hcju/stacksqa huggingface dataset:
We will use the criterion of Algebra, Lemma \ref{algebra-lemma-noetherian-dim-1-Jacobson}. Let $\mathfrak p \subset A$ be a prime ideal such that $\mathfrak p_f = \mathfrak p A_f$ is prime and not maximal. We have to show that $A_f/\mathfrak p_f = (A/\mathfrak p)_f$ has infinitely many prime ideals. After replacing $A$ by $A/\mathfrak p$ we may assume $A$ is a domain, $\dim A_f > 0$, and our goal is to show that $\Spec(A_f)$ is infinite. Since $\dim A_f > 0$ we can find a nonzero prime ideal $\mathfrak q \subset A$ not containing $f$. Choose a maximal ideal $\mathfrak m \subset A$ containing $\mathfrak q$. Since $(A, I)$ is a Zariski pair, we see $I \subset \mathfrak m$. Hence $\mathfrak m \not = \mathfrak q$ and $\dim(A_\mathfrak m) > 1$. Hence $\Spec((A_\mathfrak m)_f) \subset \Spec(A_f)$ is infinite by Algebra, Lemma \ref{algebra-lemma-Noetherian-local-domain-dim-2-infinite-opens} and we win.
Essential surjectivity holds by Algebra, Lemma \ref{algebra-lemma-lift-etale}. If $B$, $B'$ are \'etale over $A$ and $B/IB \to B'/IB'$ is a morphism of $A/I$-algebras, then we can lift this by Algebra, Lemma \ref{algebra-lemma-smooth-strong-lift}. Finally, suppose that $f, g : B \to B'$ are two $A$-algebra maps with $f \mod I = g \mod I$. Choose an idempotent $e \in B \otimes_A B$ generating the kernel of the multiplication map $B \otimes_A B \to B$, see Algebra, Lemmas \ref{algebra-lemma-diagonal-unramified} and \ref{algebra-lemma-unramified} (to see that \'etale is unramified). Then $(f \otimes g)(e) \in IB$. Since $IB$ is locally nilpotent (Algebra, Lemma \ref{algebra-lemma-locally-nilpotent}) this implies $(f \otimes g)(e) = 0$ by Algebra, Lemma \ref{algebra-lemma-lift-idempotents}. Thus $f = g$. \medskip\noindent It is clear that $I$ is contained in the Jacobson radical of $A$. Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_0h_0$ be a factorization of $\overline{f} = f \bmod I$ with $g_0, h_0 \in A/I[T]$ monic generating the unit ideal in $A/I[T]$. By Lemma \ref{lemma-lift-factorization-monic} there exists an \'etale ring map $A \to A'$ which induces an isomorphism $A/I \to A'/IA'$ such that the factorization lifts to a factorization into monic polynomials over $A'$. By the above we have $A = A'$ and the factorization is over $A$.
Fix $n$. Let $a \in A$ be an element which maps to $1$ in $A_n$. By Algebra, Lemma \ref{algebra-lemma-locally-nilpotent-unit} we see that $a$ maps to a unit in $A_m$ for all $m \geq n$. Hence $a$ is a unit in $A$. Thus by Algebra, Lemma \ref{algebra-lemma-contained-in-radical} the ideal $I_n$ is contained in the Jacobson radical of $A$. Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_nh_n$ be a factorization of $\overline{f} = f \bmod I_n$ with $g_n, h_n \in A_n[T]$ monic generating the unit ideal in $A_n[T]$. By Lemma \ref{lemma-locally-nilpotent-henselian} we can successively lift this factorization to $f \bmod I_m = g_m h_m$ with $g_m, h_m$ monic in $A_m[T]$ for all $m \geq n$. At each step we have to verify that our lifts $g_m, h_m$ generate the unit ideal in $A_n[T]$; this follows from the corresponding fact for $g_n, h_n$ and the fact that $\Spec(A_n[T]) = \Spec(A_m[T])$ because the kernel of $A_m \to A_n$ is locally nilpotent. As $A = \lim A_m$ this finishes the proof.
