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Proof. Since $p$ is flat and finitely presented, the image of $p$ is open. Up to replacing $X$ by the open image of $p$, assume that $p$ is surjective.

The argument is essentially a Bertini hyperplane argument. First of all, up to replacing $Y$ by a disjoint union of open affine subschemes, assume that $Y$ is affine. Denote by $Y_{\text{equi}}\subset Y$ the maximal open subset on which $p$ is (locally) equidimensional, i.e., at every point of every fiber of $p$, the irreducible components of the fiber containing the point all have equal dimensions. The restriction of $p$ to $Y_{\text{equi}}$ is still surjective. Thus, without loss of generality, assume that $p$ is (locally) equidimensional. Then on every connected component $Y_\alpha$ of $Y$, the fiber dimension of $p$ is constant. We construct $Z$ as a disjoint union of schemes $Z_\alpha$ where $i_\alpha:Z_\alpha\to Y_\alpha$ is a disjoint union of locally closed immersions such that $p_\alpha\circ i_\alpha$ is a quasi-finite, flat morphism with image equal to the image of $p_\alpha$.

Let $e:Y\hookrightarrow \mathbb{A}^n_X$ be a closed immersion of $X$-schemes. For every point in the image of $p$, $x:\text{Spec}\ k \to X,$ the fiber $Y_x=p^{-1}(x)$ has only finitely many associated points. For every linear polynomial $t_x$ in the coordinate ring $k[\mathbb{A}^n_k]=k[t_1,\dots,t_n]$ that vanishes at none of these finitely many associated points, multiplication by $t_x$ is injective on $\mathcal{O}_{Y_x}$. Let $t\in\mathcal{O}_{X,x}[t_1,\dots,t_n]$ be an element mapping to $t_x$. Denote by $H \subseteq \mathbb{A}^{n}_{\mathcal{O}_{X,x}}$ the zero scheme of $t$.

Proof. Since $p$ is flat and finitely presented, the image of $p$ is open. Up to replacing $X$ by the open image of $p$, assume that $p$ is surjective. Edit. Also, since $X$ is quasi-compact, it has a finite cover by open affines, $X_\beta$. For every collection $i_\beta:Z_\beta \to Y\times_X X_\beta$ as in the proposition, for the disjoint union $Z=\sqcup_\beta Z_\beta$ and for the unique morphism $i:Z\to Y$ whose restriction to each $Z_\beta$ equals $i_\beta$, also $i:Z\to Y$ satisfies the conditions in the proposition. Thus, it suffices to prove the result when $X$ is affine.

The argument is essentially a Bertini hyperplane argument. First of all, up to replacing $Y$ by a disjoint union of open affine subschemes, assume that $Y$ is affine. Denote by $Y_{\text{equi}}\subset Y$ the maximal open subset on which $p$ is (locally) equidimensional, i.e., at every point of every fiber of $p$, the irreducible components of the fiber containing the point all have equal dimensions. Edit. This open subset contains the open complement $U$ of the closed union $C$ of the finitely many intersections $Y_i\cap Y_j$ of irreducible components $Y_i$ and $Y_j$ of $Y$ such that the (constant) relative dimensions of $Y_i$ and $Y_j$ over $X$ are not equal. For every point $x\in X$, there exists an irreducible component $Y_i$ such that the fiber $(Y_i)_x$ has maximal dimension. Since $p$ is flat, $U\cap (Y_i)_x$ is dense in $(Y_i)_x$: if $(Y_j)_x\cap (Y_i)_x$ is nonempty, and if $\text{dim}(Y_j)_x\neq\text{dim}(Y_i)_x$, then $(Y_j)_x$ has smaller relative dimension, so $(Y_i)_x\cap (Y_j)_x$ is nowhere dense in $(Y_i)_x$.

Thus, the restriction of $p$ to $Y_{\text{equi}}$ is still surjective. Thus, without loss of generality, assume that $p$ is (locally) equidimensional. Then on every connected component $Y_\alpha$ of $Y$, the fiber dimension of $p$ is constant. We construct $Z$ as a disjoint union of schemes $Z_\alpha$ where $i_\alpha:Z_\alpha\to Y_\alpha$ is a disjoint union of locally closed immersions such that $p_\alpha\circ i_\alpha$ is a quasi-finite, flat morphism with image equal to the image of $p_\alpha$.

Let $e:Y\hookrightarrow \mathbb{A}^n_X$ be a closed immersion of $X$-schemes. For every point in the image of $p$, $x:\text{Spec}\ k \to X,$ the fiber $Y_x=p^{-1}(x)$ has only finitely many associated points.
Edit. Since $Y_x$ has dimension $d>0$, there exist points $y\in Y$ that specialize to none of these finitely many associated points. By prime avoidance, there exists an element $t_x\in k[\mathbb{A}^n_k]=k[t_1,\dots,t_n]$ that is in the prime ideal of $y$ yet in none of the associated primes. For every linear polynomial $t_x$ in the coordinate ring $k[\mathbb{A}^n_k]=k[t_1,\dots,t_n]$ that vanishes at none of these finitely many associated points, Multiplication by $t_x$ is injective on $\mathcal{O}_{Y_x}$. Let $t\in\mathcal{O}_{X,x}[t_1,\dots,t_n]$ be an element mapping to $t_x$. Denote by $H \subseteq \mathbb{A}^{n}_{\mathcal{O}_{X,x}}$ the zero scheme of $t$.

