Proof. We may assume that $\Lambda = \{z_0 = \dots = z_{n-h} = 0\}$ and

$$ \begin{align*}X_d = \left\lbrace \sum_{m_0+\dots+m_{n-h} = m}z_0^{m_0}\dots z_{n-h}^{m_{n-h}}A_{m_0,\dots,m_{n-h}}(z_0,\dots,z_{n+1}) = 0\right\rbrace\subset\mathbb{P}^{n+1}\end{align*} $$

with $A_{m_0,\dots ,m_{n-h}}\in k[z_0,\dots ,z_{n+1}]_{d-m}$ . The blowup of $\mathbb {P}^{n+1}$ along $\Lambda $ is the simplicial toric variety $\mathcal {T}$ with Cox ring

$$ \begin{align*}\operatorname{\mathrm{Cox}}(\mathcal{T})\cong k[x_0,\dots,x_{n-h},y_0,\dots,y_{h+1}]\end{align*} $$

$\mathbb {Z}^2$ -grading given, with respect to a fixed basis $(H_1,H_2)$ of $\operatorname {\mathrm {Pic}}(\mathcal {T})$ , by the following matrix:

$$ \begin{align*}\left(\begin{array}{cccccccc} x_0 & \dots & x_{n-h} & y_0 & y_1 & \dots & y_{h+1}\\ \hline 1 & \dots & 1 & 0 & 1 & \dots & 1\\ -1 & \dots & -1 & 1 & 0 & \dots & 0 \end{array}\right) \end{align*} $$

and irrelevant ideal $(x_0,\dots ,x_{n-h})\cap (y_0,\dots ,y_{h+1})$ . Substituting in the above matrix the first row with the sum of the rows and then swapping the rows an multiplying the top row by $-1$ we get to the following grading matrix

(2.5) $$ \begin{align} \left(\begin{array}{cccccccc} x_0 & \dots & x_{n-h} & y_0 & y_1 & \dots & y_{h+1}\\ \hline 1 & \dots & 1 & -1 & 0 & \dots & 0\\ 0 & \dots & 0 & 1 & 1 & \dots & 1 \end{array}\right) \end{align} $$

and hence $\mathcal {T} = \mathcal {T}_{1,0,\dots ,0}$ . The blow-down morphism is given by

$$ \begin{align*}\begin{array}{cccc} \phi: & \mathcal{T}_{1,0,\dots,0} & \longrightarrow & \mathbb{P}^{n+1}\\ & (x_0,\dots,x_{n-h},y_0,\dots,y_{h+1}) & \mapsto & [x_0y_0:\dots :x_{n-h}y_0:y_1:\dots :y_{h+1}] \end{array} \end{align*} $$

and the exceptional divisor is $E = \{y_0 = 0\}$ . Hence, the strict transform of $X_d$ is defined by

(2.6) $$ \begin{align} \widetilde{X}_d = \left\lbrace \sum_{m_0+\dots+m_{n-h} = m}x_0^{m_0}\dots x_{n-h}^{m_{n-h}}A_{m_0,\dots,m_{n-h}}(x_0y_0,\dots,x_{n-h}y_0,y_1,\dots,y_{h+1}) = 0\right\rbrace\subset\mathcal{T}_{1,0,\dots,0}, \end{align} $$

and the claim on the multidegree follows. Note that (2.5) yields that $E \cong \mathbb {P}^{n-h}_{(x_0,\dots ,x_{n-h})}\times \mathbb {P}^h_{(y_1,\dots ,y_{h+1})}$ and hence $\widetilde {E} = \widetilde {X}_d\cap E \subset \mathbb {P}^{n-h}_{(x_0,\dots ,x_{n-h})}\times \mathbb {P}^h_{(y_1,\dots ,y_{h+1})}$ is a divisor of bidegree $(m,d-m)$ .

Proof. Since $X_{Z,\eta }$ is integral after replacing Z by an open subset, we may assume that $X_Z$ is irreducible.

Now, note that the dominant morphism $f_{|Z}:Z\rightarrow Y$ yields a rational section of $\widetilde {f}:X_Z\rightarrow Z$ . So $X_{Z,\eta }$ has a $k(Z)$ -rational point, and hence, it is unirational over $k(Z)$ . Therefore, $X_Z$ is unirational, and hence, X is unirational as well.

Proof. Write $D\subset \mathcal {T}_{a_0,\dots ,a_{h+1}}$ as in (2.2) and dehomogenize with respect to $x_{n-h}$ and $y_{h+1}$ to get an affine hypersurface

$$ \begin{align*}X = \{f(x_0,\dots,x_{n-h-1},y_0,\dots,y_h) = 0\}\subset\mathbb{A}^{n+1}_{(x_0,\dots,x_{n-h-1},y_0,\dots,y_h)}. \end{align*} $$

Now, we introduce a new variable that we will keep denoting by $x_{n-h}$ and homogenize f in order to get a polynomial $\overline {f}(x_0,\dots ,x_{n-h-1},y_0,\dots ,y_h,x_{n-h})$ which is homogeneous of degree $\delta _{2,0,\dots ,0}+2$ . Note that $\overline {f}$ has the following form

(2.12) $$ \begin{align} \overline{f} = \sum_{i_0+\dots +i_{n-h} = \delta_{2,0,\dots,0}}x_0^{i_0}\dots x_{n-h}^{i_{n-h}}A_{i_0,\dots,i_{n-h}}(y_0,\dots,y_h,x_{n-h}) \end{align} $$

with $A_{i_0,\dots ,i_{n-h}}\in k[y_0,\dots ,y_h,x_{n-h}]_2$ for all $0\leq i_0\leq \dots \leq i_{n-h}\leq \delta _{2,0,\dots ,0}$ . The hypersurface

$$ \begin{align*}X_{\delta_{2,0,\dots,0}+2}^n = \{\overline{f}(x_0,\dots,x_{n-h-1},y_0,\dots,y_h,x_{n-h}) = 0\}\subset\mathbb{P}^{n+1} \end{align*} $$

is birational to X, and hence, it is birational to D as well. To conclude, set

$$ \begin{align*}\Lambda = \{x_0 = \dots = x_{n-h} = 0\},\: \Lambda' = \{y_0 = \dots = y_{h} = x_{n-h} = 0\} \end{align*} $$

and note that (2.12) yields $\operatorname {\mathrm {mult}}_{\Lambda }X_{\delta _{2,0,\dots ,0}+2}^n = \delta _{2,0,\dots ,0}$ and $\operatorname {\mathrm {mult}}_{\Lambda '}X_{\delta _{2,0,\dots ,0}+2}^n = 2$ .

