if } p> 0 \text{ and } p \mid d, \\ \sum\limits_{i = 0}^n x_i^d & \text{otherwise,} \end{cases} \end{align*} $$

where p is the characteristic of k. We will prove that $Z_0$ satisfies the condition (1), (2), and (3) above.

Consider the following polynomials in $k[x_0,\dots ,x_n]$

where g is defined in (6.1) and the $c_j$ ’s are defined in (6.2). Then, by construction,

$$ \begin{align*}\rho h + x_0^{d-m-n} F = f_0^{(0)} + a_0^{(0)} + a_{r+1}^{(s)} y_{r+1}^m + \sum\limits_{j=1}^r \sum\limits_{i=1}^m a_{i,j}^{(0)} \cdot y_j^i \in k[x_0,\dots,x_n,y_1,\dots,y_{r+1}] \end{align*} $$

is of the form (7.1). The polynomial $f_0^{(0)}$ is irreducible because the polynomial h is irreducible and $\rho $ is a transcendental parameter over the prime field of k, which is algebraically independent from $\pi $ . Hence condition (1) holds. Condition (2) is clearly satisfied as $a_{i,j}^{(0)} = 0$ for $1 \leq i \leq m-1$ . In particular, we see from the definition of $a_{m,j}^{(0)}$ above that

(7.3) $$ \begin{align} e_j^{(0)} = \left\lfloor \frac{d-m-\deg(c_j)}{m} \right\rfloor. \end{align} $$

Corollary 6.2 shows that condition (3) holds as well, where we note that we can choose $h^{(0)} = 1$ . This concludes Step 2.

By Steps 1 and 2 above, we can apply the double cone construction (see Section 5) as long as at least one of the $e_j$ ’s defined in (2) is positive. In each step, we reduce one of them by $1$ . Hence, the number of steps is equal to the sum

$$ \begin{align*}\sum\limits_{j=1}^{r} e_j^{(0)} = \sum\limits_{j=1}^{r} \left\lfloor \frac{d-m-\deg(c_j)}{m} \right\rfloor. \end{align*} $$

This sum becomes maximal when r is maximal, that is, $r = 2^n -2$ . Then the sum reads

$$ \begin{align*} \sum\limits_{j=1}^{2^n-2} e_j^{(0)} = \sum\limits_{l=1}^{n-1} \binom{n}{l} \left\lfloor \frac{d-m-l}{m} \right\rfloor. \end{align*} $$

Let now $X_d$ be a very general hypersurface of degree d and dimension N over k. Up to replacing k by a larger algebraically closed field (which does not affect the torsion order by Lemma 3.6), we can by Lemma 2.5 assume that $X_d$ degenerates to $Z_s$ . By choosing N general hyperplane sections through a closed point of $W_s$ , we can assume that there exists a closed subset W in the total space of the degeneration which has relative dimension $0$ and whose restriction to the special fibre $Z_s$ is contained in $W_s$ . Applying Lemma 3.8 to the degeneration with W as closed subset yields that

$$ \begin{align*}\operatorname{Tor}^{{\mathbb{Z}}/m}(X_d,W_{X_d}) = m, \end{align*} $$

where is a closed nonempty zero-dimensional subset of $X_d$ . Thus, Lemma 3.6 implies that m divides $\operatorname {Tor}(X_d)$ , which finishes the proof of the theorem.

Proof. The sum in question can be rewritten as follows

$$ \begin{align*} \sum\limits_{l=1}^n \binom{n}{l} \left\lfloor \frac{n-l}{m} \right\rfloor &= \frac{1}{2} \sum\limits_{l=1}^{n-1} \binom{n}{l} \left\lfloor \frac{n-l}{m} \right\rfloor + \frac{1}{2} \sum\limits_{l=1}^{n-1} \binom{n}{n-l} \left\lfloor \frac{n-l}{m} \right\rfloor \\ &= \frac{1}{2} \sum\limits_{l=1}^{n-1} \binom{n}{l} \left\lfloor \frac{n-l}{m} \right\rfloor + \frac{1}{2} \sum\limits_{l=1}^{n-1} \binom{n}{l} \left\lfloor \frac{l}{m} \right\rfloor \\ &= \frac{1}{2} \sum\limits_{l=1}^{n-1} \binom{n}{l} \left(\left\lfloor \frac{n-l}{m} \right\rfloor + \left\lfloor \frac{l}{m} \right\rfloor\right). \end{align*} $$

