Why are projective morphisms closed?

A) If you want to extend your rational map $f: X\to Y$ defined outside of $p$ across $p$, it is true that you can asssume that $Y=\mathbb P^n$ by embedding $Y$ into $\mathbb P^n$ as a closed set. This however has nothing to do with the completeness of $\mathbb P^n$ but follows from topology. Indeed, if $f$ sends $X\setminus \{p\}$ into $Y$, it will send the point $p$, which is in the closure of $X\setminus \{p\}$, into the closure of $f(X\setminus \{p\})$ and so in particular into $Y$.

B) Still completeness of $\mathbb P^n$ is a fundamental result that is proved here. This is a hand-out from a course given in 2009 by Mike Artin: I can't think of anybody more competent than him but I don't know if he wrote the note himself or if it was scribed by a student. Anyway, here is a link to the whole course, with many hand-outs, notes and exercises.

C) Addendum Not only don't we need completeness of $\mathbb P^n$ but the extension result is quite intuitive. In the complex case we would say that locally around $p=0$ the rational map can be written $z\mapsto F(z)=(f_0(z):...:f_n(z))$ where each $f_i(z)$ is identically zero or meromorphic of the form $f_i(z)=z^{n_i}u_i(z)$ for some $i\in \mathbb Z$ and $u_i(0)\neq 0$. If $n_0$ , say, is the smallest of the $n_i$'s ($n_0 \leq n_i$) then by multiplying all the homogeneous coordinates of $F(z)$ by $f_0 ^{-1}(z)=z^{-n_0} u_0^{-1}(z)$ we obtain $F(z)=(1:g_1(z):...:g_n(z))$ where the $g_i$'s are now holomorphic near $0$ so that the conclusion follows: $F$ can clearly be extended regularly across $p=0$.

In the purely algebraic case we adapt this idea by replacing $z$ by a uniformizing parameter near $p$, the $u_i$'s by units in $\mathcal O_p^\ast$ and the exponents $n_i$ of $z$ by the valuations of $f_i\in \mathcal O_p$, which is a discrete valuation ring thanks to the assumed regularity of $p$.

I have written this sketch because the technicalities of some proofs in the literature might obscure the actual naturality of this extension result.

We use the theory of specializations developed in Weil's Foundations of Algebraic Geometry, Chapter II. The main result of the theory of specializations(Lemma 7) below can be proved by "valuation theory + the axiom of choice". However, our proof(which is essentially the same as Weil's) does not use the axiom of choice.

Notation Let $X$ be a topological space. Let $E$ be a subset of $X$. We denote by $cl(E)$ the closure of $E$ in $X$.

Lemma 1 Let $X$ be a topological space. Let $x \in X$. Then $cl(\{x\})$ is irreducible.

Proof: Clear.

Definition 1 Let $X$ be a topological space. Suppose there exists $x \in X$ such that $X = cl(\{x\})$. We say $x$ is a generic point of $X$.

Definition 2 Let $X$ be a topological space. Let $x, y \in X$. Suppose $y \in cl(\{x\})$. Then we say $y$ is a specialization of $x$ and write $x \rightarrow y$.

We fix an algebraically closed field $\Omega$ which has infinite trancendence dimension over the prime subfield. Let $k$ be a subfield of $\Omega$ such that tr.dim $\Omega/k = \infty$. Let $E$ be a subset of a polynomial ring $k[X_1,\dots,X_n]$. We denote by $V(E)$ the common zeros of $E$ in $\Omega^n$. It is easy to see that by taking subsets of the form $V(E)$ as closed sets, we can define a topology on $\Omega^n$. We call this topology $k$-topology. A closed(resp. open) subset of $\Omega^n$ with respect to $k$-topology is called $k$-closed(resp. $k$-open) subset.

Let $E$ be a subset of $\Omega^n$. We denote by $I(E)$ the set $\{f \in k[X_1,\dots,X_n] |\ f(x) = 0$ for all $x \in E\}$. $I(E)$ is an ideal of $k[X_1,\dots,X_n]$. It is easy to see that $cl(E) = V(I(E))$. In particular, $cl(\{x\}) = V(I(\{x\}))$ for $x \in \Omega^n$. Let $x, y \in \Omega^n$. If $y$ is a specialization of $x$ with respect $k$-topology, we write $x \rightarrow_k y$ or simply $x \rightarrow y$ if there is no ambiguity. Hence $x \rightarrow_k y$ if and only if $f(y) =0$ for any $f \in k[X_1,\dots,X_n]$ such that $f(x) = 0$.