Proof. For $n = 3$ , the statement has been proven in [Reference Conte, Marchisio and MurreCMM07, Theorem 2.1]. Now, consider the incidence varieties

$$ \begin{align*}\begin{array}{ll} W_2 & = \{(y,H) \: | \: y\in H\cap Y_{2}\}\subset Y_{2}\times \mathbb{G}(n-4,n-1); \\ W_3 & = \{(y,H) \: | \: y\in H\cap Y_{3}\}\subset Y_{3}\times \mathbb{G}(n-4,n-1); \\ W_{2,3} & = \{(y,H) \: | \: y\in H\cap Y_{2,3}\} = W_2\cap W_3\subset Y_{2,3}\times \mathbb{G}(n-4,n-1), \end{array} \end{align*} $$

where $\mathbb {G}(n-4,n-1)$ is the Grassmannian of $5$ -planes in $\mathbb {P}^{n+2}$ containing $\Pi $ . Let $W_{2,\eta }, W_{3,\eta }, W_{2,3,\eta }$ be the generic fibers of the second projection onto $\mathbb {G}(n-4,n-1)$ from $W_2,W_3,W_{2,3}$ , respectively.

The $2$ -plane $\Pi \subset Y_{2}$ yields a $2$ -plane $\Pi '\subset W_{2,\eta }$ defined over $k(t_1,\dots ,t_{3(n-3)})$ , and since $Y_2$ is smooth, we have that $W_{2,\eta }$ is smooth.

Hence, $W_{2,3,\eta }\subset \mathbb {P}^5_{k(t_1,\dots ,t_{3(n-3)})}$ is a smooth complete intersection over $k(t_1,\dots ,t_{3(n-3)})$ of a quadric and a cubic satisfying the hypotheses of [Reference Conte, Marchisio and MurreCMM07, Theorem 2.1], and so $W_{2,3,\eta }$ is unirational over the function field $k(t_1,\dots ,t_{3(n-3)})$ . As in the proof of Proposition 2.7, after replacing $\mathbb {G}(n-4,n-1)$ by an open subset we may assume that $W_{2,3}$ is irreducible.

Therefore, $W_{2,3}$ is unirational over the base field k, and to conclude it is enough to observe that the first projection $W_{2,3}\rightarrow Y_{2,3}$ is dominant.

Proof. We may write $Q_0 = \{z_1 = Q = 0\}$ with $Q\in k[z_0,\dots ,z_{n+1}]_2$ . Then $X_d$ is of the form

$$ \begin{align*}X_d = \{z_1A+BQ = 0\}\end{align*} $$

with $A\in k[z_0,\dots ,z_{n+1}]_{d-1}$ and $B\in k[z_0,\dots ,z_{n+1}]_{d-2}$ . Consider the quadric $Y_2 = \{z_1u-Q = 0\}$ with $z_0,\dots ,z_{n+1},u$ homogeneous coordinates on $\mathbb {P}^{n+2}$ . Since $Q_0$ is smooth, we may assume that $Q = z_0^2+\dots + z_{n+1}^2$ . Hence, $Y_2$ is a smooth as well.

Denote by $CX_d$ , $CQ_0$ the cones, respectively, over $X_d$ and $Q_0$ with vertex $v = [0:\dots :0:1]$ . Then $CQ_0\subset Y_2$ and

$$ \begin{align*}Y_2\cap CX_d = CQ_0 \cup \{z_1u-Q = A+uB = 0\}.\end{align*} $$

Set $Y_{d-1} = \{A+uB = 0\}$ and $Y_{2,d-1} = \{z_1u-Q = A+uB = 0\}$ . Note that the tangent space of $Y_2$ at v is the hyperplane $\{z_1 = 0\}$ , the tangent cone of $Y_{d-1}$ at v is given by $\{B = 0\}$ and the tangent cone of $Y_{2,d-1}$ at v is cut out by $\{z_1 = B = 0\}$ . Hence,

$$ \begin{align*}\operatorname{\mathrm{mult}}_{v}Y_{d-1} = \operatorname{\mathrm{mult}}_{v}Y_{2,d-1} = d-2.\end{align*} $$

Therefore, if $p\in Y_{2,d-1}$ is a general point the line $\left \langle v,p\right \rangle $ intersects $Y_{2,d-1}$ just in v with multiplicity $d-2$ and in p with multiplicity one. So the projection is birational onto $\overline {\pi _{v}(Y_{2,d-1})}$ which must then be a hypersurface of degree $2(d-1)-(d-2) = d$ . To conclude, it is enough to note that since $Y_{2,d-1}$ is not a cone of vertex v and $Y_{2,d-1}\subset CX_d$ we have $\overline {\pi _{v}(Y_{2,d-1})} = X_d$ .

Proof. Set $v = [0:\dots :0:1]$ and $H = \{z_0 = z_1 = 0\}\subset \mathbb {P}^{n+2}$ . Then

$$ \begin{align*}\operatorname{\mathrm{mult}}_{v}Y_{d-1} = d-2,\: \operatorname{\mathrm{mult}}_{H}Y_{d-1} = d-3.\end{align*} $$

Take a general point $p\in H$ . The lines through p that intersect $Y_{d-1}$ at p with multiplicity at least $d-2$ are parametrized by a hypersurface $W_{d-3}$ cut out in the $(n+1)$ -dimensional projective space $\mathbb {P}(T_p\mathbb {P}^{n+2})$ of lines through p by a polynomial in $k[z_0,z_1]_{d-3}$ .

