hcju/stacksqa huggingface dataset:
We prove each of the parts in turn. \begin{enumerate} \item By Proposition \ref{proposition-localize-twice} this implies $M_\mathfrak p = 0$ for all $\mathfrak p \in \Spec(R)$, so we conclude by Lemma \ref{lemma-characterize-zero-local}. \item For each $i$ take a finite generating set $X_i$ of $M_{f_i}$. Without loss of generality, we may assume that the elements of $X_i$ are in the image of the localization map $M \rightarrow M_{f_i}$, so we take a finite set $Y_i$ of preimages of the elements of $X_i$ in $M$. Let $Y$ be the union of these sets. This is still a finite set. Consider the obvious $R$-linear map $R^Y \rightarrow M$ sending the basis element $e_y$ to $y$. By assumption this map is surjective after localizing at an arbitrary prime ideal $\mathfrak p$ of $R$, so it surjective by Lemma \ref{lemma-characterize-zero-local} and $M$ is finitely generated. \item By (2) we have a short exact sequence $$ 0 \rightarrow K \rightarrow R^n \rightarrow M \rightarrow 0 $$ Since localization is an exact functor and $M_{f_i}$ is finitely presented we see that $K_{f_i}$ is finitely generated for all $1 \leq i \leq n$ by Lemma \ref{lemma-extension}. By (2) this implies that $K$ is a finite $R$-module and therefore $M$ is finitely presented. \item By Proposition \ref{proposition-localize-twice} the assumption implies that the induced morphism on localizations at all prime ideals is an isomorphism, so we conclude by Lemma \ref{lemma-characterize-zero-local}. \item By Proposition \ref{proposition-localize-twice} the assumption implies that the induced sequence of localizations at all prime ideals is short exact, so we conclude by Lemma \ref{lemma-characterize-zero-local}. \item We will show that every ideal of $R$ has a finite generating set: For this, let $I \subset R$ be an arbitrary ideal. By Proposition \ref{proposition-localization-exact} each $I_{f_i} \subset R_{f_i}$ is an ideal. These are all finitely generated by assumption, so we conclude by (2). \item For each $i$ take a finite generating set $X_i$ of $S_{f_i}$. Without loss of generality, we may assume that the elements of $X_i$ are in the image of the localization map $S \rightarrow S_{f_i}$, so we take a finite set $Y_i$ of preimages of the elements of $X_i$ in $S$. Let $Y$ be the union of these sets. This is still a finite set. Consider the algebra homomorphism $R[X_y]_{y \in Y} \rightarrow S$ induced by $Y$. Since it is an algebra homomorphism, the image $T$ is an $R$-submodule of the $R$-module $S$, so we can consider the quotient module $S/T$. By assumption, this is zero if we localize at the $f_i$, so it is zero by (1) and therefore $S$ is an $R$-algebra of finite type. \item By the previous item, there exists a surjective $R$-algebra homomorphism $R[X_1, \ldots, X_n] \rightarrow S$. Let $K$ be the kernel of this map. This is an ideal in $R[X_1, \ldots, X_n]$, finitely generated in each localization at $f_i$. Since the $f_i$ generate the unit ideal in $R$, they also generate the unit ideal in $R[X_1, \ldots, X_n]$, so an application of (2) finishes the proof. \end{enumerate}