Lemma 2 Let $V$ be a $k$-irreducible subset of $\Omega^n$. Then $V$ has a generic point(this is one of the reasons why we need tr.dim $\Omega/k = \infty$).

Proof: Let $\mathfrak{p} = I(V)$. $\mathfrak{p}$ is a prime ideal. Let $K$ be the field of fractions of $k[X_1,\dots,X_n]/\mathfrak{p}$. Since tr.dim $K/k \le n$, there exists an $k$-embedding $\sigma\colon K \rightarrow \Omega$. Let $\bar x_i$ be the image of $x_i$ by the canonical homomorphism $k[X_1,\dots,X_n] \rightarrow k[X_1,\dots,X_n]/\mathfrak{p}$. Let $y = (\sigma(\bar x_1),\dots \sigma(\bar x_n))$. Then $\mathfrak{p} = I(\{y\})$. Hence $cl(\{y\}) = V(\mathfrak{p}) = V$. QED

We consider the set $\Omega_{\infty} = \Omega \cup \{\infty\}$, where $\infty$ is an element which does not belong to $\Omega$. Let $\rho\colon \Omega_{\infty} \rightarrow \Omega_{\infty}$ be the map defined as follows.

$\rho(x) = 1/x$ if $x \in \Omega - \{0\}$.

$\rho(0) = \infty$.

$\rho(\infty) = 0$.

Clearly $\rho$ is a bijection.

Let $I = \{1,\dots,n\}$. For $i \in I$, let $\rho_i\colon \Omega_{\infty}^n \rightarrow \Omega_{\infty}^n$ be the map defined by $\rho_i(x_1,\dots,x_i,\dots,x_n) = (x_1,\dots, \rho(x_i), \dots,x_n)$. Clearly $\rho_i$ is a bijection and $\rho_i\rho_j = \rho_j\rho_i$ for any $i, j \in I$. Let $G$ be the subgroup of the symmetric group on $\Omega_{\infty}^n$ generated by the set $\{\rho_1,\dots,\rho_n\}$. Let $J$ be a subset of $I$. We define $\rho_J = \prod_{i\in J} \rho_i$ Clearly every element of $G$ can be uniquely written as $\rho_J$ for some $J \subset I$. Hence $|G| = 2^n$.

Lemma 3 Let $x, x' \in \Omega^n$. Let $\sigma \in G$. Let $y = \sigma(x)$, $y' = \sigma(x')$. Suppose $y$ and $y'$ are contained in $\Omega^n$. Then $x \rightarrow_k x'$ if and only if $y \rightarrow_k y'$.

Proof: Without loss of generality, we can assume that $\sigma = \rho_J$, where $J = \{1,\dots,m\}$. Suppose $x \rightarrow_k x'$. It suffices to prove that $y \rightarrow_k y'$. Note that none of $x_1,\dots,x_m$ and $x_1',\dots, x_m'$ are $0$, since $y$ and $y'$ are contained in $\Omega^n$.

Suppose $f(y) = 0$ for $f \in k[X_1,\dots,X_n]$. Then $g = (X_1\dots X_m)^k f(1/X_1,\dots,1/X_m, X_{m+1},\dots,X_n) \in k[X_1,\dots,X_n]$ for some integer $k > 0$. Since $f(y) = 0$, $g(x) = 0$. Since $x \rightarrow_k x'$, $g(x') = 0$. Hence $f(y') = 0$. Hence $y \rightarrow y'$. QED

Definition 3 Let $x, x' \in \Omega_{\infty}^n$. Suppose there exists $\sigma \in G$ such that $\sigma(x), \sigma(x') \in \Omega^n$ and $\sigma(x) \rightarrow_k \sigma(x')$. Then we say $x'$ is a specialization of $x$ over $k$ and write $x \rightarrow_k x'$. By Lemma 3, this does not depend on the choice of $\sigma$. Lemma 1 also shows that, if $x, x' \in \Omega^n$ and $x \rightarrow_k x'$ in the new definition, $x \rightarrow_k x'$ in the old definition(Definition 2).

Lemma 4 $0$ has only one specialization over $k$, namely $0$. Hence $\infty$ has only one specialization over $k$, namely $\infty$.

Proof: $I(\{0\}) = (X)$. Hence $V((X)) = 0$. Hence $cl(\{0\}) = \{0\}$. QED

Lemma 5 Let $x =(x_1,\dots,x_n), x' = (x_1',\dots,x_n') \in \Omega_\infty^n$. Suppose $x \rightarrow_k x'$. If $x_i = \infty$ for some $i$, then $x_i' = \infty$.

