hcju/stacksqa huggingface dataset:
Let $K^\bullet_i$, $i \in I$ be a family of objects of $D(\mathcal{A})$ indexed by a set $I$. We claim that the termwise direct sum $\bigoplus_{i \in I} K^\bullet_i$ is a direct sum in $D(\mathcal{A})$. Namely, let $I^\bullet$ be a K-injective complex. Then we have \begin{align*} \Hom_{D(\mathcal{A})}(\bigoplus\nolimits_{i \in I} K^\bullet_i, I^\bullet) & = \Hom_{K(\mathcal{A})}(\bigoplus\nolimits_{i \in I} K^\bullet_i, I^\bullet) \\ & = \prod\nolimits_{i \in I} \Hom_{K(\mathcal{A})}(K^\bullet_i, I^\bullet) \\ & = \prod\nolimits_{i \in I} \Hom_{D(\mathcal{A})}(K^\bullet_i, I^\bullet) \end{align*} as desired. This is sufficient since any complex can be represented by a K-injective complex by Theorem \ref{theorem-K-injective-embedding-grothendieck}. To construct the product, choose a K-injective resolution $K_i^\bullet \to I_i^\bullet$ for each $i$. Then we claim that $\prod_{i \in I} I_i^\bullet$ is a product in $D(\mathcal{A})$. This follows from Derived Categories, Lemma \ref{derived-lemma-product-K-injective}.
Observe that $RF$ exists as $\mathcal{A}$ has enough K-injectives (Theorem \ref{theorem-K-injective-embedding-grothendieck} and Derived Categories, Lemma \ref{derived-lemma-K-injective-defined}). The statement means that if $K = R\lim K_n$, then $RF(K) = R\lim RF(K_n)$. See Derived Categories, Definition \ref{derived-definition-derived-limit} for notation. Since $RF$ is an exact functor of triangulated categories it suffices to see that $RF$ commutes with countable products of objects of $D(\mathcal{A})$. In the proof of Lemma \ref{lemma-derived-products} we have seen that products in $D(\mathcal{A})$ are computed by taking products of K-injective complexes and moreover that a product of K-injective complexes is K-injective. Moreover, in Derived Categories, Lemma \ref{derived-lemma-products} we have seen that products in $D(\mathcal{B})$ are computed by taking termwise products. Since $RF$ is computed by applying $F$ to a K-injective representative and since we've assumed $F$ commutes with countable products, the lemma follows.
Using the lemma we can first choose $K^\bullet$ and $F$. Then we can choose $(K')^\bullet$ and $F'$ which work for $\{(E')^i\}_{i \in \mathbf{Z}}$ and the maps $(E')^i \to E$. Using Lemma \ref{lemma-K-injective-embedding-filtration} we can assume $K^\bullet$ is a K-injective complex. Then we can choose a map of complexes $(K')^\bullet \to K^\bullet$ corresponding to the given identifications $(K')^\bullet \cong E \cong K^\bullet$. We can additionally choose a termwise injective map $(K')^\bullet \to J^\bullet$ with $J^\bullet$ acyclic and K-injective. (To do this first map $(K')^\bullet$ to the cone on the identity and then apply Theorem \ref{theorem-K-injective-embedding-grothendieck}.) Then $(K')^\bullet \to K^\bullet \times J^\bullet$ and $K^\bullet \to K^\bullet \times J^\bullet$ are both termwise injective and quasi-isomorphisms (as the product represents $E$ by Lemma \ref{lemma-derived-products}). Then we can simply take the images of the filtrations on $K^\bullet$ and $(K')^\bullet$ under these maps to conclude.