Let $e:Y\hookrightarrow \mathbb{A}^n_X$ be a closed immersion of $X$-schemes. For every point in the image of $p$, $x:\text{Spec}\ k \to X,$ the fiber $Y_x=p^{-1}(x)$ has only finitely many associated points. For every linear polynomial $t_x$ in the coordinate ring $k[\mathbb{A}^n_k]=k[t_1,\dots,t_n]$ that vanishes at none of these finitely many associated points, multiplication by $t_x$ is injective on $\mathcal{O}_{Y_x}$. Let $t\in\mathcal{O}_{X,x}[t_1,\dots,t_n]$ be an element mapping to $t_x$. Denote by $H$ the zero scheme of $t$.

By the local flatness criterion, the intersection $H\cap Y$ is flat at $x$, cf. Theorem 22.5, p. 176, "Commutative ring theory", H. Matsumura, Cambridge studies in mathematics, vol 8. By the usual arguments, there exists an open affine neighborhood of $x$ such that $t$ lifts to a section of the structure sheaf of $\mathbb{A}^n_X$ over this open. Define $H$ to be the zero scheme of $t$ on this open. By openness of the flat locus, etc., there exists an open subscheme $U$ of $Y$ containing $H\cap Y_x$ such that $H\cap U$ is flat over $X$. Choosing $t_x$ so that $H\cap Y_x$ is nonempty, $U\cap H \to X$ has image containing $p$, and it is equidimensional of dimension $d-1$ (this last by Krull's Hauptidealsatz).

Thus, by the induction hypothesis applied to $U\cap H \to X$, there exists a locally closed immersion $i_x:Z_x\to U\cap H$ such that $p\circ i_x$ is quasi-finite and flat with image $V_x$ containing $x$. As we vary $x$, the open images $V_x$ cover $X$. Since we assumed that $X$ is quasi-compact, finitely many of these opens suffice to cover $X$. Define $i:Z\to Y$ to be the disjoint union of these finitely many locally closed immersions $i_x$. Since the composition $p\circ i$ is quasi-finite and flat on each of the finitely many connected components, it is quasi-finite and flat. By construction, the image equals the image of $p$. Thus, the lemma is proved by induction on $d$. QED

Let $e:Y\hookrightarrow \mathbb{A}^n_X$ be a closed immersion of $X$-schemes. For every point in the image of $p$, $x:\text{Spec}\ k \to X,$ the fiber $Y_x=p^{-1}(x)$ has only finitely many associated points. For every linear polynomial $t_x$ in the coordinate ring $k[\mathbb{A}^n_k]=k[t_1,\dots,t_n]$ that vanishes at none of these finitely many associated points, multiplication by $t_x$ is injective on $\mathcal{O}_{Y_x}$. Let $t\in\mathcal{O}_{X,x}[t_1,\dots,t_n]$ be an element mapping to $t_x$. Denote by $H \subseteq \mathbb{A}^{n}_{\mathcal{O}_{X,x}}$ the zero scheme of $t$.

By the local flatness criterion, the intersection $H\cap Y$ is flat over $\mathcal{O}_{X,x}$, cf. Theorem 22.5, p. 176, "Commutative ring theory", H. Matsumura, Cambridge studies in mathematics, vol 8. By the usual arguments, there exists an open affine neighborhood $W \subset X$ of $x$ such that $t$ lifts to a section $t_{W}$ of the structure sheaf of $\mathbb{A}^n_W$. Define $H_{W} \subset \mathbb{A}^{n}_{W}$ to be the zero scheme of $t_{W}$. By openness of the flat locus, etc., there exists an open subscheme $U$ of $Y$ containing $H\cap Y_x$ such that $H_{W}\cap U$ is flat over $X$. Choosing $t_x$ so that $H\cap Y_x$ is nonempty, $U\cap H_{W} \to X$ has image containing $p$, and it is equidimensional of dimension $d-1$ (this last by Krull's Hauptidealsatz).