Choose $h_1, \ldots, h_n \in S$ such that $\sum h_i g_i = 1$. \medskip\noindent Proof of (1). For each $i$ choose a finite list of elements $x_{i, j} \in S_{g_i}$, $j = 1, \ldots, m_i$ which generate $S_{g_i}$ as an $R$-algebra. Write $x_{i, j} = y_{i, j}/g_i^{n_{i, j}}$ for some $y_{i, j} \in S$ and some $n_{i, j} \ge 0$. Consider the $R$-subalgebra $S' \subset S$ generated by $g_1, \ldots, g_n$, $h_1, \ldots, h_n$ and $y_{i, j}$, $i = 1, \ldots, n$, $j = 1, \ldots, m_i$. Since localization is exact (Proposition \ref{proposition-localization-exact}), we see that $S'_{g_i} \to S_{g_i}$ is injective. On the other hand, it is surjective by our choice of $y_{i, j}$. The elements $g_1, \ldots, g_n$ generate the unit ideal in $S'$ as $h_1, \ldots, h_n \in S'$. Thus $S' \to S$ viewed as an $S'$-module map is an isomorphism by Lemma \ref{lemma-cover}. \medskip\noindent Proof of (2). We already know that $S$ is of finite type. Write $S = R[x_1, \ldots, x_m]/J$ for some ideal $J$. For each $i$ choose a lift $g'_i \in R[x_1, \ldots, x_m]$ of $g_i$ and we choose a lift $h'_i \in R[x_1, \ldots, x_m]$ of $h_i$. Then we see that $$ S_{g_i} = R[x_1, \ldots, x_m, y_i]/J_i + (1 - y_ig'_i) $$ where $J_i$ is the ideal of $R[x_1, \ldots, x_m, y_i]$ generated by $J$. Small detail omitted. By Lemma \ref{lemma-finite-presentation-independent} we may choose a finite list of elements $f_{i, j} \in J$, $j = 1, \ldots, m_i$ such that the images of $f_{i, j}$ in $J_i$ and $1 - y_ig'_i$ generate the ideal $J_i + (1 - y_ig'_i)$. Set $$ S' = R[x_1, \ldots, x_m]/(\sum h'_ig'_i - 1, f_{i, j};  i = 1, \ldots, n, j = 1, \ldots, m_i) $$ There is a surjective $R$-algebra map $S' \to S$. The classes of the elements $g'_1, \ldots, g'_n$ in $S'$ generate the unit ideal and by construction the maps $S'_{g'_i} \to S_{g_i}$ are injective. Thus we conclude as in part (1).
It suffices to show that the localization of the sequence at any maximal ideal $\mathfrak m$ is exact, see Lemma \ref{lemma-characterize-zero-local}. Since $f_1, \ldots, f_n$ generate the unit ideal, there is an $i$ such that $f_i \not \in \mathfrak m$. After renumbering we may assume $i = 1$. Note that $(M_{f_i})_\mathfrak m = (M_\mathfrak m)_{f_i}$ and $(M_{f_if_j})_\mathfrak m = (M_\mathfrak m)_{f_if_j}$, see Proposition \ref{proposition-localize-twice-module}. In particular $(M_{f_1})_\mathfrak m = M_\mathfrak m$ and $(M_{f_1 f_i})_\mathfrak m = (M_\mathfrak m)_{f_i}$, because $f_1$ is a unit. Note that the maps in the sequence are the canonical ones coming from Lemma \ref{lemma-universal-property-localization-module} and the identity map on $M$. Having said all of this, after replacing $R$ by $R_\mathfrak m$, $M$ by $M_\mathfrak m$, and $f_i$ by their image in $R_\mathfrak m$, and $f_1$ by $1 \in R_\mathfrak m$, we reduce to the case where $f_1 = 1$. \medskip\noindent Assume $f_1 = 1$. Injectivity of $\alpha$ is now trivial. Let $m = (m_i) \in \bigoplus_{i = 1}^n M_{f_i}$ be in the kernel of $\beta$. Then $m_1 \in M_{f_1} = M$. Moreover, $\beta(m) = 0$ implies that $m_1$ and $m_i$ map to the same element of $M_{f_1f_i} = M_{f_i}$. Thus $\alpha(m_1) = m$ and the proof is complete.
By Lemma \ref{lemma-disjoint-decomposition} we have $U = D(e)$ and $V = D(1-e)$ for some idempotent $e$. By Lemma \ref{lemma-standard-covering} we see that $R \cong R_e \times R_{1 - e}$ (since clearly $R_{e(1-e)} = 0$ so the glueing condition is trivial; of course it is trivial to prove the product decomposition directly in this case). The lemma follows.