We will give two proofs of this lemma. \medskip\noindent The first proof will use the adjoint functor theorem, see Categories, Theorem \ref{categories-theorem-adjoint-functor}. Observe that that $G : \mathcal{A} \to \text{Mod}_R$ is left exact and sends products to products. Hence $G$ commutes with limits. To check the set theoretical condition in the theorem, suppose that $M$ is an object of $\text{Mod}_R$. Choose a suitably large cardinal $\kappa$ and denote $E$ a set of objects of $\mathcal{A}$ such that every object $A$ with $|A| \leq \kappa$ is isomorphic to an element of $E$. This is possible by Lemma \ref{lemma-set-iso-classes-bounded-size}. Set $I = \coprod_{A \in E} \text{Hom}_R(M, G(A))$. We think of an element $i \in I$ as a pair $(A_i, f_i)$. Finally, let $A$ be an arbitrary object of $\mathcal{A}$ and $f : M \to G(A)$ arbitrary. We are going to think of elements of $\Im(f) \subset G(A) = \Hom_\mathcal{A}(U, A)$ as maps $u : U \to A$. Set $$ A' = \Im(\bigoplus\nolimits_{u \in \Im(f)} U \xrightarrow{u} A) $$ Since $G$ is left exact, we see that $G(A') \subset G(A)$ contains $\Im(f)$ and we get $f' : M \to G(A')$ factoring $f$. On the other hand, the object $A'$ is the quotient of a direct sum of at most $|M|$ copies of $U$. Hence if $\kappa = |\bigoplus_{|M|} U|$, then we see that $(A', f')$ is isomorphic to an element $(A_i, f_i)$ of $E$ and we conclude that $f$ factors as $M \xrightarrow{f_i} G(A_i) \to G(A)$ as desired. \medskip\noindent The second proof will give a construction of $F$ which will show that ``$F(M) = M \otimes_R U$'' in some sense. Namely, for any $R$-module $M$ we can choose a resolution $$ \bigoplus\nolimits_{j \in J} R \to \bigoplus\nolimits_{i \in I} R \to M \to 0 $$ Then we define $F(M)$ by the corresponding exact sequence $$ \bigoplus\nolimits_{j \in J} U \to \bigoplus\nolimits_{i \in I} U \to F(M) \to 0 $$ This construction is independent of the choice of the resolution and is functorial; we omit the details. For any $A$ in $\mathcal{A}$ we obtain an exact sequence $$ 0 \to \Hom_\mathcal{A}(F(M), A) \to \prod\nolimits_{i \in I} G(A) \to \prod\nolimits_{j \in J} G(A) $$ which is isomorphic to the sequence $$ 0 \to \Hom_R(M, A) \to \Hom_R(\bigoplus\nolimits_{i \in I} R, G(A)) \to \Hom_R(\bigoplus\nolimits_{j \in J} R, G(A)) $$ which shows that $F$ is the left adjoint to $G$.
Choose a map $R^{\oplus n} \to M$ and consider the corresponding map $U^{\oplus n} \to F(M)$. Consider a map $v : U \to U^{\oplus n}$ such that the composition $U \to U^{\oplus n} \to F(M) \to A$. Then this arrow $v : U \to U^{\oplus n}$ is an element $v$ of $R^{\oplus n}$ mapping to zero in $G(A)$. Since $f$ is injective, we conclude that $v$ maps to zero in $M$ which means that $U \to U^{\oplus n} \to F(M)$ is zero by construction of $F(M)$ in the proof of Lemma \ref{lemma-gabriel-popescu-left-adjoint}. Since $U$ is a generator we conclude that $$ \Ker(U^{\oplus n} \to F(M) \to A) = \Ker(U^{\oplus n} \to F(M)) $$ To finish the proof we choose a surjection $\bigoplus_{i \in I} R \to M$ and we consider the corresponding surjection $$ \pi : \bigoplus\nolimits_{i \in I} U \longrightarrow F(M) $$ To prove $f'$ is injective it suffices to show that $\Ker(\pi) = \Ker(f' \circ \pi)$ as subobjects of $\bigoplus_{i \in I} U$. However, now we can write $\bigoplus_{i \in I} U$ as the filtered colimit of its subobjects $\bigoplus_{i \in I'} U$ where $I' \subset I$ ranges over the finite subsets. Since filtered colimits are exact by AB5 for $\mathcal{A}$, we see that $$ \Ker(\pi) = \colim_{I' \subset I\text{ finite}} \left(\bigoplus\nolimits_{i \in I'} U\right) \bigcap \Ker(\pi) $$ and $$ \Ker(f' \circ \pi) = \colim_{I' \subset I\text{ finite}} \left(\bigoplus\nolimits_{i \in I'} U\right) \bigcap \Ker(f' \circ \pi) $$ and we get equality because the same is true for each $I'$ by the first displayed equality above.