Now, consider the cone $C\overline {Q}$ over the $(n-2)$ -dimensional quadric $\overline {Q} = \{z_0 = z_1 = A_1 = 0\}$ , and take a general point $p \in C\overline {Q}$ . Note that $C\overline {Q}\subset Y_2$ . The lines through p that are contained in $Y_2$ are parametrized by a quadric hypersurface $W_2\subset \mathbb {P}(T_pY_2)$ .

Let $\mathcal {F} = \mathcal {T}_{Y_2|C\overline {Q}}$ be the restriction of the tangent sheaf of $Y_2$ to $C\overline {Q}$ . Summing up, there is a subvariety $W_{2,d-3}\subset \mathbb {P}(\mathcal {F})$ with a surjective morphism $\rho :W_{2,d-3}\rightarrow C\overline {Q}$ whose fiber over a general point of $p\in C\overline {Q}$ is a complete intersection of a quadric and a hypersurface of degree $d-3$ in $\mathbb {P}^{n}$ . Hence, $\dim (W_{2,d-3}) = 2n-3$ .

By hypothesis, $\overline {Q}$ has a point. Let $\overline {C}$ be a conic in $\overline {Q}$ through this point and S the cone over $\overline {C}$ with vertex v. Then $S\subset C\overline {Q}$ is a rational surface. Set $W = \rho ^{-1}(S)$ and $W_s = \rho ^{-1}(s)$ for $s\in S$ . A general point $w\in W$ represents a pair $(s,l_s)$ , where $s\in S$ and $l_s$ is a line through s which is contained in $Y_2$ and intersects $Y_{d-1}$ with multiplicity $d-2$ at s. Since $\deg (Y_{d-1}) = d-1$ the line $l_s$ intersects $Y_{d-1}$ just at one more point $x_{(s,l_s)}\in Y_2 \cap Y_{d-1} = Y_{2,d-1}$ and we get a rational map

If W is unirational, in order to prove that $\psi $ is dominant it is enough to prove that the induced map

between the algebraic closures is dominant. Take a general point $p \in \overline {Y}_{2,d-1}$ , and assume that $x_{(s,l_s)} = p$ . Then $l_s$ lies in the tangent space of $Y_2$ at p which is given by $\{L = 0\}$ . Such tangent space intersects $\overline {S}$ in a conic, and further intersecting with $W_{d-3}$ we see that there are finitely may points $s\in \overline {S}$ such that $x_{(s,l_s)} = p$ for some $l_s\in W_s$ . Furthermore, if $x_{(s,l_s)} = x_{(s,l_s')}$ , then $l_s = l_s'$ . Hence, $\overline {\psi }$ is generically finite and since $\dim (W) = \dim (Y_{2,d-1})$ we conclude that $\overline {\psi }$ is dominant.

Proof. We may write $\Lambda = \{z_0 = z_1 = 0\}$ and

$$ \begin{align*}X_d = \{z_0^{d-2}A_1 +z_{0}^{d-3}z_1A_2+\dots +z_1^{d-2}A_{d-1} = 0\}\end{align*} $$

with $A_i \in k[z_0,\dots ,z_{n+1}]_2$ . Note that $X_d$ contains the smooth $(n-1)$ -dimensional quadric $Q_p = \{z_1 = A_1 = 0\}$ . Hence, by Lemma 3.2 there is an irreducible complete intersection $Y_{2,d-1}\subset \mathbb {P}^{n+2}$ which is birational to $X_d$ .

Consider the quadric $Y_2 = \{z_1u-A_1 = 0\}\subset \mathbb {P}^{n+2}$ with homogeneous coordinates $z_0,\dots ,z_{n+1},u$ , and let $CX_d$ be the cone over $X_d$ as in the proof of Lemma 3.2. The intersection $Y_2\cap CX_d$ has two components: the cone $CQ_p$ over $Q_p$ and the degree $d-1$ hypersurface

$$ \begin{align*}Y_{d-1} = \{z_0^{d-2}u+z_0^{d-3}A_2+z_0^{d-4}z_1A_3+\dots +z_1^{d-3}A_{d-1} = 0\}.\end{align*} $$

Let $W\rightarrow S$ be the fibration constructed in Lemma 3.3 starting from the complete intersection $Y_{2,d-1} = Y_2\cap Y_{d-1}$ . If W is unirational, Lemma 3.3 yields a dominant rational map

. Finally, let

be the dominant rational map in Lemma 3.2. By considering the composition

we get a dominant rational map

, and hence $X_d$ is unirational.

Proof. By Lemma 3.3 with $d = 4$ , there exists a variety W with a morphism onto a rational surface S whose general fiber is a complete intersection of a quadric and a hyperplane. Hence, W has a structure of quadric bundle $W\rightarrow S$ over S with $(n-2)$ -dimensional quadrics as fibers.

By the proof of Lemma 3.3, S is a cone over a conic and it is contained in $Y_2 = \{uz_1 - A = 0\}$ . In particular, any line in S through its vertex is contained in $Y_2$ . Moreover, S is contained in the intersection of $Y_3$ with its tangent cone at v. Hence, the lines in S through its vertex yield a section of the quadric bundle $W\rightarrow S$ and so W is rational. Finally, to conclude it is enough to apply Lemma 3.3.

Proof. Up to a change of coordinates, we may assume that $\Lambda = \{z_0 = z_1 = 0\}$ , $X_4$ is given by

$$ \begin{align*}X_4 = \{z_0^2A+z_0z_1B+z_1^2C = 0\}\subset\mathbb{P}^{n+1} \end{align*} $$

and $Q_{\Lambda } = \{z_1 = A = 0\}$ . The intersection $\{z_1u-A = z_0^2A+z_0z_1B+z_1^2C = 0\}\subset \mathbb {P}^{n+2}_{(z_0,\dots ,z_{n+1},u)}$ has two components: $\{z_1 = A = 0\}$ and

$$ \begin{align*}Y_{2,3} = \{z_1u-A = uz_0^2+z_0B+z_1^2C = 0\}\subset\mathbb{P}^{n+2}_{(z_0,\dots,z_{n+1},u)}. \end{align*} $$

By Lemma 3.2 and its proof, there exists a birational map . To conclude, it is enough to note that $Y_{2,3}$ is a complete intersection of the form covered by Proposition 3.5.

