hcju/stacksqa huggingface dataset:
Let $a \leq b$ be such that $H^i(K) = 0$ for $i \not \in [a, b]$. By induction on $b - a$ we will prove there exists a covering $\{U_i \to X\}$ such that $K|_{U_i}$ can be represented by a complex $\underline{M^\bullet}_{U_i}$ with $M^p$ a finite type $\Lambda$-module and $M^p = 0$ for $p \not \in [a, b]$. If $b = a$, then this is clear. In general, we may replace $X$ by the members of a covering and assume that $H^b(K)$ is constant, say $H^b(K) = \underline{M}$. By Modules on Sites, Lemma \ref{sites-modules-lemma-locally-constant-finite-type} the module $M$ is a finite $\Lambda$-module. Choose a surjection $\Lambda^{\oplus r} \to M$ given by generators $x_1, \ldots, x_r$ of $M$. \medskip\noindent By a slight generalization of Lemma \ref{lemma-kill-cohomology-class-on-covering} (details omitted) there exists a covering $\{U_i \to X\}$ such that $x_i \in H^0(X, H^b(K))$ lifts to an element of $H^b(U_i, K)$. Thus, after replacing $X$ by the $U_i$ we reach the situation where there is a map $\underline{\Lambda^{\oplus r}}[-b] \to K$ inducing a surjection on cohomology sheaves in degree $b$. Choose a distinguished triangle $$ \underline{\Lambda^{\oplus r}}[-b] \to K \to L \to \underline{\Lambda^{\oplus r}}[-b + 1] $$ Now the cohomology sheaves of $L$ are nonzero only in the interval $[a, b - 1]$, agree with the cohomology sheaves of $K$ in the interval $[a, b - 2]$ and there is a short exact sequence $$ 0 \to H^{b - 1}(K) \to H^{b - 1}(L) \to \underline{\Ker(\Lambda^{\oplus r} \to M)} \to 0 $$ in degree $b - 1$. By Modules on Sites, Lemma \ref{sites-modules-lemma-kernel-finite-locally-constant} we see that $H^{b - 1}(L)$ is locally constant of finite type. By induction hypothesis we obtain an isomorphism $\underline{M^\bullet} \to L$ in $D(\mathcal{C}, \underline{\Lambda})$ with $M^p$ a finite $\Lambda$-module and $M^p = 0$ for $p \not \in [a, b - 1]$. The map $L \to \Lambda^{\oplus r}[-b + 1]$ gives a map $\underline{M^{b - 1}} \to \underline{\Lambda^{\oplus r}}$ which locally is constant (Modules on Sites, Lemma \ref{sites-modules-lemma-morphism-locally-constant}). Thus we may assume it is given by a map $M^{b - 1} \to \Lambda^{\oplus r}$. The distinguished triangle shows that the composition $M^{b - 2} \to M^{b - 1} \to \Lambda^{\oplus r}$ is zero and the axioms of triangulated categories produce an isomorphism $$ \underline{M^a \to \ldots \to M^{b - 1} \to \Lambda^{\oplus r}} \longrightarrow K $$ in $D(\mathcal{C}, \Lambda)$.
We'll prove this as an application of Lemma \ref{lemma-locally-constant}. Note that $H^i(K \otimes_\Lambda^\mathbf{L} L)$ is the same as $H^i(\tau_{\geq i - 1}K \otimes_\Lambda^\mathbf{L} \tau_{\geq i - 1}L)$. Thus we may assume $K$ and $L$ are bounded. By Lemma \ref{lemma-locally-constant} we may assume that $K$ and $L$ are represented by complexes of locally constant sheaves of $\Lambda$-modules of finite type. Then we can replace these complexes by bounded above complexes of finite free $\Lambda$-modules. In this case the result is clear.
Recall that the locally constant sheaves of $\Lambda$-modules of finite type form a weak Serre subcategory of all $\underline{\Lambda}$-modules, see Modules on Sites, Lemma \ref{sites-modules-lemma-kernel-finite-locally-constant}. Thus the subcategory of $D(\mathcal{C}, \Lambda)$ consisting of complexes whose cohomology sheaves are locally constant sheaves of $\Lambda$-modules of finite type forms a strictly full, saturated triangulated subcategory of $D(\mathcal{C}, \Lambda)$, see Derived Categories, Lemma \ref{derived-lemma-cohomology-in-serre-subcategory}. Next, consider the distinguished triangles $$ K \otimes_\Lambda^\mathbf{L} \underline{I^n/I^{n + 1}} \to K \otimes_\Lambda^\mathbf{L} \underline{\Lambda/I^{n + 1}} \to K \otimes_\Lambda^\mathbf{L} \underline{\Lambda/I^n} \to K \otimes_\Lambda^\mathbf{L} \underline{I^n/I^{n + 1}}[1] $$ and the isomorphisms $$ K \otimes_\Lambda^\mathbf{L} \underline{I^n/I^{n + 1}} = \left(K \otimes_\Lambda^\mathbf{L} \underline{\Lambda/I}\right) \otimes_{\Lambda/I}^\mathbf{L} \underline{I^n/I^{n + 1}} $$ Combined with Lemma \ref{lemma-locally-constant-tensor-product} we obtain the result.