By Algebra, Lemma \ref{algebra-lemma-radical-completion} the ideal $I$ is contained in the Jacobson radical of $A$. Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_0h_0$ be a factorization of $\overline{f} = f \bmod I$ with $g_0, h_0 \in A/I[T]$ monic generating the unit ideal in $A/I[T]$. By Lemma \ref{lemma-locally-nilpotent-henselian} we can successively lift this factorization to $f \bmod I^n = g_n h_n$ with $g_n, h_n$ monic in $A/I^n[T]$ for all $n \geq 1$. As $A = \lim A/I^n$ this finishes the proof.
Observe that $A \to B$ is quasi-finite at every prime of the closed subset $T = \Spec(C_1) \subset \Spec(B)$ (this follows by looking at fibre rings, see Algebra, Definition \ref{algebra-definition-quasi-finite}). Consider the diagram of topological spaces $$ \xymatrix{ \Spec(B) \ar[rr]_\phi \ar[rd]_\psi & & \Spec(B') \ar[ld]^{\psi'} \\ & \Spec(A) } $$ By Algebra, Theorem \ref{algebra-theorem-main-theorem} for every $\mathfrak p \in T$ there is a $h_\mathfrak p \in B'$, $h_\mathfrak p \not \in \mathfrak p$ such that $B'_h \to B_h$ is an isomorphism. The union $U = \bigcup D(h_\mathfrak p)$ gives an open $U \subset \Spec(B')$ such that $\phi^{-1}(U) \to U$ is a homeomorphism and $T \subset \phi^{-1}(U)$. Since $T$ is open in $\psi^{-1}(V(I))$ we conclude that $\phi(T)$ is open in $U \cap (\psi')^{-1}(V(I))$. Thus $\phi(T)$ is open in $(\psi')^{-1}(V(I))$. On the other hand, since $C_1$ is finite over $A/I$ it is finite over $B'$. Hence $\phi(T)$ is a closed subset of $\Spec(B')$ by Algebra, Lemmas \ref{algebra-lemma-going-up-closed} and \ref{algebra-lemma-integral-going-up}. We conclude that $\Spec(B'/IB') \supset \phi(T)$ is open and closed. By Algebra, Lemma \ref{algebra-lemma-disjoint-implies-product} we get a corresponding product decomposition $B'/IB' = C'_1 \times C'_2$. The map $B'/IB' \to B/IB$ maps $C'_1$ into $C_1$ and $C'_2$ into $C_2$ as one sees by looking at what happens on spectra (hint: the inverse image of $\phi(T)$ is exactly $T$; some details omitted). Pick a $g \in B'$ mapping to $(1, 0)$ in $C'_1 \times C'_2$ such that $D(g) \subset U$; this is possible because $\Spec(C'_1)$ and $\Spec(C'_2)$ are disjoint and closed in $\Spec(B')$ and $\Spec(C'_1)$ is contained in $U$. Then $B'_g \to B_g$ defines a homeomorphism on spectra and an isomorphism on local rings (by our choice of $U$ above). Hence it is an isomorphism, as follows for example from Algebra, Lemma \ref{algebra-lemma-characterize-zero-local}. Finally, it follows that $C'_1 = C_1$ and the proof is complete.
For any integral ring map $A \to B$ we see that $V(IB) = V(JB)$. Hence idempotents of $B/IB$ and $B/JB$ are in bijective correspondence (Algebra, Lemma \ref{algebra-lemma-disjoint-decomposition}). It follows that $B \to B/IB$ induces a bijection on sets of idempotents if and only if $B \to B/JB$ induces a bijection on sets of idempotents. Thus we conclude by Lemma \ref{lemma-characterize-henselian-pair}.
Assume (1). Let $B$ be an integral $A$-algebra. Consider the ring maps $$ B \to B/IB \to B/JB $$ By Lemma \ref{lemma-characterize-henselian-pair} we find that both arrows induce bijections on idempotents. Hence so does the composition. Whence $(A, J)$ is a henselian pair by Lemma \ref{lemma-characterize-henselian-pair}. \medskip\noindent Conversely, assume (2) holds. Then $(A/I, J/I)$ is a henselian pair by Lemma \ref{lemma-integral-over-henselian-pair}. Let $B$ be an integral $A$-algebra. Consider the ring maps $$ B \to B/IB \to B/JB $$ By Lemma \ref{lemma-characterize-henselian-pair} we find that the composition and the second arrow induce bijections on idempotents. Hence so does the first arrow. It follows that $(A, I)$ is a henselian pair (by the lemma again).