Proof. Since $x \rightarrow_k x'$, $x_i \rightarrow_k x_i'$. Hence, by Lemma 4, $x_i' = \infty$. QED

Lemma 6 Let $x, x' \in \Omega^n$, $y \in \Omega$. Suppose $x \rightarrow_k x'$. Then there exists $y' \in \Omega_{\infty}$ such that $(x, y) \rightarrow_k (x', y')$.

Proof: If there exists $y'$ in an finite extension of $k(x')$ such that $(x, y) \rightarrow_k (x', y')$, we are done. Suppose there exist no $y'$ in any finite extensuon of $k(x')$ such that $(x, y) \rightarrow_k (x', y')$. Then, by Lemma 4 of my answer to this question, $y \neq 0$ and $(x, 1/y) \rightarrow_k (x', 0)$. Hence $(x, y) \rightarrow_k (x', \infty)$. QED

Lemma 7(Fundamental theorem of the theory of specializations) Let $x, x' \in \Omega_\infty^n$, $y \in \Omega_\infty^m$. Suppose $x \rightarrow_k x'$. Then there exists $y' \in \Omega_{\infty}^m$ such that $(x, y) \rightarrow_k (x', y')$.

Proof: By using induction on $m$, it suffices to prove the lemma when $m = 1$. Since $x \rightarrow_k x'$, there exists $\sigma \in G$ such that $\sigma(x), \sigma(x') \in \Omega^n$ and $\sigma(x) \rightarrow_k \sigma(x')$. Since $(\sigma(x), \infty) \rightarrow_k (\sigma(x'), \infty)$, we may assume that $y \neq \infty$. Hence the assertion follows from Lemma 6. QED

Lemma 8(Weil, Foundations of algebraic geometry, Prop. 10, Sec. 3, Chap. II) Suppose $x \rightarrow_k x'$, where $x, x' \in \Omega_\infty^n$ Let $y = (y_1,\dots,y_m) \in \Omega^m$. Suppose $y \neq 0$. Let $I = {1,\dots,m}$. Then there exists $i_0 \in I$ such that $(x, z) \rightarrow_k (x, z')$, where $z = (y_1/y_{i_0},\dots,y_m/y_{i_0})$ and $z' \in \Omega^m$.

Proof: Without loss of generality, we may assume that $y_i \neq 0$ for all $i \in I$. Suppose no such $i_0$ exists. Let $u_{i,j} = y_i/y_j$ for all $(i, j) \in I\times I$. Let $u = (u_{i,j}) \in \Omega^{m^2}$. By Lemma 7, there exists $v = (v_{i,j}) \in \Omega_{\infty}^{m^2}$ such that $(x, u) \rightarrow_k (x, v)$. By the assumption, for each $j_0 \in I$ there exists $i \in I$ such that $v_{i,j_0} = \infty$. Since $u_{j_0,i} = 1/u_{i,j_0}, v_{j_0,i} = 0$. Hence we have a map $\phi\colon I \rightarrow I$ such that $v_{j, \phi(j)} = 0$. Let $j_0 = 1$ and $j_{k+1} = \phi(j_k)$ for $k = 0,1,2,\dots$ Since $I$ is finite, there exists $p, r$ such that $j_{p+r} = j_p$. Then $v_{j_p,j_{p+1}} = \cdots = v_{j_{p+r-1}, j_p} = 0$. But this is a contradiction, because $u_{j_p,j_{p+1}}\cdots u_{j_{p+r-1}, j_p} = 1$. QED

Let $\mathbf{P}^n(\Omega)$ be the $n$-projective space over $\Omega$. Let $E$ be a set of homogeneous polynomials in $k[X_0,\dots,X_n]$. We denote by $V_+(E)$ the common zeros of $E$ in $\mathbf{P}^n(\Omega)$. It is easy to see that by taking subsets of the form $V_+(E)$ as closed sets, we can define a topology on $\mathbf{P}^n(\Omega)$. We call this topology $k$-topology. A closed(resp. open) subset of $\mathbf{P}^n(\Omega)$ with respect to $k$-topology is called $k$-closed(resp. $k$-open) subset. It is easy to see that $\mathbf{P}^n(\Omega)$ is a $k$-variety defined in this question.