To show that $M_{f_1} \to M_1$ is an isomorphism, it suffices to show that its localization at every prime $\mathfrak p'$ of $R_{f_1}$ is an isomorphism, see Lemma \ref{lemma-characterize-zero-local}. Write $\mathfrak p' = \mathfrak p R_{f_1}$ for some prime $\mathfrak p \subset R$, $f_1 \not \in \mathfrak p$, see Lemma \ref{lemma-standard-open}. Since localization is exact (Proposition \ref{proposition-localization-exact}), we see that \begin{align*} (M_{f_1})_{\mathfrak p'} & = M_\mathfrak p \\ & = \Ker\left( \bigoplus\nolimits_{1 \leq i \leq n} M_{i, \mathfrak p} \longrightarrow \bigoplus\nolimits_{1 \leq i, j \leq n} ((M_i)_{f_j})_\mathfrak p \right) \\ & = \Ker\left( \bigoplus\nolimits_{1 \leq i \leq n} M_{i, \mathfrak p} \longrightarrow \bigoplus\nolimits_{1 \leq i, j \leq n} (M_{i, \mathfrak p})_{f_j} \right) \end{align*} Here we also used Proposition \ref{proposition-localize-twice-module}. Since $f_1$ is a unit in $R_\mathfrak p$, this reduces us to the case where $f_1 = 1$ by replacing $R$ by $R_\mathfrak p$, $f_i$ by the image of $f_i$ in $R_\mathfrak p$, $M$ by $M_\mathfrak p$, and $f_1$ by $1$. \medskip\noindent Assume $f_1 = 1$. Then $\psi_{1j} : (M_1)_{f_j} \to M_j$ is an isomorphism for $j = 2, \ldots, n$. If we use these isomorphisms to identify $M_j = (M_1)_{f_j}$, then we see that $\psi_{ij} : (M_1)_{f_if_j} \to (M_1)_{f_if_j}$ is the canonical identification. Thus the complex $$ 0 \to M_1 \to \bigoplus\nolimits_{1 \leq i \leq n} (M_1)_{f_i} \longrightarrow \bigoplus\nolimits_{1 \leq i, j \leq n} (M_1)_{f_if_j} $$ is exact by Lemma \ref{lemma-cover-module}. Thus the first map identifies $M_1$ with $M$ in this case and everything is clear.
If some element $x$ of ${\mathfrak p}R_{\mathfrak p}$ is not nilpotent, then $D(x) \not = \emptyset$, see Lemma \ref{lemma-Zariski-topology}. This contradicts the minimality of $\mathfrak p$. If $R$ is reduced, then ${\mathfrak p}R_{\mathfrak p} = 0$ and hence it is a field.
By Lemma \ref{lemma-minimal-prime-reduced-ring} each of the rings $R_\mathfrak p$ is a field. In particular, the kernel of the ring map $R \to R_\mathfrak p$ is $\mathfrak p$. By Lemma \ref{lemma-Zariski-topology} we have $\bigcap_{\mathfrak p} \mathfrak p = (0)$. Hence (2) and (1) are true. If $x y = 0$ and $y \not = 0$, then $y \not \in \mathfrak p$ for some minimal prime $\mathfrak p$. Hence $x \in \mathfrak p$. Thus every zerodivisor of $R$ is contained in $\bigcup_{\mathfrak p\text{ minimal}} \mathfrak p$. Conversely, suppose that $x \in \mathfrak p$ for some minimal prime $\mathfrak p$. Then $x$ maps to zero in $R_\mathfrak p$, hence there exists $y \in R$, $y \not \in \mathfrak p$ such that $xy = 0$. In other words, $x$ is a zerodivisor. This finishes the proof of (3) and the lemma.
There are natural maps $Q(R) \to R_{\mathfrak q_i}$ since any nonzerodivisor is contained in $R \setminus \mathfrak q_i$. Hence a natural map $Q(R) \to R_{\mathfrak q_1} \times \ldots \times R_{\mathfrak q_t}$. For any nonminimal prime $\mathfrak p \subset R$ we see that $\mathfrak p \not \subset \mathfrak q_1 \cup \ldots \cup \mathfrak q_t$ by Lemma \ref{lemma-silly}. Hence $\Spec(Q(R)) = \{\mathfrak q_1, \ldots, \mathfrak q_t\}$ (as subsets of $\Spec(R)$, see Lemma \ref{lemma-spec-localization}). Therefore $\Spec(Q(R))$ is a finite discrete set and it follows that $Q(R) = A_1 \times \ldots \times A_t$ with $\Spec(A_i) = \{q_i\}$, see Lemma \ref{lemma-disjoint-implies-product}. Moreover $A_i$ is a local ring, which is a localization of $R$. Hence $A_i \cong R_{\mathfrak q_i}$.