The existence and adjointness of the functors follows from Theorems \ref{theorem-gabriel-popescu} and \ref{theorem-K-injective-embedding-grothendieck} and Derived Categories, Lemmas \ref{derived-lemma-K-injective-defined}, \ref{derived-lemma-right-derived-exact-functor}, and \ref{derived-lemma-derived-adjoint-functors}. The statement $F \circ RG = \text{id}$ follows because we can compute $RG$ on an object of $D(\mathcal{A})$ by applying $G$ to a suitable representative complex $I^\bullet$ (for example a K-injective one) and then $F(G(I^\bullet)) = I^\bullet$ because $F \circ G = \text{id}$. Fully faithfulness of $RG$ follows from this by Categories, Lemma \ref{categories-lemma-adjoint-fully-faithful}.
Let $R, F, G, RG$ be as in Lemma \ref{lemma-gabriel-popescu} and consider the functor $H \circ F : D(\text{Mod}_R) \to \textit{Ab}$. Observe that since $F$ is a left adjoint it sends direct sums to direct sums and hence $H \circ F$ transforms direct sums into products. On the other hand, the derived category $D(\text{Mod}_R)$ is generated by a single compact object, namely $R$. By Derived Categories, Lemma \ref{derived-lemma-brown} we see that $H \circ F$ is representable, say by $L \in D(\text{Mod}_R)$. Choose a distinguished triangle $$ M \to L \to RG(F(L)) \to M[1] $$ in $D(\text{Mod}_R)$. Then $F(M) = 0$ because $F \circ RG = \text{id}$. Hence $H(F(M)) = 0$ hence $\Hom(M, L) = 0$. It follows that $L \to RG(F(L))$ is the inclusion of a direct summand, see Derived Categories, Lemma \ref{derived-lemma-split}. For $A$ in $D(\mathcal{A})$ we obtain \begin{align*} H(A) & = H(F(RG(A)) \\ & = \Hom(RG(A), L) \\ & \to \Hom(RG(A), RG(F(L))) \\ & = \Hom(F(RG(A)), F(L)) \\ & = \Hom(A, F(L)) \end{align*} where the arrow has a left inverse functorial in $A$. In other words, we find that $H$ is the direct summand of a representable functor. Since $D(\mathcal{A})$ is Karoubian (Derived Categories, Lemma \ref{derived-lemma-projectors-have-images-triangulated}) we conclude.
The proof is straightforward, but let us first think about a special case. If $M$ is finite, then the claim is that for any inductive system $\{B_\beta\}$ with injections between them, parametrized by a limit ordinal, any map $M \to \colim B_\beta$ factors through one of the $B_\beta$. And this we proved in Lemma \ref{lemma-out-of-finite}. \medskip\noindent Now we start the proof in the general case. We need only show that the map (\ref{equation-compare}) is a surjection. Let $f : M \to \colim B_\beta$ be a map. Consider the subobjects $\{f^{-1}(B_\beta)\}$ of $M$, where $B_\beta$ is considered as a subobject of the colimit $B = \bigcup_\beta B_\beta$. If one of these, say $f^{-1}(B_\beta)$, fills $M$, then the map factors through $B_\beta$. \medskip\noindent So suppose to the contrary that all of the $f^{-1}(B_\beta)$ were proper subobjects of $M$. However, we know that $$ \bigcup f^{-1}(B_\beta) = f^{-1}\left(\bigcup B_\beta\right) = M. $$ Now there are at most $\kappa$ different subobjects of $M$ that occur among the $f^{-1}(B_\alpha)$, by hypothesis. Thus we can find a subset $S \subset \alpha$ of cardinality at most $\kappa$ such that as $\beta'$ ranges over $S$, the $f^{-1}(B_{\beta'})$ range over \emph{all} the $f^{-1}(B_\alpha)$. \medskip\noindent However, $S$ has an upper bound $\widetilde{\alpha} < \alpha$ as $\alpha$ has cofinality bigger than $\kappa$. In particular, all the $f^{-1}(B_{\beta'})$, $\beta' \in S$ are contained in $f^{-1}(B_{\widetilde{\alpha}})$. It follows that $f^{-1}(B_{\widetilde{\alpha}}) = M$. In particular, the map $f$ factors through $B_{\widetilde{\alpha}}$.