Lemma 9 Let $X$ be a topological space. Let $U$ be an open subset of $X$. Let $x, y \in U$. Then $x \rightarrow y$ in $X$ if and only if $x \rightarrow y$ in $U$.

Proof: Suppose $x \rightarrow y$ in $X$. Let $V'$ be an open neighborhood of $y$ in $U$. Then there exists an open subset $V$ of $X$ such that $V' = V \cap U$. Since $x \in V$, $x \in V$. Hence $x \in V'$. Hence $x \rightarrow y$ in $U$.

Conversely suppose $x \rightarrow y$ in $U$. Let $V$ be an open subset of $X$ such that $y \in V$. Then $V \cap U$ is an open neighborhood of $y$ in $U$. Hence $x \in V \cap U \subset V$. Hence $x \rightarrow y$ in $X$. QED

Lemma 10 Let $X = \mathbf{P}^n(\Omega)$. We regard $X$ as a $k$-variety. Let $Y$ be a $k$-variety. Then $X\times Y$ is a $k$-variety as defined in this question. Let $(x, y) \in X\times Y$. Suppose $y \rightarrow y'$ in $Y$. Then there exists $x' \in X$ such that $(x, y) \rightarrow (x', y')$ in $X\times Y$.

Proof. Let $V$ be an affine open neighborhood of $y'$ in $Y$. Since $y \rightarrow y'$, $y \in V$. Hence, by Lemma 9, we may assume $Y$ is an affine $k$-variety. Then the assertion follows from Lemma 8. QED

Lemma 11 Let $X$ be an irrreducible topological space. Let $U$ be a non-empty open subset of $X$. Suppose $U$ has a generic point(Definition 1) $x$. Then $x$ is a generic point of $X$.

Proof: Let $y \in X$. Let $V$ be an open neighborhood of $y$ in $X$. Since $X$ is irreducible, $V \cap U = \emptyset$. Hence $x \in V \cap U \subset V$. Hence $x \rightarrow y$ in $X$. QED

Lemma 12 Let $f\colon X \rightarrow Y$ be a continuous map of topological spaces. Suppose $X$ is irreducible and has a generic point $x$. Then $f(x)$ is a generic point of $cl(f(X))$.

Proof: Let $y = f(x)$. Since $cl(\{y\}) \subset cl(f(X))$, it suffices to prove that $cl(f(X)) \subset cl(\{y\})$. Let $y' \in cl(f(X))$. It suffices to prove that $y \rightarrow y'$. Let $V$ be a neighborhood of $y'$ in $Y$. There exists $x' \in X$ such that $f(x') \in V$. Since $f$ is continuous, there exists a neighborhood $U$ of $x'$ such that $f(U) \subset V$. Since $x \in U$, $y = f(x) \in V$. Hence $y \rightarrow y'$ as desired. QED

Lemma 13 Let $X = \mathbf{P}^n(\Omega)$. Let $Y$ be a $k$-variety. Then the projection map $\pi\colon X\times Y \rightarrow Y$ is a closed map.

Proof: Let $Z$ be a closed subset of $X\times Y$. It suffices to prove that $\pi(Z)$ is closed. Since $X\times Y$ is a Noethrian space, $Z = Z_1\cup\cdots\cup Z_r$, where each $Z_i$ is a closed irreducible subset. Then $\pi(Z) = \pi(Z_1)\cup\cdots\cup \pi(Z_r)$. Hence $cl(\pi(Z)) = cl(\pi(Z_1)\cup\cdots\cup cl(\pi(Z_r))$. Hence we may assume that $Z$ is irreducible. By Lemma 2 and Lemma 11, $Z$ has a generic point $(x, y)$. By Lemma 11, $y$ is a generic point of $cl(\pi(Z))$. Let $y' \in cl(\pi(Z))$. Then $y \rightarrow y'$. By Lemma 10, there exists $x' \in X$ such that $(x, y) \rightarrow (x', y')$ in $X\times Y$. Since $(x',y') \in Z, y' \in \pi(Z)$. Hence $cl(\pi(Z)) = \pi(Z)$. QED

Lemma 14 Let $f\colon X \rightarrow Y$ be a morphism of $k$-varieties(see this question). Let $x \in X$ and $y = f(x)$. Let $y' \in Y$ be such that $cl(\{y\}) = cl(\{y'\})$. Then there exists $x' \in X$ such that $cl({x}) = cl({x'})$ and $f(x') = y'$.