For every $j \in J$, the projection $\prod_{j \in J} A_j \rightarrow A_j$ is an integral ring map, so Lemma \ref{lemma-integral-over-henselian-pair} proves that each $(A_j, I_j)$ is Henselian if $(\prod_{j \in J} A_j, \prod_{j\in J} I_j)$ is Henselian. \medskip\noindent Conversely, suppose that each $(A_j, I_j)$ is a Henselian pair. Then every $1 + x$ with $x \in \prod_{j \in J} I_j$  is a unit in $\prod_{j \in J} A_j$ because it is so componentwise by Algebra, Lemma \ref{algebra-lemma-contained-in-radical} and Definition \ref{definition-henselian-pair}. Thus, by Algebra, Lemma \ref{algebra-lemma-contained-in-radical} again, $\prod_{j \in J} I_j$ is contained in the Jacobson radical of $\prod_{j \in J} A_j$. Continuing to work componentwise, it likewise follows that for every monic $f \in (\prod_{j \in J} A_j)[T]$ and every factorization $\overline{f} = g_0h_0$ with monic $g_0, h_0 \in (\prod_{j \in J} A_j / \prod_{j \in J} I_j)[T] = (\prod_{j \in J} A_j/I_j)[T]$ that generate the unit ideal in $(\prod_{j \in J} A_j / \prod_{j \in J} I_j)[T]$, there exists a factorization $f = gh$ in $(\prod_{j \in J} A_j)[T]$ with $g$, $h$ monic and reducing to $g_0$, $h_0$. In conclusion, according to Definition \ref{definition-henselian-pair} $(\prod_{j \in J} A_j, \prod_{j\in J} I_j)$ is a Henselian pair.
By Categories, Lemma \ref{categories-lemma-limits-products-equalizers}, we only need to consider products and equalizers. For products, the claim follows from Lemma \ref{lemma-product-henselian-pairs}. Thus, consider an equalizer diagram $$ \xymatrix{ (A, I) \ar[r] & (A', I') \ar@<1ex>[r]^{\varphi} \ar@<-1ex>[r]_{\psi} & (A'', I'')   } $$ in which the pairs $(A', I')$ and $(A'', I'')$ are henselian. To check that the pair $(A, I)$ is also henselian, we will use the Gabber's criterion in Lemma \ref{lemma-characterize-henselian-pair}. Every element of $1 + I$ is a unit in $A$ because, due to the uniqueness of the inverses of units, this may be checked in $(A', I')$. Thus $I$ is contained in the Jacobson radical of $A$, see Algebra, Lemma \ref{algebra-lemma-contained-in-radical}. Thus, let  $$ f(T) = T^{N - 1}(T - 1) + a_{N - 1} T^{N - 1} + \dotsb + a_1 T + a_0 $$ be a polynomial in $A[T]$ with $a_{N - 1}, \dotsc, a_0 \in I$ and $N \ge 1$. The image of $f(T)$ in $A'[T]$ has a unique root $\alpha' \in 1 + I'$ and likewise for the further image in $A''[T]$. Thus, due to the uniqueness, $\varphi(\alpha') = \psi(\alpha')$, to the effect that $\alpha'$ defines a root of $f(T)$ in $1 + I$, as desired.
If $u \in 1 + I$ then for some $j \in J$ we see that $u$ is the image of some $u_j \in 1 + I_j$. Then $u_j$ is invertible in $A_j$ by Algebra, Lemma \ref{algebra-lemma-contained-in-radical} and the assumption that $I_j$ is contained in the Jacobson radical of $A_j$. Hence $u$ is invertible in $A$. Thus $I$ is contained in the Jacobson radical of $A$ (by the lemma). \medskip\noindent Let $f \in A[T]$ be a monic polynomial and let $\overline{f} = g_0 h_0$ be a factorization with $g_0, h_0 \in A/I[T]$ monic generating the unit ideal in $A/I[T]$. Write $1 = g_0 g'_0 + h_0 h'_0$ for some $g'_0, h'_0 \in A/I[T]$. Since $A = \colim A_j$ and $A/I = \colim A_j/I_j$ are filterd colimits we can find a $j \in J$ and $f_j \in A_j$ and a factorization $\overline{f}_j = g_{j, 0} h_{j, 0}$ with $g_{j, 0}, h_{j, 0} \in A_j/I_j[T]$ monic and $1 = g_{j, 0} g'_{j, 0} + h_{j, 0} h'_{j, 0}$ for some $g'_{j, 0}, h'_{j, 0} \in A_j/I_j[T]$ with $f_j, g_{j, 0}, h_{j, 0}, g'_{j, 0}, h'_{j, 0}$ mapping to $f, g_0, h_0, g'_0, h'_0$. Since $(A_j, I_j)$ is a henselian pair, we can lift $\overline{f}_j = g_{j, 0} h_{j, 0}$ to a factorization over $A_j$ and taking the image in $A$ we obtain a corresponding factorization in $A$. Hence $(A, I)$ is henselian.