Proof. By Lemma 3.2, $X_4$ is birational to a complete intersection $Y_{2,3} = Y_2\cap Y_3$ , where $Y_2$ is a smooth quadric. By the proof of Lemma 3.2, we have that since $Q_0$ contains a $2$ -plane $Y_2$ contains a $2$ -plane as well. Hence, $Y_2$ is a smooth quadric containing a $2$ -plane and we conclude by Lemma 3.1.

Proof. Slightly modifying the proof of Lemma 3.2, we see that in this case $X_4$ is birational to a complete intersection $Y_2\cap Y_2'\subset \mathbb {P}^{n+2}$ of two quadrics, and hence, the claim follows from [Reference Colliot-Thélène, Sansuc and Swinnerton-DyerCTSSD87, Proposition 2.3].

Proof. Note that $X_d$ passes through the point $p = [1:0:\dots :0]$ , and the rational map

is birational with birational inverse

where $L = L(w_1,\dots ,w_{n+1})$ . Note that $\varphi ^{-1}$ contracts the divisor $\{L = 0\}$ to the point p. The strict transform of $X_d$ via $\varphi ^{-1}$ is given by

$$ \begin{align*}\widetilde{X}_d = \{w_0^{d-2}(w_0w_1L+LQ) &+ w_0^{d-3}(w_0^2w_1^2A_1^0 + w_0w_1LA_1^1+L^2A_1^2)+ \dots\\ &+L^{d-3}(w_0^2w_1^2A_{d-2}^0 + w_0w_1LA_{d-2}^1+L^2A_{d-2}^2)=0\}, \end{align*} $$

which we rewrite as

$$ \begin{align*}\widetilde{X}_d = \{w_0^{d-1}(w_1L+w_1^2A_1^0)& +w_0^{d-2}L(Q+w_1A_1^1+w_1^2A_2^0)+\dots\\ & +w_0L^{d-2}(A_{d-3}^2+w_1A_{d-2}^1) + L^{d-1}A_{d-2}^2 = 0\}. \end{align*} $$

Finally, substituting $w_0 = \frac {x_0}{x_1}L$ we get the equation cutting out the divisor $Y_{(d-1,2)}\subset \mathbb {P}^1_{(x_0,x_1)}\times \mathbb {P}^{n}_{(w_1:\dots : w_{n+1})}$ in the statement.

Proof. Consider the rational map

(3.11)

where $\eta _i = B_ix_0^{2}+B_{i+1}x_0x_1+\dots +B_{3}x_0^{i-1}x_1^{3-i}$ for $i = 1,2,3$ . By [Reference OttemOtt15, Theorem 1.1 (ii)] the rational map

yields a small transformation

, where $Y_{(3,2)}^{+}\subset \mathbb {P}^{2}_{(u_1,u_2,u_{3})}\times \mathbb {P}^{n}_{(w_1,\dots , w_{n+1})}$ is cut out by the minors of order three of the following matrix:

(3.12) $$ \begin{align} M_{(u_1,u_2,u_{3})} = \left( \begin{array}{ccc} 0 & u_1 & B_0\\ -u_1 & u_2 & B_1\\ -u_{2} & u_{3} & B_{2}\\ -u_{3} & 0 & B_{3} \end{array} \right). \end{align} $$

Consider the point $p = ([1:0],[0:\dots :0:1])\in Y_{(3,2)}$ , its image

$$ \begin{align*}q &= ([B_1(0,\dots,0,1),B_{2}(0,\dots,0,1)],B_{3}(0,\dots,0,1)])\\ &= ([Q(0,\dots,0,1),A_{1}^2(0,\dots,0,1),A_{2}^2(0,\dots,0,1)]) \end{align*} $$

via $\eta $ and set $\overline {u}_1 = Q(0,\dots ,0,1),\overline {u}_{2} = A_{1}^2(0,\dots ,0,1),\overline {u}_{3} = A_{2}^2(0,\dots ,0,1)$ . Let $F_{\overline {u}}$ be the fiber of the first projection $\pi _1:\mathbb {P}^{2}_{(u_1,u_2,u_{3})}\times \mathbb {P}^{n}_{(w_1,\dots , w_{n+1})}\rightarrow \mathbb {P}^{2}_{(u_1,u_2,u_{3})}$ over $\overline {u} = [\overline {u}_1:\overline {u}_2:\overline {u}_{3}]$ . Then

$$ \begin{align*}F_{\overline{u}} = \{\operatorname{\mathrm{rank}}(M_{(\overline{u}_1,\overline{u}_2,\overline{u}_{3})}) < 3\}\subset \mathbb{P}^{n}_{(w_1,\dots, w_{n+1})} \end{align*} $$

is a complete intersection of two quadrics. Note that $q\in F_{\overline {u}}$ and since the $\overline {u}_i$ are general, $F_{\overline {u}}$ is smooth. Therefore, if $n\geq 4$ [Reference Colliot-Thélène, Sansuc and Swinnerton-DyerCTSSD87, Proposition 2.3] yields the unirationality of $F_{\overline {u}}$ . The strict transform of $F_{\overline {u}}$ via $\eta ^{+}$ is given by

$$ \begin{align*}\widetilde{F}_{\overline{u}} = \left\lbrace \operatorname{\mathrm{rank}}\left(\begin{array}{ccc} \overline{u}_1 & \overline{u}_2 & \overline{u}_3 \\ B_1x_0^2+B_2x_0x_1+B_3x_1^2 & B_2x_0^2+B_3x_0x_1 & B_3x_0^2 \end{array}\right) < 2\right\rbrace \subset \mathbb{P}^1_{(x_0,x_1)}\times \mathbb{P}^{n}_{(w_1,\dots, w_{n+1})}. \end{align*} $$

So $\widetilde {F}_{\overline {u}}$ is unirational and maps dominantly onto $\mathbb {P}^1_{(x_0,x_1)}$ . Finally, to conclude it is enough to note that $Y_{(3,2)}\rightarrow \mathbb {P}^1_{(x_0,x_1)}$ is a fibration in quadric hypersurfaces and to apply Proposition 2.7 and Remark 2.8.