Proof: Let $X'$ be the reduced scheme of finite type over $k$ corresponding to $X$ by this question. Then the underlying set of $X$ can be idenitified with $Hom_k(Spec(\Omega), X')$. Then the assertion follows from this question. QED

Definition 4 Let $X$ be a $k$-variety. Let $x \in X$. We denote by $k(x)$ the residue field of the local ring $\mathcal{O}_x$. We say $x$ is algebraic if $k(x)$ is algebraic over $k$.

Lemma 15 Let $f\colon X \rightarrow Y$ be a morphism of $k$-varieties. Let $X_0$(resp. $Y_0$) be the set of algebraic points of $X$(resp. $Y$). Then $f(X_0) \subset Y_0$.

Proof: Let $x \in X_0$ and $y = f(x)$. $f$ induces a $k$-homomorphism $\mathcal{O}_y \rightarrow \mathcal{O}_x$. Hence it induces a $k$-homomorphism $k(y) \rightarrow k(x)$. Since $k(x)$ is algebraic over $k$, $k(y)$ is algebraic over $k$. QED

Lemma 16 Let $x$ be a point of $k$-variety $X$. Then there exists an algebraic point $x_0 \in X$ such that $x \rightarrow x_0$.

Proof: By Lemma 9, we may assume that $X$ is affine. Let $A$ be the affine coordinate ring of $X$. Let $P = \{f \in A|\ f(x) = 0\}$. Let $\mathfrak{m}$ be a maximal ideal such that $P \subset \mathfrak{m}$. Let $K = A/\mathfrak{m}$. Since $K$ is algebraic over $k$, there exists a $k$-embedding $\phi\colon K \rightarrow \Omega$. $\phi$ induces a $k$-homomorphism $\alpha\colon A \rightarrow \Omega$. $\alpha$ can be idenitified with a point $x_0 \in X$. Clearly $x_0$ is algebraic. Since $Ker(\alpha) = \mathfrak{m}$, $x \rightarrow x_0$. QED

Lemma 17 Let $f\colon X \rightarrow Y$ be a morphism of $k$-varieties. Let $X_0$(resp. $Y_0$) be the set of algebraic points of $X$(resp. $Y$). By Lemma 15, $f$ induces a map $f_0\colon X_0 \rightarrow Y_0$. Suppose $f$ is a closed map. Then so is $f_0$.

Proof: Let $Z_0$ be a closed subset of $X_0$. There exists a closed subset $Z$ of $X$ such that $Z \cap X_0 = Z_0$. It suffices to prove that $f(Z) \cap Y_0 = f(Z_0)$. Clearly $f(Z_0) \subset f(Z) \cap Y_0$. Let $y_0 \in f(Z) \cap Y_0$. It suffices to prove that $y_0 \in f(Z_0)$. There exists $z \in Z$ such that $f(z) = y_0$. By Lemma 16, there exists $z_0 \in Z_0$ such that $z \rightarrow z_0$. Then $y_0 = f(z) \rightarrow f(z_0)$. Since $y_0$ is algebraic, $cl(\{y_0\}) = cl(\{f(z_0)\})$. Hence, by Lemma 14, there exists $w_0 \in X$ such that $cl({z_0}) = cl({w_0})$ and $f(w_0) = y_0$. Since $Z$ is closed, $w_0 \in Z$. Since $z_0$ is algebraic, so is $w_0$. Hence $w_0 \in Z_0$. Hence $y_0 \in f(Z_0)$. QED

Proposition Let $k$ be an algebraically closed field. Let $X = \mathbf{P}^n(k)$ be a projective $n$-space over $k$. Let $Y$ be a variety over $k$ in the sense of Serre's FAC. Then the porojection map $\pi\colon X\times Y \rightarrow Y$ is a closed map.

Proof: Let $Z$ be a closed subset of $X\times Y$. Let $(U_i)$ be a finite affine open cover of $Y$. Since $(X \times U_i)$ is an affine open cover of $X\times Y$ and $\pi(Z) \cap U_i = \pi(Z\cap (X\times U_i))$, we may assume that $Y$ is affine. Then $Y$ is identified with a Zariski closed subset of $k^m$. There exists an ideal $I$ of $k[X_1,\dots,X_m]$ such that $Y$ is the set of common zeros of $I$ in $k^m$. Let $Y'$ be the set of common zeros of $I$ in $\Omega^m$, where $\Omega$ is an algebraically closed field such that tr.dim $\Omega/k = \infty$.Since $Y'$ is a $k$-closed subset of $\Omega^m$, $Y'$ is an affine $k$-variety. Let $X' = \mathbf{P}^n(\Omega)$. Since $X\times Y$ is the set of algebraic points of a $k$-variety $X'\times Y'$, the assertion follows from Lemma 13 and Lemma 17. QED

Let $k$ be an algebraically closed field. Let $X$ be a reduced separated scheme of finite type over $k$. Let $X_0$ be the set of closed points of $X$. By this question, $(X_0, \mathcal{O}_X|X_0)$ is a variety in the sense of Serre's FAC.