By Lemma \ref{lemma-integral-over-henselian-pair} we see that $(A/\mathfrak p, I + \mathfrak p/\mathfrak p)$ is a henselian pair. Thus it suffices to prove: If $(A, I)$ is a henselian pair and $A$ is a domain, then $\Spec(A/I) = V(I)$ is connected. If not, then $A/I$ has a nontrivial idempotent by Algebra, Lemma \ref{algebra-lemma-characterize-spec-connected}. By Lemma \ref{lemma-characterize-henselian-pair} this would imply $A$ has a nontrivial idempotent. This is a contradiction.
In the proof of Lemma \ref{lemma-henselization} we have seen that $A^h$ is a filtered colimit of \'etale $A$-algebras $B$ such that $A/I \to B/IB$ is an isomorphism and we have seen that $I^h = IA^h$. As an \'etale ring map is flat (Algebra, Lemma \ref{algebra-lemma-etale}) we conclude that $A \to A^h$ is flat by Algebra, Lemma \ref{algebra-lemma-colimit-flat}. Since each $A \to B$ is flat we find that the maps $A/I^n \to B/I^nB$ are isomorphisms as well (for example by Algebra, Lemma \ref{algebra-lemma-lift-basis}). Taking the colimit we find that $A/I^n = A^h/I^nA^h$ as desired.
Let $(A^h, \mathfrak m^h)$ be the henselization of the pair $(A, \mathfrak m)$ constructed in Lemma \ref{lemma-henselization}. Then $\mathfrak m^h = \mathfrak m A^h$ is a maximal ideal by Lemma \ref{lemma-henselization-flat} and since it is contained in the Jacobson radical, we conclude $A^h$ is local with maximal ideal $\mathfrak m^h$. Having said this there are two ways to finish the proof. \medskip\noindent First proof: observe that the construction in the proof of Algebra, Lemma \ref{algebra-lemma-henselization} as a colimit is the same as the colimit used to construct $A^h$ in Lemma \ref{lemma-henselization}. Second proof: Both the henselization $A \to S$ and $A \to A^h$ of Lemma \ref{lemma-henselization} are local ring homomorphisms, both $S$ and $A^h$ are filtered colimits of \'etale $A$-algebras, both $S$ and $A^h$ are henselian local rings, and both $S$ and $A^h$ have residue fields equal to $\kappa(\mathfrak m)$ (by Lemma \ref{lemma-henselization-flat} for the second case). Hence they are canonically isomorphic by Algebra, Lemma \ref{algebra-lemma-uniqueness-henselian}.
The first statement is an immediate consequence of Lemma \ref{lemma-henselization-flat} and in fact holds without assuming $A$ is Noetherian. In the proof of Lemma \ref{lemma-henselization} we have seen that $A^h$ is a filtered colimit of \'etale $A$-algebras $B$ such that $A/I \to B/IB$ is an isomorphism. For each such $A \to B$ the induced map $A^\wedge \to B^\wedge$ is an isomorphism (see proof of Lemma \ref{lemma-henselization-flat}). By Algebra, Lemma \ref{algebra-lemma-completion-flat} the ring map $B \to A^\wedge = B^\wedge = (A^h)^\wedge$ is flat for each $B$. Thus $A^h \to A^\wedge = (A^h)^\wedge$ is flat by Algebra, Lemma \ref{algebra-lemma-colimit-rings-flat}. Since $I^h = IA^h$ is contained in the Jacobson radical of $A^h$ and since $A^h \to A^\wedge$ induces an isomorphism $A^h/I^h \to A/I$ we see that $A^h \to A^\wedge$ is faithfully flat by Algebra, Lemma \ref{algebra-lemma-ff}. By Algebra, Lemma \ref{algebra-lemma-completion-Noetherian-Noetherian} the ring $A^\wedge$ is Noetherian. Hence we conclude that $A^h$ is Noetherian by Algebra, Lemma \ref{algebra-lemma-descent-Noetherian}.