Note that if $ \mathfrak p \in V(I)$, then $I \subset \mathfrak p$. Hence, clearly $\overline{\{\mathfrak p\}} = V(\mathfrak p)$. In particular $V(\mathfrak p)$ is the closure of a singleton and hence irreducible. The second assertion implies the third. To show the second, let $V(I) \subset \Spec(R)$ with $I$ a radical ideal. If $I$ is not prime, then choose $a, b\in R$, $a, b\not \in I$ with $ab\in I$. In this case $V(I, a) \cup V(I, b) = V(I)$, but neither $V(I, b) = V(I)$ nor $V(I, a) = V(I)$, by Lemma \ref{lemma-Zariski-topology}. Hence $V(I)$ is not irreducible.
Since $W$ is quasi-compact we may write it as a finite union of standard affine opens $D(g_i)$, $i = 1, \ldots, n$. Since $\mathfrak p \not \in W$ we have $g_i \in \mathfrak p$ for all $i$. By Lemma \ref{lemma-minimal-prime-reduced-ring} each $g_i$ is nilpotent in $R_{\mathfrak p}$. Hence we can find an $f \in R$, $f \not \in \mathfrak p$ such that for all $i$ we have $f g_i^{n_i} = 0$ for some $n_i > 0$. Then $D(f)$ works.
First proof. It is clear that (5), (6), and (7) are equivalent. It is clear that (4) and (8) are equivalent as every quasi-compact open is a finite union of standard opens. The implication (7) $\Rightarrow$ (4) follows from Lemma \ref{lemma-standard-open-containing-maximal-point}. Assume (4) holds. Let $\mathfrak p, \mathfrak p'$ be distinct primes of $R$. Choose an $f \in \mathfrak p'$, $f \not \in \mathfrak p$ (if needed switch $\mathfrak p$ with $\mathfrak p'$). Then $\mathfrak p' \not \in D(f)$ and $\mathfrak p \in D(f)$. By (4) the open $D(f)$ is also closed. Hence $\mathfrak p$ and $\mathfrak p'$ are in disjoint open neighbourhoods whose union is $X$. Thus $X$ is Hausdorff and totally disconnected. Thus (4) $\Rightarrow$ (2) and (3). If (3) holds then there cannot be any specializations between points of $\Spec(R)$ and we see that (5) holds. If $X$ is Hausdorff then every point is closed, so (2) implies (6). Thus (2), (3), (4), (5), (6), (7) and (8) are equivalent. Any profinite space is Hausdorff, so (1) implies (2). If $X$ satisfies (2) and (3), then $X$ (being quasi-compact by Lemma \ref{lemma-quasi-compact}) is profinite by Topology, Lemma \ref{topology-lemma-profinite}. \medskip\noindent Second proof. Besides the equivalence of (4) and (8) this follows from Lemma \ref{lemma-spec-spectral} and purely topological facts, see Topology, Lemma \ref{topology-lemma-characterize-profinite-spectral}.
The first assertion is an immediate consequence of Example \ref{example-oka-family-finitely-generated} and Proposition \ref{proposition-oka}. For the second, suppose that there exists an ideal $I \subset R$ which is not finitely generated. The union of a totally ordered chain $\left\{I_\alpha\right\}$ of ideals that are not finitely generated is not finitely generated; indeed, if $I = \bigcup I_\alpha$ were generated by $a_1, \ldots, a_n$, then all the generators would belong to some $I_\alpha $ and would consequently generate it. By Zorn's lemma, there is an ideal maximal with respect to being not finitely generated. By the first part this ideal is prime.
The first part follows from Example \ref{example-oka-family-principal} and Proposition \ref{proposition-oka}. For the second, suppose that there exists an ideal $I \subset R$ which is not principal. The union of a totally ordered chain $\left\{I_\alpha\right\}$ of ideals that not principal is not principal; indeed, if $I = \bigcup I_\alpha$ were generated by $a$, then $a$ would belong to some $I_\alpha $ and $a$ would generate it. By Zorn's lemma, there is an ideal maximal with respect to not being principal. This ideal is necessarily prime by the first part.