Proof. The equation of $X_4\subset \mathbb {P}^{n+1}$ can be written as in Lemma 3.9 for $d = 4$ . Hence, the claim follows from Lemma 3.9 and Proposition 3.10.

Proof. By Lemma 3.9, it is enough to prove that a general divisor $Y_{3,2}\subset \mathbb {P}^1\times \mathbb {P}^3$ as in Proposition 3.10 has dense k-points. Consider the $2$ -plane $H = \{x_1 = w_1 = 0\}\subset Y_{3,2}$ and the rational map

induced by (3.11). Note that $\eta '$ maps H dominantly onto $\mathbb {P}^{2}_{(u_1,u_2,u_{3})}$ . Take a general point $p\in H$ , its image $q = \eta '(p)$ and a general line $L = \{\alpha _1u_1+\alpha _2u_2+\alpha _3u_3 = 0\}\subset \mathbb {P}^{2}_{(u_1,\dots ,u_{3})}$ through q. Set $S_{L,p} = \eta ^{\prime -1}(L)$ . Since $Y_{(3,2)}$ is general $\eta '$ restricts to a morphism $\eta ^{\prime }_{|S_{L,p}}:S_{L,p}\rightarrow \mathbb {P}^1$ . A straightforward computation shows that $S_{L,p}$ is smooth, and by the argument in the second part of the proof of Proposition 3.10, the general fiber of $\eta ^{\prime }_{|S_{L,p}}:S_{L,p}\rightarrow \mathbb {P}^1$ is a smooth curve of genus one.

Furthermore, $C = H\cap S_{L,p} = \{\alpha _1B_1+\alpha _2B_2+\alpha _3B_3 = x_1 = w_1 = 0\}$ . Hence, since $Y_{3,2}$ is general, C is a smooth conic with a point $p\in C$ . Note that a general fiber of $\eta ^{\prime }_{|S_{L,p}}:S_{L,p}\rightarrow \mathbb {P}^1$ intersects a general fiber of the quadric bundle $Y_{3,2}\rightarrow \mathbb {P}^1$ in a zero-dimensional scheme of degree eight. So, $C\subset S_{L,p}$ is a rational $4$ -section of $\eta ^{\prime }_{|S_{L,p}}$ .

Consider the fiber product $T = C\times _{L}S_{L,p}$ . Since $p,L$ and $Y_{3,2}$ are general, the ramification divisor of $\eta ^{\prime }_{|C}:C\rightarrow \mathbb {P}^1$ is disjoint from the singular fibers of $\eta ^{\prime }_{|S_{L,p}}$ . Hence, T is smooth and the proof of [Reference Harris and TschinkelHT00, Theorem 8.1] goes through. So the set $S_{L,p}(k)$ of the k-rational points of $S_{L,p}$ is Zariski dense in $S_{L,p}$ . Finally, letting the point p vary in H and the line L vary among the lines passing through $q = \eta '(p)$ we get the claim.

Proof. The case $h = n-1$ comes from Proposition 3.13. Let be the projection from $H = \left \langle p,\Lambda \right \rangle $ , $\widetilde {\pi }_{H}:\widetilde {X}_4\rightarrow \mathbb {P}^{n-h-1}$ , and $F_p\cong \mathbb {P}^{n-h-1}$ the fiber over p of the blowup $\widetilde {X}_4\rightarrow X_4$ along $H\cap X_4$ . Note that since $X_4$ is irreducible the blowup $\widetilde {X}_4$ is also irreducible.

The generic fiber $\widetilde {X}_{4,\eta }$ of $\widetilde {\pi }_{H}:\widetilde {X}_4\rightarrow \mathbb {P}^{n-h-1}$ is a quartic hypersurface $\widetilde {X}_{4,\eta }\subset \mathbb {P}^{h+2}_{k(t_1,\dots ,t_{n-h-1})}$ with double points along an h-plane and with a $k(t_1,\dots ,t_{n-h-1})$ -rational point induced by $F_p$ .

Therefore, Proposition 3.13 yields that $\widetilde {X}_{4,\eta }$ is unirational over $k(t_1,\dots ,t_{n-h-1})$ and since $\widetilde {X}_4$ is irreducible $\widetilde {X}_4$ is unirational.

Now, assume k to be $C_r$ . The exceptional divisor $\widetilde {E}$ in Lemma 2.4 is a divisor of bidegree $(2,2)$ in $\mathbb {P}^{n-h}\times \mathbb {P}^h$ , and since $X_4$ is general $\widetilde {E}$ maps dominantly onto $\mathbb {P}^{n-h}$ . Furthermore, since both the fibrations on $\widetilde {E}$ have quadric hypersurfaces as fibers and $s+1> 2^r$ by Remark 2.10 the divisor $\widetilde {E}$ has a point. To conclude, it is enough to apply [Reference MassarentiMas22, Theorem 1.8] and Proposition 2.7.

Proof. It is enough to substitute $z_0 = \frac {x_0}{x_1}z_1$ in the equation of $X_d$ and to clear the denominators of the resulting polynomial.

Proof. We may assume that $\Lambda = \{z_0 = z_1 = 0\}$ and $q = [1:0:\dots :0]$ so that the equation of $X_d\subset \mathbb {P}^{n+1}$ can be written as in Lemma 3.18. First, consider the case $d = 4$ . By Lemma 3.18 $X_4$ is birational to a divisor $Y_{(2,2)}\subset \mathbb {P}^1_{(x_0,x_1)}\times \mathbb {P}^{n}_{(z_1,\dots , z_{n+1})}$ of bidegree $(2,2)$ . The point $p\in X_d\setminus \Lambda $ yields a point $p'\in Y_{(2,2)}$ . Let H be a general $2$ -plane in $\mathbb {P}^{n}_{(z_1,\dots , z_{n+1})}$ through the projection of $p'$ and set $\Gamma = \mathbb {P}^1_{(x_0,x_1)}\times H$ . Then $W = Y_{(2,2)}\cap \Gamma $ is a divisor of bidegree $(2,2)$ in $\mathbb {P}^1_{(x_0,x_1)} \times \mathbb {P}^2$ which by [Reference Kollár and MellaKM17, Corollary 8] is unirational. Since W maps dominantly onto $\mathbb {P}^1_{(x_0,x_1)}$ , to conclude it is enough to apply Proposition 2.7.