Please compare with Proposition \ref{proposition-modules-are-small}. We need only show that the map (\ref{equation-compare}) is a surjection. Let $f : M \to \colim B_\beta$ be a map. Consider the subobjects $\{f^{-1}(B_\beta)\}$ of $M$, where $B_\beta$ is considered as a subobject of the colimit $B = \bigcup_\beta B_\beta$. If one of these, say $f^{-1}(B_\beta)$, fills $M$, then the map factors through $B_\beta$. \medskip\noindent So suppose to the contrary that all of the $f^{-1}(B_\beta)$ were proper subobjects of $M$. However, because $\mathcal{A}$ has AB5 we have $$ \colim f^{-1}(B_\beta) = f^{-1}\left(\colim B_\beta\right) = M. $$ Now there are at most $\kappa$ different subobjects of $M$ that occur among the $f^{-1}(B_\alpha)$, by hypothesis. Thus we can find a subset $S \subset \alpha$ of cardinality at most $\kappa$ such that as $\beta'$ ranges over $S$, the $f^{-1}(B_{\beta'})$ range over \emph{all} the $f^{-1}(B_\alpha)$. \medskip\noindent However, $S$ has an upper bound $\widetilde{\alpha} < \alpha$ as $\alpha$ has cofinality bigger than $\kappa$. In particular, all the $f^{-1}(B_{\beta'})$, $\beta' \in S$ are contained in $f^{-1}(B_{\widetilde{\alpha}})$. It follows that $f^{-1}(B_{\widetilde{\alpha}}) = M$. In particular, the map $f$ factors through $B_{\widetilde{\alpha}}$.
For an object $Y$ of $\mathcal{D}$ consider the contravariant functor $$ D(\mathcal{A}) \to \textit{Ab},\quad W \mapsto \Hom_\mathcal{D}(F(W), Y) $$ This is a cohomological functor as $F$ is exact and transforms direct sums into products as $F$ transforms direct sums into direct sums. Thus by Lemma \ref{lemma-brown} we find an object $X$ of $D(\mathcal{A})$ such that $\Hom_{D(\mathcal{A})}(W, X) = \Hom_\mathcal{D}(F(W), Y)$. The existence of the adjoint follows from Categories, Lemma \ref{categories-lemma-adjoint-exists}. Exactness follows from Derived Categories, Lemma \ref{derived-lemma-adjoint-is-exact}.
We may choose a simple submodule $M \subset A$ and then the $k$-algebra $K = \text{End}_A(M)$ is a skew field, see Lemma \ref{lemma-simple-module}. By Lemma \ref{lemma-rieffel} we see that $A = \text{End}_K(M)$. Since $K$ is a skew field and $M$ is finitely generated (since $\dim_k(M) < \infty$) we see that $M$ is finite free as a left $K$-module. It follows immediately that $A \cong \text{Mat}(n \times n, K^{op})$.
Choose a simple $A$-module $M$. Set $L = \text{End}_A(M)$. Then $L$ is a skew field with center $k$ which acts on the left on $M$, see Lemmas \ref{lemma-simple-module} and \ref{lemma-simple-module-unique}. Then $M$ has two $B \otimes_k L^{op}$-module structures defined by $m \cdot_1 (b \otimes l) = lmf(b)$ and $m \cdot_2 (b \otimes l) = lmg(b)$. The $k$-algebra $B \otimes_k L^{op}$ is simple by Lemma \ref{lemma-tensor-simple}. Since $B$ is simple, the existence of a $k$-algebra homomorphism $B \to A$ implies that $B$ is finite. Thus $B \otimes_k L^{op}$ is finite simple and we conclude the two $B \otimes_k L^{op}$-module structures on $M$ are isomorphic by Lemma \ref{lemma-simple-module-unique}. Hence we find $\varphi : M \to M$ intertwining these operations. In particular $\varphi$ is in the commutant of $L$ which implies that $\varphi$ is multiplication by some $x \in A$, see Lemma \ref{lemma-simple-module-unique}. Working out the definitions we see that $x$ is a solution to our problem.