We have (1) $\Rightarrow$ (2) because $\Spec(R)$ is quasi-compact, see Lemma \ref{lemma-quasi-compact}. We have (2) $\Rightarrow$ (3) because standard opens form a basis for the topology. Proof of (3) $\Rightarrow$ (1). Let $U = \bigcup_{i = 1\ldots n} D(f_i)$. To show that $U$ is retrocompact in $\Spec(R)$ it suffices to show that $U \cap V$ is quasi-compact for any quasi-compact open $V$ of $\Spec(R)$. Write $V = \bigcup_{j = 1\ldots m} D(g_j)$ which is possible by (2) $\Rightarrow$ (3). Each standard open is homeomorphic to the spectrum of a ring and hence quasi-compact, see Lemmas \ref{lemma-standard-open} and \ref{lemma-quasi-compact}. Thus $U \cap V = (\bigcup_{i = 1\ldots n} D(f_i)) \cap (\bigcup_{j = 1\ldots m} D(g_j)) = \bigcup_{i, j} D(f_i g_j)$ is a finite union of quasi-compact opens hence quasi-compact. To finish the proof note that (4) is equivalent to (3) by  Lemma \ref{lemma-Zariski-topology}.
We first show that the inverse image of any quasi-compact open $U \subset \Spec(R)$ is quasi-compact. By Lemma \ref{lemma-qc-open} we may write $U$ as a finite open of standard opens. Thus by Lemma \ref{lemma-spec-functorial} we see that $f^{-1}(U)$ is a finite union of standard opens. Hence $f^{-1}(U)$ is quasi-compact by Lemma \ref{lemma-qc-open} again. The second assertion now follows from Topology, Lemma \ref{topology-lemma-inverse-images-constructibles}.
By Lemma \ref{lemma-qc-open} the subset $D(f)$ and the complement of $V(g_1, \ldots, g_m)$ are retro-compact open. Hence $D(f) \cap V(g_1, \ldots, g_m)$ is a constructible subset and so is any finite union of such. Conversely, let $T \subset \Spec(R)$ be constructible. By Topology, Definition \ref{topology-definition-constructible}, we may assume that $T = U \cap V^c$, where $U, V \subset \Spec(R)$ are retrocompact open. By Lemma \ref{lemma-qc-open} we may write $U = \bigcup_{i = 1, \ldots, n} D(f_i)$ and $V = \bigcup_{j = 1, \ldots, m} D(g_j)$. Then $T = \bigcup_{i = 1, \ldots, n} \big(D(f_i) \cap V(g_1, \ldots, g_m)\big)$.
The spectrum of a finite product of rings is the disjoint union of the spectra, see Lemma \ref{lemma-spec-product}. Hence if $T = T_1 \cup T_2$ and the result holds for $T_1$ and $T_2$, then the result holds for $T$. By Lemma \ref{lemma-constructible} we may assume that $T = D(f) \cap V(g_1, \ldots, g_m)$. In this case $T$ is the image of the map $\Spec((R/(g_1, \ldots, g_m))_f) \to \Spec(R)$, see Lemmas \ref{lemma-standard-open} and \ref{lemma-spec-closed}.
We repeatedly use Lemma \ref{lemma-qc-open} without mention. Let $U, V$ be quasi-compact open in $\Spec(S)$. We will show that the image of $U \cap V^c$ is constructible. Under the identification $\Spec(S) = D(f)$ of Lemma \ref{lemma-standard-open} the sets $U, V$ correspond to quasi-compact opens $U', V'$ of $\Spec(R)$. Hence it suffices to show that $U' \cap (V')^c$ is constructible in $\Spec(R)$ which is clear.
If $I = (f_1, \ldots, f_m)$, then we see that $V(I)$ is the complement of $\bigcup D(f_i)$, see Lemma \ref{lemma-Zariski-topology}. Hence it is constructible, by Lemma \ref{lemma-qc-open}. Denote the map $R \to S$ by $f \mapsto \overline{f}$. We have to show that if $\overline{U}, \overline{V}$ are retrocompact opens of $\Spec(S)$, then the image of $\overline{U} \cap \overline{V}^c$ in $\Spec(R)$ is constructible. By Lemma \ref{lemma-qc-open} we may write $\overline{U} = \bigcup D(\overline{g_i})$. Setting $U = \bigcup D({g_i})$ we see $\overline{U}$ has image $U \cap V(I)$ which is constructible in $\Spec(R)$. Similarly the image of $\overline{V}$ equals $V \cap V(I)$ for some retrocompact open $V$ of $\Spec(R)$. Hence the image of $\overline{U} \cap \overline{V}^c$ equals $U \cap V(I) \cap V^c$ as desired.