Now, let $d = 5$ . Then by Lemma 3.18 $X_5$ is birational to a divisor $Y_{(3,2)}\subset \mathbb {P}^1_{(x_0,x_1)}\times \mathbb {P}^{n}_{(z_1,\dots , z_{n+1})}$ of bidegree $(3,2)$ . This divisor is not of the same form of those considered in Proposition 3.10. However, the argument in the proof of Proposition 3.10 goes through smoothly and shows $Y_{(3,2)}$ is unirational.

Proof. If $n-h = 1$ , then the projection $\pi _{2|D}:D\rightarrow \mathbb {P}^h$ is birational. Assume $n-h\geq 2$ . The generic fiber of $\pi _{2|D}:D\rightarrow \mathbb {P}^h$ is an $(n-h-1)$ -dimensional projective space over the function field of $\mathbb {P}^h$ . So we can find an open subset $U\subset \mathbb {P}^h$ such that $\pi _{2|D}^{-1}(U)$ is birational to $U\times \mathbb {P}^{n-h-1}$ . Finally, since D is irreducible we conclude that D is rational.

Proof. Consider the blowup $\widetilde {X}_4$ along $\Lambda $ in Lemma 2.4. Under our hypotheses, $X_4$ cannot be singular at the general point of $\Lambda $ . So the exceptional divisor $\widetilde {E}\subset \mathbb {P}^{n-h}\times \mathbb {P}^h$ has bidegree $(1,3)$ , and Lemma 3.21 yields that $\widetilde {E}$ is unirational. The projection $\pi :\widetilde {X}_4\rightarrow \mathbb {P}^{n-h}$ is a fibration in h-dimensional cubic hypersurfaces. Therefore, thanks to Proposition 2.7 and Remark 2.8 to conclude it is enough to show that $\pi _{|\widetilde {E}}:\widetilde {E}\rightarrow \mathbb {P}^{n-h}$ is dominant.

Since $X_4$ is not a cone over a quartic of smaller dimension by the expression of $\widetilde {X}_4$ in (2.6), we see that $\pi _{|\widetilde {E}}:\widetilde {E}\rightarrow \mathbb {P}^{n-h}$ is not dominant if and only if the equation of $X_d$ is of the following form

$$ \begin{align*}X_d = \left\{\sum_{i=0}^{n-h}z_i(C_i(z_0,\dots,z_{n-h})+c_iP) = 0\right\}\subset\mathbb{P}^{n+1}, \end{align*} $$

where $C_i\in k[z_0,\dots ,z_{n-h}]_{d-1}$ , $c_i\in k$ and $P\in k[z_0,\dots ,z_{n+1}]_{d-1}$ . In particular,

$$ \begin{align*}Z = \{z_0 = \dots = z_{n-h} = P = 0\}\subset \Lambda\cap\operatorname{\mathrm{Sing}}(X_d)\end{align*} $$

and since $\dim (Z)\geq h-1$ we get a contradiction.

Proof. Note that $n> 2h-1$ yields that a general quartic containing a fixed h-plane is smooth. The claim follows from Theorem 3.22.

Proof. Let $\mathcal {C}'$ be the generic fiber of $\mathcal {C}\rightarrow W$ . Then $\mathcal {C}'$ is a cubic hypersurface over the function field $F = k(t_1,\dots ,t_{\dim (W)})$ . Since $m+2> 3^{r+\dim (W)}$ , Remark 2.10 yields that $\mathcal {C}'$ has a point. Hence, by [Reference KollárKol02, Theorem 1] $\mathcal {C}'$ is unirational over F and so since $\mathcal {C}$ is irreducible $\mathcal {C}$ is unirational.

Proof. Since $N-1> 2^r - 2$ by Remark 2.10 $Q^{N-1}$ has a point. Hence, $Q^{N-1}$ is rational and in particular the set of its rational points is dense. Take a point $x_0 \in Q^{N-1}$ .

The tangent space $T_{x_0}Q^{N-1}$ cuts out on $Q^{N-1}$ a cone with vertex $x_0$ over an $(N-3)$ -dimensional quadric $Q^{N-3}$ . Since $N-3> 2^r - 2$ Remark 2.10 yields the existence of a point $x_1 \in Q^{N-3}$ . The line $\left \langle x_0,x_1\right \rangle $ is therefore contained in $Q^{N-1}$ .

Now, $T_{x_1}Q^{N-3}\cap Q^{N-3}$ is a cone with vertex $x_1$ over an $(N-5)$ -dimensional quadric $Q^{N-5}$ which again by Remark 2.10 has a point $x_2\in Q^{N-5}$ . The $2$ -plane $\left \langle x_0,x_1,x_2\right \rangle $ is contained in $Q^{N-1}$ .

Proceeding recursively, we have that $T_{x_{s-1}}Q^{N-2(s-1)-1}\cap Q^{N-2(s-1)-1}$ is a cone with vertex $x_{s-1}$ over an $(N-2s-1)$ -dimensional quadric $Q^{N-2s-1}$ . Since $N-2s+1> 2^r$ , the quadric $Q^{N-2s-1}$ has a point $x_s$ and the s-plane $\left \langle x_0,\dots ,x_s\right \rangle $ is then contained in $Q^{N-1}$ .

Proof. Note that since X is smooth $Q_i$ must be smooth along X. Let H be a general c-plane containing $\Lambda $ . Then H intersects $Q_i$ along $\Lambda \cup H_i$ where $H_i$ is a $(c-1)$ -plane. Hence, we get c linear subspaces of dimension $c-1$ of $H\cong \mathbb {P}^c$ .

Since H is general, these $(c-1)$ -planes intersect in a point $x_H = H_1\cap \dots \cap H_c \in X$ . To conclude, it is enough to parametrize X with the $\mathbb {P}^{N-c}$ of c-planes containing $\Lambda $ .