By Categories, Lemma \ref{categories-lemma-adjoint-exact} we see that $(A^h, I^h)$ is the colimit of the system $(A_i^h, I_i^h)$ in the category of henselian pairs. Thus for a henselian pair $(B, J)$ we have $$ \Mor((A^h, I^h), (B, J)) = \lim \Mor((A_i^h, I_i^h), (B, J)) = \Mor(\colim (A_i^h, I_i^h), (B, J)) $$ Here the colimit is in the category of pairs. Since the colimit is filtered we obtain $\colim (A_i^h, I_i^h) = (\colim A_i^h, \colim I_i^h)$ in the category of pairs; details omitted. Again using the colimit is filtered, this is a henselian pair (Lemma \ref{lemma-filtered-colimits-henselian}). Hence by the Yoneda lemma we find $(A^h, I^h) = (\colim A_i^h, \colim I_i^h)$.
Let $(A', IA')$ be the pair produced by Lemma \ref{lemma-henselization} starting with the pair $(A, I)$, see Lemma \ref{lemma-henselization-flat}. Let $(A'', JA'')$ be the pair produced by Lemma \ref{lemma-henselization} starting with the pair $(A, J)$. By Lemma \ref{lemma-change-ideal-henselian-pair} we see that $(A', JA')$ is a henselian pair and $(A'', IA'')$ is a henselian pair. By the universal property of the construction we obtain unique $A$-algebra maps $A'' \to A'$ and $A' \to A''$. The uniqueness shows that these are mutually inverse.
By Lemma \ref{lemma-henselization-change-ideal} we may assume $J = IB$. By Lemma \ref{lemma-integral-over-henselian-pair} the pair $(A^h \otimes_A B, I^h(A^h \otimes_A B))$ is henselian. By the universal property of $(B^h, IB^h)$ we obtain a map $B^h \to A^h \otimes_A B$. We omit the proof that this map is the inverse of the map in the lemma.
This follows from the fact that we can lift the finite projective $R/I$-module $\overline{P}$ to a finite projective module $P'$ over some $R'$ \'etale over $R$ with $R/I = R'/IR'$, see Lemma \ref{lemma-lift-projective-module}. Then, since $(R, I)$ is a henselian pair, the \'etale ring map $R \to R'$ has a section $\tau : R' \to R$ (Lemma \ref{lemma-characterize-henselian-pair}). Setting $P = P' \otimes_{R', \tau} R$ finishes the proof.