It suffices to show that the image of a standard open $D(f)$, $f\in R[x]$ is quasi-compact open. The image of $D(f)$ is the image of $\Spec(R[x]_f) \to \Spec(R)$. Let $\mathfrak p \subset R$ be a prime ideal. Let $\overline{f}$ be the image of $f$ in $\kappa(\mathfrak p)[x]$. Recall, see Lemma \ref{lemma-in-image}, that $\mathfrak p$ is in the image if and only if $R[x]_f \otimes_R \kappa(\mathfrak p) = \kappa(\mathfrak p)[x]_{\overline{f}}$ is not the zero ring. This is exactly the condition that $f$ does not map to zero in $\kappa(\mathfrak p)[x]$, in other words, that some coefficient of $f$ is not in $\mathfrak p$. Hence we see: if $f = a_d x^d + \ldots + a_0$, then the image of $D(f)$ is $D(a_d) \cup \ldots \cup D(a_0)$.
Write $g = ux^d + a_{d-1}x^{d-1} + \ldots + a_0$, where $d$ is the degree of $g$, and hence $u \in R^*$. Consider the ring $A = R[x]/(g)$. It is, as an $R$-module, finite free with basis the images of $1, x, \ldots, x^{d-1}$. Consider multiplication by (the image of) $f$ on $A$. This is an $R$-module map. Hence we can let $P(T) \in R[T]$ be the characteristic polynomial of this map. Write $P(T) = T^d + r_{d-1} T^{d-1} + \ldots + r_0$. We claim that $r_0, \ldots, r_{d-1}$ have the desired property. We will use below the property of characteristic polynomials that $$ \mathfrak p \in V(r_0, \ldots, r_{d-1}) \Leftrightarrow \text{multiplication by }f\text{ is nilpotent on } A \otimes_R \kappa(\mathfrak p). $$ This was proved in Lemma \ref{lemma-characteristic-polynomial-prime}. \medskip\noindent Suppose $\mathfrak q\in D(f) \cap V(g)$, and let $\mathfrak p = \mathfrak q \cap R$. Then there is a nonzero map $A \otimes_R \kappa(\mathfrak p) \to \kappa(\mathfrak q)$ which is compatible with multiplication by $f$. And $f$ acts as a unit on $\kappa(\mathfrak q)$. Thus we conclude $\mathfrak p \not \in  V(r_0, \ldots, r_{d-1})$. \medskip\noindent On the other hand, suppose that $r_i \not\in \mathfrak p$ for some prime $\mathfrak p$ of $R$ and some $0 \leq i \leq d - 1$. Then multiplication by $f$ is not nilpotent on the algebra $A \otimes_R \kappa(\mathfrak p)$. Hence there exists a prime ideal $\overline{\mathfrak q} \subset A \otimes_R \kappa(\mathfrak p)$ not containing the image of $f$. The inverse image of $\overline{\mathfrak q}$ in $R[x]$ is an element of $D(f) \cap V(g)$ mapping to $\mathfrak p$.
Let $\xi \in X$ be a point of $f(E)$. Choose a point $\eta \in E$ mapping to $\xi$. Let $\mathfrak p \subset R$ be the prime corresponding to $\xi$ and let $\mathfrak q \subset S$ be the prime corresponding to $\eta$. Consider the diagram $$ \xymatrix{ \eta \ar[r] \ar@{|->}[d] & E \cap Y' \ar[r] \ar[d] & Y' = \Spec(S/\mathfrak q) \ar[r] \ar[d] & Y \ar[d] \\ \xi \ar[r] & f(E) \cap X' \ar[r] & X' = \Spec(R/\mathfrak p) \ar[r] & X } $$ By Lemma \ref{lemma-affine-map-quasi-compact} the set $E \cap Y'$ is constructible in $Y'$. It follows that we may replace $X$ by $X'$ and $Y$ by $Y'$. Hence we may assume that $R \subset S$ is an inclusion of domains, $\xi$ is the generic point of $X$, and $\eta$ is the generic point of $Y$. By Lemma \ref{lemma-generic-finite-presentation} combined with Chevalley's theorem (Theorem \ref{theorem-chevalley}) we see that there exist dense opens $U \subset X$, $V \subset Y$ such that $f(V) \subset U$ and such that $f : V \to U$ maps constructible sets to constructible sets. Note that $E \cap V$ is constructible in $V$, see Topology, Lemma \ref{topology-lemma-open-immersion-constructible-inverse-image}. Hence $f(E \cap V)$ is constructible in $U$ and contains $\xi$. By Topology, Lemma \ref{topology-lemma-generic-point-in-constructible} we see that $f(E \cap V)$ contains a dense open $U' \subset U$.