Proof. For $s = 0$ , the claim follows from Remark 2.10. We proceed by induction on s. Since

$$ \begin{align*}N-(s-1)(c+1)+1> N-s(c+1)+1 > 2^rc \end{align*} $$

through any point of X, there is an $(s-1)$ -plane. Fix one of these $(s-1)$ -planes, and denote it by $\Lambda ^{s-1}\subset X$ . The s-planes in $\mathbb {P}^{N}$ containing $\Lambda ^{s-1}$ are parametrized by $\mathbb {P}^{N-s}$ .

Arguing as in the proof of Lemma 4.2, we see that requiring such an s-plane to be contained in $Q_i$ yields s linear equations given by the tangent spaces of $Q_i$ at s general points of $\Lambda ^{s-1}$ plus a quadratic equation induced by $Q_i$ itself.

Hence, the points of $\mathbb {P}^{N-s}$ corresponding to s-planes contained in X are parametrized by a subvariety cut out by $sc$ linear equations and c quadratic equations that is an intersection Y of c quadrics in $\mathbb {P}^{N-s-sc}$ . Since $\dim (Y) = N-s(c+1)-c> 2^rc-1-c \geq -1$ and $N-s(c+1)+1> 2^rc$ , Remark 2.10 yields that Y has a point and so there is an s-plane through $\Lambda ^{s-1}$ contained in X.

Proof. By Proposition 4.4 with $s = c-1$ , the complete intersection X contains a $(c-1)$ -plane, and hence, the claim follows from Lemma 4.3.

Proof. Note that $X^{n}_5$ is a hypersurface of the form (2.12) in Proposition 2.11. Hence, $X^{n}_5$ has multiplicity three along $\Lambda = \{x_0 = \dots = x_{n-h} = 0\}$ , and multiplicity two along $\Lambda ' = \{y_0 = \dots = y_h = x_{n-h} = 0\}$ .

First, consider $\text {(i)}$ . Fix a point $p \in \Lambda '$ , say $p = [1:0:\dots : 0]$ . The lines through p intersecting $X_5^n$ at p with multiplicity at least four are parametrized by the variety

$$ \begin{align*}Y = \{A_{3,\dots,0} = x_1A_{2,1,0,\dots,0} + \dots + x_{n-h}A_{2,0,\dots,0,1} = 0\}\subset\mathbb{P}^{n}_{(x_1,\dots,x_{n-h},y_0,\dots,y_h)}.\end{align*} $$

Assume Y to be unirational. Associating to a general point $y\in Y$ , representing a line $l_y$ , the fifth intersection point of $l_y$ and $X_5^n$ we get a rational map

Set $Y_{\psi } = \overline {\psi (Y)}\subset X_5^n$ , and let

be the restriction to $X_5^n$ of the projection from $\Lambda $ . We want to prove that

is dominant. Indeed, if so $Y_{\psi }$ would be a unirational variety transverse to the quadric fibration induced by the projection from $\Lambda $ and dominating $\mathbb {P}^{n-h}$ , and Proposition 2.7 would imply that $X_5^n$ is unirational.

Since $Y_{\psi }$ is unirational it is enough to prove that the induced map between the algebraic closures is dominant. Consider the complete intersection

$$ \begin{align*}Z = Y\cap X^n_5 = \{A_{3,\dots,0} = x_1A_{2,1,0,\dots,0} + \dots + x_{n-h}A_{2,0,\dots,0,1} = \overline{f}= 0\}\subset \mathbb{P}^{n+1}_{(x_0,\dots,x_{n-h},y_0,\dots,y_h)}. \end{align*} $$

A general point of $z\in Z\subset X^n_5$ represents a line $l_z$ intersecting $X^n_5$ with multiplicity four at p and multiplicity one at z. Hence, $Z\subset Y_{\psi }$ .

Fix a general point $q\in \mathbb {P}^{n-h}$ , and consider

$$ \begin{align*}Z_q = \overline{\pi}_{|Y_{\psi}}^{-1}(q)\cap Z.\end{align*} $$

Note that since $q\in \mathbb {P}^{n-h}$ is defined by $n-h$ equations, $Z\subset \mathbb {P}^{n+1}_{(x_0,\dots ,x_{n-h},y_0,\dots ,y_h)}$ is cut out by three equations and by hypothesis $n+1-(n-h)-3 = h-2> 0$ . So $Z_q$ is nonempty. Then $\overline {\pi }_{|W_{\psi }}$ is dominant when restricted to $\overline {Z}$ , and hence, it is dominant.

Now, we prove that Y is unirational. In $\mathbb {P}^{n}_{(x_1,\dots ,x_{n-h},y_0,\dots ,y_h)}$ , fix the point $p' = [1:0:\dots :0]$ . The quadric

$$ \begin{align*}\{x_1 = \dots = x_{n-h} = A_{2,1,0,\dots,0}\}\subset\mathbb{P}^{h}_{(y_0,\dots,y_h)}\end{align*} $$

parametrizes lines that are contained in

$$ \begin{align*}Y_3 = \{x_1A_{2,1,0,\dots,0} + \dots + x_{n-h}A_{2,0,\dots,0,1}\}.\end{align*} $$

Indeed, the tangent cone of $Y_3$ at $p'$ is defined by $\{A_{2,1,0,\dots ,0} = 0\}$ and $\{x_1 = \dots = x_{n-h}= 0\}\subset Y_3$ . So these lines intersect $Y_3$ with multiplicity three at $p'$ and at another point in the linear space $\{x_1 = \dots = x_{n-h}= 0\}$ .

Set $Y_2 = \{A_{3,\dots ,0} = 0\}$ , and let W be the cone over $W'$ with vertex $p'$ . Since $p'$ is in the vertex of $Y_2$ , we have that

$$ \begin{align*}W\subset Y = Y_2\cap Y_3.\end{align*} $$

Furthermore, since $W'$ contains a line by Lemma 4.3 it is rational, and hence, W is rational as well.

As in the proof of Proposition 3.4, we construct a quadric bundle $\widetilde {\mathcal {Q}}\rightarrow W$ with $(n-4)$ -dimensional fibers whose general point $(w,l_w)$ represents a point $w\in W$ and a line $l_w$ which is contained in $Y_2$ and intersects $Y_3$ with multiplicity two at w.