The estimates in (7.4) follow now from the observation

$$ \begin{align*}\left(\left\lfloor\frac{n}{m}\right\rfloor - 1 \right) \leq \left(\left\lfloor \frac{n-l}{m} \right\rfloor + \left\lfloor \frac{l}{m} \right\rfloor\right) \leq \left\lfloor \frac{n}{m} \right\rfloor \end{align*} $$

for all $0 \leq l \leq n$ together with the summation formula for binomial coefficients.

Proof. We check (7.5) by the following computation:

$$ \begin{align*} \sum\limits_{l=1}^n \binom{n}{l} \left\lfloor \frac{n-l}{2} \right\rfloor &= \sum\limits_{l=1}^n \frac{n-l}{2} \binom{n}{l} - \frac{1}{2}\sum\limits_{\substack{l=1 \\ n -l \text{ odd}}}^n \binom{n}{l} \\ &= \frac{n}{2} (2^{n} - 1) - \frac{1}{2}\sum\limits_{l=1}^n l \binom{n}{l} - \frac{1}{2} \sum\limits_{\substack{l=1 \\ n-l \text{ odd}}}^n \left[\binom{n-1}{l-1} + \binom{n-1}{l}\right] \\ &= \frac{n}{2} (2^{n} - 1) - n 2^{n-2} - 2^{n-2} + \left(\frac{1}{4} - (-1)^n \frac{1}{4}\right) \\ &= (n-1)2^{n-2} - \left\lfloor \frac{n}{2} \right\rfloor. \end{align*} $$

We turn to (7.6) and prove first a combinatorial formula for a lacunary sum of binomial coefficients, see, for example, [Reference Riordan39, Section 4, Problem 8, p.161]. Let $\xi $ be a primitive third root of unity in ${\mathbb {C}}$ , then the following holds

$$ \begin{align*}\begin{aligned} 3 \sum\limits_{l \equiv r \ (3)} \binom{n}{l} &= \sum\limits_{l = 0}^n \binom{n}{l} (1 + \xi^{l-r} + \xi^{{2l-2r}}) \\ &= 2^n + \xi^{-r}(1+\xi)^{n} + \xi^{-2r} (1+\xi^2)^n \\ &= 2^n + (-1)^n (\xi^{n+r})^2 + (-1)^n\xi^{n+r}. \end{aligned} \end{align*} $$

Write $n = 3a + b$ for integers $a \in {\mathbb {Z}}_{\geq 0}$ and $b \in \{0,1,2\}$ . Then we get

$$ \begin{align*}\begin{aligned} \sum\limits_{l=1}^n \binom{n}{l} \left\lfloor \frac{n-l}{3} \right\rfloor &= \sum\limits_{l=1}^n \frac{n-l}{3} \binom{n}{l} - \frac{1}{3}\sum\limits_{\substack{n-l \equiv 1 \ (3) \\ l \neq 0}} \binom{n}{l} - \frac{2}{3}\sum\limits_{\substack{n-l \equiv 2 \ (3) \\ l \neq 0}} \binom{n}{l} \\ &= \frac{n}{3} (2^{n-1}-1) - \frac{1}{3}\sum\limits_{l \equiv 1 \ (3)} \binom{n}{l} - \frac{2}{3}\sum\limits_{l \equiv 2 \ (3)} \binom{n}{l} + \frac{b}{3}\\ &= \frac{n-2}{3} 2^{n-1} - \frac{n}{3} + \frac{b}{3} - \frac{(-1)^n}{9} \left(\xi^{2b+2} + \xi^{b+1} + 2 \xi^{2b+1} + 2 \xi^{b+2}\right). \\ \end{aligned} \end{align*} $$

A simple computation shows that

$$ \begin{align*}\frac{b}{3} - \frac{(-1)^n}{9} \left(\xi^{2b+2} + \xi^{b+1} + 2 \xi^{2b+1} + 2 \xi^{b+2}\right) = \delta, \end{align*} $$

which proves (7.6) and thus the lemma.