If $J \subset S$ is an ideal, then $\sqrt{\varphi^{-1}(J)} = \varphi^{-1}(\sqrt{J})$. This shows that (2) and (3) are equivalent. The implication (3) $\Rightarrow$ (4) is immediate. If $I \subset R$ is a radical ideal, then Lemma \ref{lemma-Zariski-topology} guarantees that $I = \bigcap_{I \subset \mathfrak p} \mathfrak p$. Hence (4) $\Rightarrow$ (2). By Lemma \ref{lemma-in-image} we have $\mathfrak p = \varphi^{-1}(\mathfrak p S)$ if and only if $\mathfrak p$ is in the image. Hence (1) $\Leftrightarrow$ (4). Thus (1), (2), (3), and (4) are equivalent. \medskip\noindent Assume (1) holds. Let $R \to R'$ be a ring map. Let $\mathfrak p' \subset R'$ be a prime ideal lying over the prime $\mathfrak p$ of $R$. To see that $\mathfrak p'$ is in the image of $\Spec(R' \otimes_R S) \to \Spec(R')$ we have to show that $(R' \otimes_R S) \otimes_{R'} \kappa(\mathfrak p')$ is not zero, see Lemma \ref{lemma-in-image}. But we have $$ (R' \otimes_R S) \otimes_{R'} \kappa(\mathfrak p') = S \otimes_R \kappa(\mathfrak p) \otimes_{\kappa(\mathfrak p)} \kappa(\mathfrak p') $$ which is not zero as $S \otimes_R \kappa(\mathfrak p)$ is not zero by assumption and $\kappa(\mathfrak p) \to \kappa(\mathfrak p')$ is an extension of fields.
Let $K$ be the field of fractions of the domain $R$. Assume that $R \to S$ is injective. Since localization is exact we see that $K \to S \otimes_R K$ is injective. Hence there is a prime mapping to $(0)$ by Lemma \ref{lemma-in-image}. \medskip\noindent Note that $(0)$ is dense in $\Spec(R)$, so that the last condition implies the second. \medskip\noindent Suppose the second condition holds. Let $f \in R$, $f \not = 0$. As $R$ is a domain we see that $V(f)$ is a proper closed subset of $R$. By assumption there exists a prime $\mathfrak q$ of $S$ such that $\varphi(f) \not \in \mathfrak q$. Hence $\varphi(f) \not = 0$. Hence $R \to S$ is injective.
Let $I = \Ker(R \to S)$. Note that $\sqrt{(0)} = \bigcap_{\mathfrak q \subset S} \mathfrak q$, see Lemma \ref{lemma-Zariski-topology}. Hence $\sqrt{I} = \bigcap_{\mathfrak q \subset S} R \cap \mathfrak q$. Thus $V(I) = V(\sqrt{I})$ is the closure of the image of $\Spec(S) \to \Spec(R)$. This shows that (1) is equivalent to (3). It is clear that (2) implies (3). Finally, assume (1). We may replace $R$ by $R/I$ and $S$ by $S/IS$ without affecting the topology of the spectra and the map. Hence the implication (1) $\Rightarrow$ (2) follows from Lemma \ref{lemma-injective-minimal-primes-in-image}.
Let $M$ be a finite $R$-module. By Lemma \ref{lemma-trivial-filter-finite-module} we can find a finite filtration of $M$ whose successive quotients are of the form $R/I$. Since any ideal is finitely generated, each of the quotients $R/I$ is finitely presented. Hence $M$ is finitely presented by Lemma \ref{lemma-extension}. This proves (1). To see (2) note that any ideal of $R[x_1, \ldots, x_n]$ is finitely generated by Lemma \ref{lemma-Noetherian-permanence}.