Associating to $(w,l_w)$ the third point of intersection of $l_w$ and $Y_3$ , we get a rational map and arguing as in the proof of Proposition 3.4 we see that such rational map is dominant.

Now, we consider $\text {(ii)}$ . Note that

$$ \begin{align*}\widetilde{Q}' = \{x_1 = \dots = x_{n-h} = A_{3,\dots,0} = 0\}\subset X^n_5\subset\mathbb{P}^{n+1} \end{align*} $$

is an h-dimensional quadric cone over $\widetilde {Q}$ with vertex $[1:0:\dots :0]$ . Since $\widetilde {Q}$ contains a $2$ -plane $\widetilde {Q}'$ contains a $3$ -plane $H\subset X^n_5$ . If $x\in H$ is a general point the lines through x intersecting $X^n_5$ with multiplicity four at x are parametrized by a complete intersection of a quadric and a cubic in $\mathbb {P}^{n-1}$ . Hence, we get a fibration $\mathcal {Y}_{2,3}\rightarrow H$ whose fiber over a general $x\in H$ is a complete intersection $Y_{2,3,x}$ of a quadric $Y_{2,x}\subset \mathbb {P}^{n-1}$ and a cubic $Y_{3,x}\subset \mathbb {P}^{n-1}$ . Note that since $X_5^n$ is general among the quintics satisfying $(ii)$ for $x\in H$ both $Y_{2,x}$ and $Y_{2,3,x}$ are smooth.

The generic fiber of $\mathcal {Y}_{2,3}\rightarrow H$ is then a complete intersection $\mathcal {Y}_{2,3,k(H)} = \mathcal {Y}_{2,k(H)}\cap \mathcal {Y}_{3,k(H)}$ satisfying the hypotheses of Lemma 3.1. Indeed, by considering the $2$ -plane parametrizing lines in H through a general $x\in H$ we get a $2$ -plane over $k(H)$ contained in $\mathcal {Y}_{2,k(H)}$ .

Therefore, by Lemma 3.1 $\mathcal {Y}_{2,3,k(H)}$ is unirational over $k(H)$ . As usual after replacing H by an open subset, we may assume that $\mathcal {Y}_{2,3}$ is irreducible and so we get that $\mathcal {Y}_{2,3}$ is unirational. Now, a general point of $\mathcal {Y}_{2,3}$ represents a pair $(x,l_x)$ with $x\in H$ general and $l_x$ a line intersecting $X_5^n$ with multiplicity four at x. As usual, associating to $(x,l_x)\in \mathcal {Y}_{2,3}$ the fifth point of intersection of $l_x$ with $X_5^n$ we get a rational map

Note that $\dim (\mathcal {Y}_{2,3}) = 3+(n-3) = n$ . Arguing as in the last part of the proof of Proposition 3.4, we see that $\phi $ is generically finite and therefore it is dominant.

Proof. By Proposition 2.11, D is birational to a quintic hypersurface $X^n_{5}\subset \mathbb {P}^{n+1}$ of the form considered in Proposition 4.6.

Consider $\text {(i)}$ . Since, $h> 2^{r+1}+2$ by Proposition 4.4 the complete intersection $W'$ in Proposition 4.6 contains a line, and hence, Proposition 4.6 yields that $X^n_{5}$ is unirational.

For $\text {(ii)}$ , note that since $h> 2^{r}+3$ Lemma 4.2 implies that the quadric $\widetilde {Q}$ in Proposition 4.6 contains a $2$ -plane.

Now, consider $\text {(iii)}$ . The projection $\pi _2:D\rightarrow \mathbb {P}^{h+1}$ endows D with a structure of fibration in $(n-h-1)$ -dimensional cubic hypersurfaces. Take a general line $L\subset \mathbb {P}^{h+1}$ , and set $\mathcal {C}_L = \pi _2^{-1}(L)$ . Then $\mathcal {C}_L$ is a fibration in $(n-h-1)$ -dimensional cubic hypersurfaces over $\mathbb {P}^1$ and since $n-h-1> 3^{r+1}-2$ Lemma 4.1 yields that $\mathcal {C}_L$ is unirational. Finally, to conclude it is enough to note that $\pi _{|\mathcal {C}_L}:\mathcal {C}_L\rightarrow \mathbb {P}^{n-h}$ is dominant and to apply Proposition 2.7.

Proof. By Lemma 2.4, the exceptional divisor $\widetilde {E}\subset \widetilde {X}_5$ is a divisor of bidegree $(3,2)$ in $\mathbb {P}^{n-h}\times \mathbb {P}^h$ . Since $X_5$ is general, $\widetilde {E}$ maps dominantly onto $\mathbb {P}^{n-h}$ , and hence, to conclude we apply Proposition 2.7 and Theorem 4.7. For the second statement, it is enough to argue as in the previous case on a divisor of bidegree $(2,3)$ in $\mathbb {P}^{n-h}\times \mathbb {P}^h$ .

Proof. The exceptional divisor $\widetilde {E}\subset \widetilde {X}_5$ is a general divisor of bidegree $(3,2)$ in $\mathbb {P}^{1}\times \mathbb {P}^{n-1}$ . By [Reference MassarentiMas22, Corollary 4.13, Lemma 4.18] and Remark 2.10, under our hypotheses $\widetilde {E}$ has a point and hence [Reference MassarentiMas22, Theorem 1.8] yields that $\widetilde {E}$ is unirational. To conclude, it is enough to argue as in the proof of Theorem 4.10 applying Propositions 2.7.

Proof. In this case, the exceptional divisor $\widetilde {E}\subset \widetilde {X}_d$ is a general divisor of bidegree $(d-2,2)$ in $\mathbb {P}^{n-h}\times \mathbb {P}^{h}$ . Note that the discriminant of the quadric bundle $\widetilde {E}\rightarrow \mathbb {P}^{n-h}$ has degree $(h+1)(d-2)$ . Hence, the claim follows from [Reference MassarentiMas22, Theorem 1.7] and Propositions 2.7.