Throughout this proof we use the results of Lemma \ref{lemma-simple-module-unique} freely. Choose a simple $A$-module $M$. Set $L = \text{End}_A(M)$. Then $L$ is a skew field with center $k$ which acts on the left on $M$ and $A = \text{End}_L(M)$. Then $M$ is a right $B \otimes_k L^{op}$-module and $C = \text{End}_{B \otimes_k L^{op}}(M)$. Since the algebra $B \otimes_k L^{op}$ is simple by Lemma \ref{lemma-tensor-simple} we see that $C$ is simple (by Lemma \ref{lemma-simple-module-unique} again). \medskip\noindent Write $B \otimes_k L^{op} = \text{Mat}(m \times m, K)$ for some skew field $K$ finite over $k$. Then $C = \text{Mat}(n \times n, K^{op})$ if $M$ is isomorphic to a direct sum of $n$ copies of the simple $B \otimes_k L^{op}$-module $K^{\oplus m}$ (the lemma again). Thus we have $\dim_k(M) = nm [K : k]$, $[B : k] [L : k] = m^2 [K : k]$, $[C : k] = n^2 [K : k]$, and $[A : k] [L : k] = \dim_k(M)^2$ (by the lemma again). We conclude that (2) holds. \medskip\noindent Part (3) follows because of (2) applied to $C \subset A$ shows that $[B : k] = [C' : k]$ where $C'$ is the centralizer of $C$ in $A$ (and the obvious fact that $B \subset C')$.
Assume (2). It suffices to show that $B \otimes_k k'$ is a matrix algebra. We know that $B \otimes_k B^{op} \cong \text{End}_k(B)$. Since $k'$ is the centralizer of $k'$ in $B^{op}$ by Lemma \ref{lemma-self-centralizing-subfield} we see that $B \otimes_k k'$ is the centralizer of $k \otimes k'$ in $B \otimes_k B^{op} = \text{End}_k(B)$. Of course this centralizer is just $\text{End}_{k'}(B)$ where we view $B$ as a $k'$ vector space via the embedding $k' \to B$. Thus the result. \medskip\noindent Assume (1). This means that we have an isomorphism $A \otimes_k k' \cong \text{End}_{k'}(V)$ for some $k'$-vector space $V$. Let $B$ be the commutant of $A$ in $\text{End}_k(V)$. Note that $k'$ sits in $B$. By Lemma \ref{lemma-when-tensor-is-equal} the classes of $A$ and $B$ add up to zero in $\text{Br}(k)$. From the dimension formula in Theorem \ref{theorem-centralizer} we see that $$ [B : k] [A : k] = \dim_k(V)^2 = [k' : k]^2 \dim_{k'}(V)^2 = [k' : k]^2 [A : k]. $$ Hence $[B : k] = [k' : k]^2$. Thus we have proved the result for the opposite to the Brauer class of $A$. However, $k'$ splits the Brauer class of $A$ if and only if it splits the Brauer class of the opposite algebra, so we win anyway.
Write $A = \text{Mat}(n \times n, K)$ for some skew field $K$ finite over $k$, see Theorem \ref{theorem-wedderburn}. By Lemma \ref{lemma-centralizer} the center of $A$ is $k \otimes_k k'$ where $k' \subset K$ is the center of $K$. Since the center of a skew field is a field, we win.
By Theorem \ref{theorem-wedderburn} we can write $A = \text{Mat}(n \times n, K)$ for some finite skew field extension $K$ of $k$. By Lemma \ref{lemma-matrix-algebras} the category of modules over $A$ is equivalent to the category of modules over $K$. Thus (1), (2), and (3) hold because every module over $K$ is free. Part (4) holds because the equivalence transforms the $K$-module $K$ to $M = K^{\oplus n}$. Using $M = K^{\oplus n}$ in (5) we see that $L = K^{op}$. The statement about the center of $L = K^{op}$ follows from Lemma \ref{lemma-matrix-algebras}. The statement about $\text{End}_L(M)$ follows from the explicit form of $M$. The formula of dimensions is clear. Part (6) follows as $N$ is isomorphic to a direct sum of copies of a simple module.