Let $B, B'$ be two $A$-algebras finite \'etale over $A$. Then $B' \to B'' = B \otimes_A B'$ is finite \'etale as well (Algebra, Lemmas \ref{algebra-lemma-etale} and \ref{algebra-lemma-base-change-integral}). Now we have $1$-to-$1$ correspondences between \begin{enumerate} \item $A$-algebra maps $B \to B'$, \item sections of $B' \to B''$, and \item idempotents $e$ of $B''$ such that $B' \to B'' \to eB''$ is an isomorphism. \end{enumerate} The bijection between (2) and (3) sends $\sigma : B'' \to B'$ to $e$ such that $(1 - e)$ is the idempotent that generates the kernel of $\sigma$ which exists by Algebra, Lemmas \ref{algebra-lemma-map-between-etale} and \ref{algebra-lemma-surjective-flat-finitely-presented}. There is a similar correspondence between $A/I$-algebra maps $B/IB \to B'/IB'$ and idempotents $\overline{e}$ of $B''/IB''$ such that $B'/IB' \to B''/IB'' \to \overline{e}(B''/IB'')$ is an isomorphism. However every idempotent $\overline{e}$ of $B''/IB''$ lifts uniquely to an idempotent $e$ of $B''$ (Lemma \ref{lemma-characterize-henselian-pair}). Moreover, if $B'/IB' \to \overline{e}(B''/IB'')$ is an isomorphism, then $B' \to eB''$ is an isomorphism too by Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK}). In this way we see that the functor is fully faithful. \medskip\noindent Essential surjectivity. Let $A/I \to C$ be a finite \'etale map. By Algebra, Lemma \ref{algebra-lemma-lift-etale} there exists an \'etale map $A \to B$ such that $B/IB \cong C$. Let $B'$ be the integral closure of $A$ in $B$.  By Lemma \ref{lemma-helper-finite-type} we have $B'/IB' = C \times C'$ for some ring $C'$ and $B'_g \cong B_g$ for some $g \in B'$ mapping to $(1, 0) \in C \times C'$. Since idempotents lift (Lemma \ref{lemma-characterize-henselian-pair}) we get $B' = B'_1 \times B'_2$ with $C = B'_1/IB'_1$ and $C' = B'_2/IB'_2$. The image of $g$ in $B'_1$ is invertible. Then $B_g = B'_g = B'_1 \times (B_2)_g$ and this implies that $A \to B'_1$ is \'etale. We conclude that $B'_1$ is finite \'etale over $A$ (integral \'etale implies finite \'etale by Algebra, Lemma \ref{algebra-lemma-characterize-finite-in-terms-of-integral} for example) and the proof is done.
By Lemma \ref{lemma-limit-henselian} the pair $(A, \Ker(A \to A_1))$ is henselian. By Lemma \ref{lemma-lift-finite-projective-module} we can choose a finite projective $A$-module $P$ and an isomorphism $P \otimes_A A_1 \to M_1$. Since $P$ is projective, we can successively lift the $A$-module map $P \to M_1$ to $A$-module maps $P \to M_2$, $P \to M_3$, and so on. Thus we obtain a map $$ P \longrightarrow M $$ Since $P$ is finite projective, we can write $A^{\oplus m} = P \oplus Q$ for some $m \geq 0$ and $A$-module $Q$. Since $A = \lim A_n$ we conclude that $P = \lim P \otimes_A A_n$. Hence, in order to show that the displayed $A$-module map is an isomorphism, it suffices to show that the maps $P \otimes_A A_n \to M_n$ are isomorphisms. From Lemma \ref{lemma-lift-projective} we see that $M_n$ is a finite projective module. By Lemma \ref{lemma-isomorphic-finite-projective-lifts} the maps $P \otimes_A A_n \to M_n$ are isomorphisms.
Let $I$ be the set of monic polynomials over $A$. For $i \in I$ denote $x_i$ a variable and $P_i$ the corresponding monic polynomial in the variable $x_i$. Then we set $$ F(A) = A[x_i; i \in I]/(P_i; i \in I) $$ As the notation suggests $F$ is a functor from the category of rings to itself. Note that $A \subset F(A)$, that $F(A)$ is free as an $A$-module, and that $F(A)$ is a filtered colimit of finite free $A$-algebras. Then we take $$ B = \colim F^n(A) $$ where the transition maps are the inclusions $F^n(A) \subset F(F^n(A)) = F^{n + 1}(A)$. Any monic polynomial with coefficients in $B$ actually has coefficients in $F^n(A)$ for some $n$ and hence has a solution in $F^{n + 1}(A)$ by construction. This implies that $B$ is absolutely integrally closed by Lemma \ref{lemma-absolutely-integrally-closed}. We omit the proof of the other properties.
By Lemma \ref{lemma-absolutely-integrally-closed-quotient-localization} we may assume $A$ is a local ring and $\mathfrak p$ is its maximal ideal. The residue field is algebraically closed by Lemma \ref{lemma-absolutely-integrally-closed-quotient-localization}. Every monic polynomial decomposes completely into linear factors hence Algebra, Definition \ref{algebra-definition-henselian} applies directly.
By assumption there exists a nonzero element $x \in R$ such that for every $f \in \mathfrak m$ we have $f^n x = 0$. Say $I = (f_1, \ldots, f_r)$. Then $x$ is in the kernel of $R \to \bigoplus R_{f_i}$. Hence we see that there exists a nonzero $y \in R$ such that $f_i y = 0$ for all $i$, see Algebra, Lemma \ref{algebra-lemma-when-injective-covering}. As $y \in \text{Ann}_R(I)$ we win.