Conversely let $V$ be a variety over $k$ in Serre's sense. By this question, there exists a reduced separated scheme $X$ of finite type over $k$ such that $V$ is identified with $(X_0, \mathcal{O}_X|X_0)$. Such $X$ is essentially unique(i.e. they are all $k$-isomorphic).

Let $V$ be a projective variety. Let $W$ be a variety in Serre's sense. Let $p\colon V\times W \rightarrow W$ be the projection map. We will prove that $p$ is a closed map. There exists a projective scheme $X$ over $k$ such that $V$ is isomoprphic to $(X_0, \mathcal{O}_X|X_0)$. Similiarly there exists a scheme $Y$ over $k$ such that $W$ is isomoprphic to $(Y_0, \mathcal{O}_Y|Y_0)$. Let $\pi\colon X\times_k Y \rightarrow Y$ be the projection. Since $X$ is proper over $k$(Hartshorne, Theorem 4.9, Chapter II, or the theorem below), $\pi$ is a closed morphism. Since $(X\times_k Y)_0 = X_0 \times Y_0$ as a set, $p$ is a closed map by this question.

Let $f\colon V \rightarrow W$ be a morphism. Let $\Gamma = \{(x, f(x)) \in V \times W|\ x \in V\}$. Let $\Delta = \{(y, y) \in W \times W|\ y \in W\}$. Consider the map $f \times id_W\colon V\times W \rightarrow W \times W$. Then $\Gamma = (f \times id_W)^{-1}(\Delta)$. Since $\Delta$ is closed in $W\times W$, $\Gamma$ is closed in $V\times W$. Hence $f(V) = p(\Gamma)$ is closed in $W$. Since any closed subset of $V$ is a projective variety, $f$ is a closed map as desired.

Theorem The projective scheme $\mathbf{P}_{\mathbf{Z}}^n$ is proper over $\mathbf{Z}$.

Proof(EGA II, theorem 5.5.3): We need to show that the projection $\mathbf{P}_{\mathbf{Z}}^n \times Y \rightarrow Y$ is a closed morphism for any scheme $Y$ over $\mathbf{Z}$. Since the problem is local on $Y$, we may assume that $Y$ is affine. Let $Y = Spec(A)$. Then $\mathbf{P}_{\mathbf{Z}}^n \times Y = \mathbf{P}_A^n = Proj(A[X_0,\dots,X_n))$. Let $Z$ be a closed subset of $Proj(A[X_0,\dots,X_n))$. Then $Z$ is isomorphic to $Proj(A[X_0,\dots,X_n]/I)$, where $I$ is a homogeneous ideal. Let $S = A[X_0,\dots,X_n]/I$. Let $f\colon Proj(S) \rightarrow Y$ be the canonical homomorphism. It suffices to prove that $f(Proj(S))$ is closed in $Y$.

Let $y \in Y$. $f^{-1}(y) = Proj(S)\times_Y Spec (k(y)) = Proj(S\otimes_A k(y))$. Hence $f^{-1}(y) = \emptyset$ if and only if $S_n\otimes_A k(y) = 0$ for all sufficiently large $n$.Since $S_n$ is an $A$-module of finite type, $S_n\otimes_A \mathcal{O}_y$ is an $\mathcal{O}_y$-module of finite type. Hence $S_n\otimes_A k(y) = 0$ is equivalent to $S_n\otimes_A \mathcal{O}_y = 0$ by Nakayama's lemma. This is equivalent to that $\mathfrak{p}_y$ does not contain $\mathfrak{a}_n$, where $\mathfrak{p}_y$ is the prime ideal corresponding to $y$ and $\mathfrak{a}_n$ is the annihilator of $A$-module $S_n$. Since $S_n = S_{n}S_1$, $\mathfrak{a}_n \subset \mathfrak{a}_{n+1}$. Let $\mathfrak{a} = \bigcup_n \mathfrak{a}_n$. Then $f(Proj(S)) = V(\mathfrak{a})$. QED