Suppose that $A'$ is finite and central over $k$. Write $A' = \text{Mat}(n \times n, K')$, see Theorem \ref{theorem-wedderburn}. Then the center of $K'$ is $k$ and we conclude that $A \otimes_k K'$ is simple by Lemma \ref{lemma-generate-two-sided-ideal}. Hence $A \otimes_k A' = \text{Mat}(n \times n, A \otimes_k K')$ is simple by Lemma \ref{lemma-matrix-algebras}.
By Lemma \ref{lemma-tensor-central-simple} the algebra $A \otimes_k A^{op}$ is simple. Hence the map $$ A \otimes_k A^{op} \longrightarrow \text{End}_k(A),\quad a \otimes a' \longmapsto (x \mapsto axa') $$ is injective. Since both sides of the arrow have the same dimension we win.
Note that by Wedderburn's theorem (Theorem \ref{theorem-wedderburn}) we can always write a finite central simple algebra as a matrix algebra over a finite central skew field. Hence it suffices to prove the third assertion. To see this it suffices to show that if $A = \text{Mat}(n \times n, K) \cong \text{Mat}(m \times m, K') = B$ then $K \cong K'$. To see this note that for a simple module $M$ of $A$ we have $\text{End}_A(M) = K^{op}$, see Lemma \ref{lemma-simple-module-unique}. Hence $A \cong B$ implies $K^{op} \cong (K')^{op}$ and we win.
Let $k \subset K$ be a finite central skew field extension. For any element $x \in K$ the subring $k[x] \subset K$ is a commutative finite integral $k$-sub algebra, hence a field, see Algebra, Lemma \ref{algebra-lemma-integral-over-field}. Since $k$ is algebraically closed we conclude that $k[x] = k$. Since $x$ was arbitrary we conclude $k = K$.
By Theorem \ref{theorem-splitting} there exists a finite central simple algebra $B$ in the Brauer class of $K$ such that $[B : k] = [k' : k]^2$. By Lemma \ref{lemma-similar} we see that $B = \text{Mat}(n \times n, K)$ for some $n$. Then $[k' : k]^2 = n^2d^2$ whence the result.
The equivalence of (1) and (2) is a consequence of the definitions, see Section \ref{section-algebras}. Assume (1). By Proposition \ref{proposition-separable-splitting-field} there exists a separable splitting field $k \subset k'$ for $A$. Of course, then a Galois closure of $k'/k$ is a splitting field also. Thus we see that (1) implies (5). It is clear that (5) $\Rightarrow$ (4) $\Rightarrow$ (3). Assume (3). Then $A \otimes_k \overline{k}$ is a finite central simple $\overline{k}$-algebra for example by Lemma \ref{lemma-matrix-algebras}. This trivially implies that $A$ is a finite central simple $k$-algebra. Finally, the equivalence of (1) and (6) is Wedderburn's theorem, see Theorem \ref{theorem-wedderburn}.