Thus, by the induction hypothesis applied to $U\cap H_{W} \to X$, there exists a locally closed immersion $i_x:Z_x\to U\cap H_{W}$ such that $p\circ i_x$ is quasi-finite and flat with image $V_x$ containing $x$. As we vary $x$, the open images $V_x$ cover $X$. Since we assumed that $X$ is quasi-compact, finitely many of these opens suffice to cover $X$. Define $i:Z\to Y$ to be the disjoint union of these finitely many locally closed immersions $i_x$. Since the composition $p\circ i$ is quasi-finite and flat on each of the finitely many connected components, it is quasi-finite and flat. By construction, the image equals the image of $p$. Thus, the lemma is proved by induction on $d$. QED

Let $e:Y\hookrightarrow \mathbb{A}^n_X$ be a closed immersion of $X$-schemes. For every point in the image of $p$, $x:\text{Spec}\ k \to X,$ the fiber $Y_x=p^{-1}(x)$ has only finitely many associated points. For every linear polynomial $t_x$ in the coordinate ring $k[\mathbb{A}^n_k]=k[t_1,\dots,t_n]$ that vanishes at none of these finitely many associated points, multiplication by $t_x$ is injective on $\mathcal{O}_{Y_p}$. Let $t\in\mathcal{O}_{X,x}[t_1,\dots,t_n]$ be an element mapping to $t_x$. Denote by $H$ the zero scheme of $t$.

There exists a norm map, $$\text{Nm}_q : \text{Pic}(q^{-1}(V))\to \text{Pic}(V), \ \ \mathcal{L} \mapsto \text{det}(q_*\mathcal{L})\otimes \text{det}(q_*\mathcal{O})^{\vee}.$$ This is described in detail in Mumford's Lectures on curves on an algebraic surface. For every connected component $V_\alpha$ of $V$, the morphism $q$ has constant finite degree $m$ over that component. For every invertible sheaf $\mathcal{M}_\alpha$ on $V_\alpha$, $\text{Nm}_q(q^*\mathcal{M}_\alpha)$ equals $\mathcal{M}_\alpha^{\otimes m}$. Thus, defining $n$ to equal the least common multiple of these integers $m$, for every invertible sheaf $\mathcal{M}$ on $V$, if $q^*\mathcal{M}$ is trivial, then also $\mathcal{M}^{\otimes n}$ is trivial. Thus, for an invertible sheaf $\mathcal{M}$ on $X$, if $q^*\mathcal{M}$ is trivial, then also $j^*\mathcal{M}^{\otimes n}$ is trivial. QED

Proof. The quotient of $\text{Ker}(p^*)$ by $\text{Pic}^0(X)[p]$ is a subgroup of the Néron-Severi group $\text{Pic}(X)/\text{Pic}^0(X)$. By the "Theorem of the Base" of Lang and Néron, the N´ron-Severi group is finitely generated. Thus, the subgroup is also finitely generated. By the proposition, $\text{Pic}^0(X)[p]$ is finitely generated. Thus, the extension group $\text{Ker}(p^*)$ is also finitely generated. QED

Let $e:Y\hookrightarrow \mathbb{A}^n_X$ be a closed immersion of $X$-schemes. For every point in the image of $p$, $x:\text{Spec}\ k \to X,$ the fiber $Y_x=p^{-1}(x)$ has only finitely many associated points. For every linear polynomial $t_x$ in the coordinate ring $k[\mathbb{A}^n_k]=k[t_1,\dots,t_n]$ that vanishes at none of these finitely many associated points, multiplication by $t_x$ is injective on $\mathcal{O}_{Y_x}$. Let $t\in\mathcal{O}_{X,x}[t_1,\dots,t_n]$ be an element mapping to $t_x$. Denote by $H$ the zero scheme of $t$.

There exists a norm map, $$\text{Nm}_q : \text{Pic}(q^{-1}(V))\to \text{Pic}(V), \ \ \mathcal{L} \mapsto \text{det}(q_*\mathcal{L})\otimes \text{det}(q_*\mathcal{O})^{\vee}.$$ This is described in detail in Mumford's Lectures on curves on an algebraic surface. For every connected component $V_\alpha$ of $V$, the morphism $q$ has constant finite degree $m_{\alpha}$ over that component. For every invertible sheaf $\mathcal{M}_\alpha$ on $V_\alpha$, $\text{Nm}_q(q^*\mathcal{M}_\alpha)$ equals $\mathcal{M}_\alpha^{\otimes m_{\alpha}}$. Thus, defining $n$ to equal the least common multiple of these integers $m_{\alpha}$, for every invertible sheaf $\mathcal{M}$ on $V$, if $q^*\mathcal{M}$ is trivial, then also $\mathcal{M}^{\otimes n}$ is trivial. Thus, for an invertible sheaf $\mathcal{M}$ on $X$, if $q^*\mathcal{M}$ is trivial, then also $j^*\mathcal{M}^{\otimes n}$ is trivial. QED

Proof. The quotient of $\text{Ker}(p^*)$ by $\text{Pic}^0(X)[p]$ is a subgroup of the Néron-Severi group $\text{Pic}(X)/\text{Pic}^0(X)$. By the "Theorem of the Base" of Lang and Néron, the Néron-Severi group is finitely generated. Thus, the subgroup is also finitely generated. By the proposition, $\text{Pic}^0(X)[p]$ is finitely generated. Thus, the extension group $\text{Ker}(p^*)$ is also finitely generated. QED