The implication (1) $\Rightarrow$ (2) is Lemma \ref{lemma-P-universally-injective}. It is clear that (3) and (4) are equivalent. We have (2) $\Rightarrow$ (3), (4) by Algebra, Lemma \ref{algebra-lemma-universally-exact-split}. Part (5) is a special case of (4). Assume (5). Let $I = (a_1, \ldots, a_n)$ be a proper finitely generated ideal of $R$. As $I \not = R$ we see that $R \to R^{\oplus n}$, $x \mapsto (xa_1, \ldots, xa_n)$ is not a split injection. Hence it has a nonzero kernel and we conclude that $\text{Ann}_R(I) \not = 0$. Thus (1) holds.
If any matrix coefficient of $\varphi$ is not in $\mathfrak m$, then we apply Algebra, Lemma \ref{algebra-lemma-add-trivial-complex} to write $\varphi$ as the sum of $1 : R \to R$ and a map $\varphi' : R^{m-1} \to R^{n-1}$. It is easy to see that the lemma for $\varphi'$ implies the lemma for $\varphi$. Thus we may assume from the outset that all the matrix coefficients of $\varphi$ are in $\mathfrak m$. \medskip\noindent Suppose $\varphi$ is injective. We may assume $m > 0$. Let $\mathfrak q \in \text{WeakAss}(R)$ so that $R_\mathfrak q$ is an auto-associated ring. Then $\varphi$ induces a injective map $R_\mathfrak q^m \to R_\mathfrak q^n$ which is universally injective by Lemmas \ref{lemma-auto-ass-implies-P} and \ref{lemma-P-universally-injective}. Thus $\varphi : \kappa(\mathfrak q)^m \to \kappa(\mathfrak q)^n$ is injective. Hence the rank of $\varphi \bmod \mathfrak q$ is $m$ and $I(\varphi \otimes \kappa(\mathfrak q))$ is not the zero ideal. Since $m$ is the maximum rank $\varphi$ can have, we conclude that $\varphi$ has rank $m$ as well (ranks of matrices can only drop after base change). Hence $I(\varphi) \cdot \kappa(\mathfrak q) = I(\varphi \otimes \kappa(\mathfrak q))$ is not zero. Thus $I(\varphi)$ is not contained in $\mathfrak q$. Thus none of the weakly associated primes of $R$ are weakly associated primes of the $R$-module $\text{Ann}_R I(\varphi)$. Thus $\text{Ann}_R I(\varphi)$ has no weakly associated primes, see Algebra, Lemma \ref{algebra-lemma-weakly-ass}. It follows from Algebra, Lemma \ref{algebra-lemma-weakly-ass-zero} that $\text{Ann}_R I(\varphi)$ is zero. \medskip\noindent Conversely, assume (2). The rank being $m$ implies $n \geq m$. Write $I(\varphi) = (f_1, \ldots, f_r)$ which is possible as $I(\varphi)$ is finitely generated. By Algebra, Lemma \ref{algebra-lemma-matrix-left-inverse} we can find maps $\psi_i : R^n \to R^m$ such that $\psi \circ \varphi = f_i \text{id}_{R^m}$. Thus $\varphi(x) = 0$ implies $f_i x = 0$ for $i = 1, \ldots, r$. This implies $x = 0$ and hence $\varphi$ is injective. \medskip\noindent For the equivalence of (1) and (3) in the Noetherian local case we refer to Algebra, Proposition \ref{algebra-proposition-what-exact}. If the ring $R$ is Noetherian but not local, then the reader can deduce it from the local case; details omitted. Another option is to redo the argument above using associated primes, using that there are finitely many of these, using prime avoidance, and using the characterization of nonzerodivisors as elements of a Noetherian ring not contained in any associated prime.
Let $\varphi^t : R^n \to R^n$ be the transpose of $\varphi$. The lemma claims that $\varphi^t$ is injective. With notation as in Lemma \ref{lemma-exact-length-1} we see that the rank of $\varphi^t$ is $n$ and that $I(\varphi) = I(\varphi^t)$. Thus we conclude by the equivalence of (1) and (2) of the lemma.