Since every Brauer class is represented by a finite central skew field over $k$, we see that the second statement follows from the first by Lemma \ref{lemma-maximal-subfield-splits}. \medskip\noindent To prove the first statement, suppose that we are given a separable subfield $k' \subset K$. Then the centralizer $K'$ of $k'$ in $K$ has center $k'$, and the problem reduces to finding a maximal subfield of $K'$ separable over $k'$. Thus it suffices to prove, if $k \not = K$, that we can find an element $x \in K$, $x \not \in k$ which is separable over $k$. This statement is clear in characteristic zero. Hence we may assume that $k$ has characteristic $p > 0$. If the ground field $k$ is finite then, the result is clear as well (because extensions of finite fields are always separable). Thus we may assume that $k$ is an infinite field of positive characteristic. \medskip\noindent To get a contradiction assume no element of $K$ is separable over $k$. By the discussion in Fields, Section \ref{fields-section-algebraic} this means the minimal polynomial of any $x \in K$ is of the form $T^q - a$ where $q$ is a power of $p$ and $a \in k$. Since it is clear that every element of $K$ has a minimal polynomial of degree $\leq \dim_k(K)$ we conclude that there exists a fixed $p$-power $q$ such that $x^q \in k$ for all $x \in K$. \medskip\noindent Consider the map $$ (-)^q : K \longrightarrow K $$ and write it out in terms of a $k$-basis $\{a_1, \ldots, a_n\}$ of $K$ with $a_1 = 1$. So $$ (\sum x_i a_i)^q = \sum f_i(x_1, \ldots, x_n)a_i. $$ Since multiplication on $A$ is $k$-bilinear we see that each $f_i$ is a polynomial in $x_1, \ldots, x_n$ (details omitted). The choice of $q$ above and the fact that $k$ is infinite shows that $f_i$ is identically zero for $i \geq 2$. Hence we see that it remains zero on extending $k$ to its algebraic closure $\overline{k}$. But the algebra $A \otimes_k \overline{k}$ is a matrix algebra, which implies there are some elements whose $q$th power is not central (e.g., $e_{11}$). This is the desired contradiction.
This follows from Algebra, Lemma \ref{algebra-lemma-associated-primes-localize}. But we can also argue directly as follows. Suppose $\mathfrak p$ is associated to $M$. Then there exists an $m \in M$ whose annihilator is $\mathfrak p$. Since localization is exact we see that $\mathfrak pA_{\mathfrak p}$ is the annihilator of $m/1 \in M_{\mathfrak p}$. Since $M_{\mathfrak p} = \mathcal{F}_x$ (Schemes, Lemma \ref{schemes-lemma-spec-sheaves}) we conclude that $x$ is associated to $\mathcal{F}$. \medskip\noindent Conversely, assume that $x$ is associated to $\mathcal{F}$, and $\mathfrak p$ is finitely generated. As $x$ is associated to $\mathcal{F}$ there exists an element $m' \in M_{\mathfrak p}$ whose annihilator is $\mathfrak pA_{\mathfrak p}$. Write $m' = m/f$ for some $f \in A$, $f \not \in \mathfrak p$. The annihilator $I$ of $m$ is an ideal of $A$ such that $IA_{\mathfrak p} = \mathfrak pA_{\mathfrak p}$. Hence $I \subset \mathfrak p$, and $(\mathfrak p/I)_{\mathfrak p} = 0$. Since $\mathfrak p$ is finitely generated, there exists a $g \in A$, $g \not \in \mathfrak p$ such that $g(\mathfrak p/I) = 0$. Hence the annihilator of $gm$ is $\mathfrak p$ and we win. \medskip\noindent If $X$ is locally Noetherian, then $A$ is Noetherian (Properties, Lemma \ref{properties-lemma-locally-Noetherian}) and $\mathfrak p$ is always finitely generated.
For every point $x \in X$ the sequence of stalks $0 \to \mathcal{F}_{1, x} \to \mathcal{F}_{2, x} \to \mathcal{F}_{3, x} \to 0$ is a short exact sequence of $\mathcal{O}_{X, x}$-modules. Hence the lemma follows from Algebra, Lemma \ref{algebra-lemma-ass}.
This is true because the set of associated primes of a finite module over a Noetherian ring is finite, see Algebra, Lemma \ref{algebra-lemma-finite-ass}. To translate from schemes to algebra use that $U$ is a finite union of affine opens, each of these opens is the spectrum of a Noetherian ring (Properties, Lemma \ref{properties-lemma-locally-Noetherian}), $\mathcal{F}$ corresponds to a finite module over this ring (Cohomology of Schemes, Lemma \ref{coherent-lemma-coherent-Noetherian}), and finally use Lemma \ref{lemma-associated-affine-open}.
If $\mathcal{F} = 0$, then $\text{Ass}(\mathcal{F}) = \emptyset$ by definition. Conversely, if $\text{Ass}(\mathcal{F}) = \emptyset$, then $\mathcal{F} = 0$ by Algebra, Lemma \ref{algebra-lemma-ass-zero}. To translate from schemes to algebra, restrict to any affine and use Lemma \ref{lemma-associated-affine-open}.