Proof of Theorem 1.1

If $d = 4$ , then the statement follows from [Reference Totaro48] for $N \leq 4$ (see also [Reference Schreieder44, Theorem 1.1]) and [Reference Pavic and Schreieder36, Theorem 1.1] for $N = 5$ , see also Example 4.6. Let now $d \geq 5$ and $N \leq \frac {d+1}{16} 2^d$ be positive integers. If $3 \leq N \leq (d-2) + 2^{d-2}-2$ , then the theorem follows from [Reference Schreieder44, Theorem 1.1], because we can then write N uniquely as $N = n + r$ for integers $n,r \geq 1$ satisfying $2 \leq n \leq d-2$ and $2^{n-1}-2 \leq r \leq 2^n - 2$ . Hence we can assume that $N \geq d-4+2^{d-2}$ . Let $n = d-2$ , $r = 2^{d-2} - 2$ and $s = N - n - r$ be non-negative integers. We claim that

$$ \begin{align*}s \leq (d-3) 2^{d-4} - \left\lfloor \frac{d-2}{2} \right\rfloor. \end{align*} $$

Indeed, otherwise we get

$$ \begin{align*}N = n+r+s> d-4 + 2^{d-2} + (d-3) 2^{d-4} - \left\lfloor \frac{d-2}{2} \right\rfloor \geq \frac{d+1}{16}2^d, \end{align*} $$

which yields a contradiction to the assumption $N \leq \frac {d+1}{16} 2^d$ . Thus, Theorem 7.1 implies by Lemma 7.4 that $\operatorname {Tor}(X_d)$ is divisible by $2$ for a very general degree d hypersurface $X_d \subset {\mathbb {P}}^{N+1}$ . In particular, $X_d$ does not admit an integral decomposition of the diagonal.

Proof of Theorem 1.2

Let $d \geq 5$ and $3 \leq N \leq \frac {d+1}{48} 2^d$ be integers. If $3 \leq N \leq (d-3) + 2^{d-3} - 2$ , then the theorem follows from [Reference Schreieder45, Theorem 7.1], because we can write $N = n+r$ for unique positive integers $n,r$ satisfying $2 \leq n \leq d-3$ and $2^{n-1}-2 \leq r \leq 2^n -2$ . Hence we can assume that $N \geq d-5 + 2^{d-3}$ . Let $n = d-3$ , $r = 2^{d-3}-2$ and $s = N -n-r$ be non-negative integers. We claim that

$$ \begin{align*}s \leq \frac{d-5}{3} 2^{d-4} - \frac{d-3}{3} + \delta, \end{align*} $$

where $\delta $ is defined as in Lemma 7.4. (Note that $n = d-3$ .) Indeed, otherwise we get

$$ \begin{align*}N = n + r + s> d-3 + 2^{d-3}-2 + \frac{d-5}{3} 2^{d-4} - \frac{d-3}{3} + \delta = \frac{d+1}{48} 2^d + \frac{2d}{3} + \delta - 4 \geq \left\lfloor\frac{d+1}{48} 2^d\right\rfloor, \end{align*} $$

which yields a contradiction to the assumption that N is an integer satisfying $N \leq \frac {d+1}{48}2^d$ . Thus, Theorem 7.1 implies by Lemma 7.4 that $3 \mid \operatorname {Tor}(X_d)$ for a very general degree d hypersurface $X_d \subset {\mathbb {P}}^{N+1}$ . In particular, $X_d$ does not admit an integral decomposition of the diagonal.

Proof of Theorem 1.3

This follows from Theorem 7.1 together with the estimate in Lemma 7.3.