Proof of (1). If the sequence is exact, then evaluating at any weakly contractible element of $\mathcal{C}$ gives an exact sequence by Lemma \ref{lemma-w-contractible}. Conversely, assume that $\mathcal{F}_1(U) \to \mathcal{F}_2(U) \to \mathcal{F}_3(U)$ is exact for all $U \in \mathcal{B}$. Let $V$ be an object of $\mathcal{C}$ and let $s \in \mathcal{F}_2(V)$ be an element of the kernel of $\mathcal{F}_2 \to \mathcal{F}_3$. By assumption there exists a covering $\{U_i \to V\}$ with $U_i \in \mathcal{B}$. Then $s|_{U_i}$ lifts to a section $s_i \in \mathcal{F}_1(U_i)$. Thus $s$ is a section of the image sheaf $\Im(\mathcal{F}_1 \to \mathcal{F}_2)$. In other words, the sequence $\mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3$ is exact. \medskip\noindent Proof of (2). This holds by Lemma \ref{lemma-is-limit-dimension} with $d = 0$. \medskip\noindent Proof of (3). Let $(\mathcal{F}_n)$ be a system as in (2) and set $\mathcal{F} = \lim \mathcal{F}_n$. If $U \in \mathcal{B}$, then $\mathcal{F}(U) = \lim \mathcal{F}_n(U)$ surjects onto $\mathcal{F}_1(U)$ as all the transition maps $\mathcal{F}_{n + 1}(U) \to \mathcal{F}_n(U)$ are surjective. Thus $\mathcal{F} \to \mathcal{F}_1$ is surjective by Sites, Definition \ref{sites-definition-sheaves-injective-surjective} and the assumption that every object has a covering by elements of $\mathcal{B}$. \medskip\noindent Proof of (4). Let $\mathcal{F}_{i, 1} \to \mathcal{F}_{i, 2} \to \mathcal{F}_{i, 3}$ be a family of exact sequences of $\mathcal{O}$-modules. We want to show that $\prod \mathcal{F}_{i, 1} \to \prod \mathcal{F}_{i, 2} \to \prod \mathcal{F}_{i, 3}$ is exact. We use the criterion of (1). Let $U \in \mathcal{B}$. Then $$ (\prod \mathcal{F}_{i, 1})(U) \to (\prod \mathcal{F}_{i, 2})(U) \to (\prod \mathcal{F}_{i, 3})(U) $$ is the same as $$ \prod \mathcal{F}_{i, 1}(U) \to \prod \mathcal{F}_{i, 2}(U) \to \prod \mathcal{F}_{i, 3}(U) $$ Each of the sequences $\mathcal{F}_{i, 1}(U) \to \mathcal{F}_{i, 2}(U) \to \mathcal{F}_{i, 3}(U)$ are exact by (1). Thus the displayed sequences are exact by Homology, Lemma \ref{homology-lemma-product-abelian-groups-exact}. We conclude by (1) again. \medskip\noindent Proof of (5). Follows from (4) and (slightly generalized) Derived Categories, Lemma \ref{derived-lemma-products}. \medskip\noindent Proof of (6) and (7). We refer to Section \ref{section-derived-limits} for a discussion of derived and homotopy limits and their relationship. By Derived Categories, Definition \ref{derived-definition-derived-limit} we have a distinguished triangle $$ R\lim K_n \to \prod K_n \to \prod K_n \to R\lim K_n[1] $$ Taking the long exact sequence of cohomology sheaves we obtain $$ H^{p - 1}(\prod K_n) \to H^{p - 1}(\prod K_n) \to H^p(R\lim K_n) \to H^p(\prod K_n) \to H^p(\prod K_n) $$ Since products are exact by (4) this becomes $$ \prod H^{p - 1}(K_n) \to \prod H^{p - 1}(K_n) \to H^p(R\lim K_n) \to \prod H^p(K_n) \to \prod H^p(K_n) $$ Now we first apply this to the case $K_n = \mathcal{F}_n[0]$ where $(\mathcal{F}_n)$ is as in (6). We conclude that (6) holds. Next we apply it to $(K_n)$ as in (7) and we conclude (7) holds.
Let $F$ be a field. By Lemma \ref{lemma-size-algebraic-extension} the cardinality of an algebraic extension of $F$ is bounded by $\max(\aleph_0, |F|)$. Choose a set $S$ containing $F$ with $|S| > \max(\aleph_0, |F|)$. Let's consider triples $(E, \sigma_E, \mu_E)$ where \begin{enumerate} \item $E$ is a set with $F \subset E \subset S$, and \item $\sigma_E : E \times E \to E$ and $\mu_E : E \times E \to E$ are maps of sets such that $(E, \sigma_E, \mu_E)$ defines the structure of a field extension of $F$ (in particular $\sigma_E(a, b) = a +_F b$ for $a, b \in F$ and similarly for $\mu_E$), and \item $F \subset E$ is an algebraic field extension. \end{enumerate} The collection of all triples $(E, \sigma_E, \mu_E)$ forms a set $I$. For $i \in I$ we will denote $E_i = (E_i, \sigma_i, \mu_i)$ the corresponding field extension to $F$. We define a partial ordering on $I$ by declaring $i \leq i'$ if and only if $E_i \subset E_{i'}$ (this makes sense as $E_i$ and $E_{i'}$ are subsets of the same set $S$) and we have $\sigma_i = \sigma_{i'}|_{E_i \times E_i}$ and $\mu_i = \mu_{i'}|_{E_i \times E_i}$, in other words, $E_{i'}$ is a field extension of $E_i$. \medskip\noindent Let $T \subset I$ be a totally ordered subset. Then it is clear that $E_T = \bigcup_{i \in T} E_i$ with induced maps $\sigma_T = \bigcup \sigma_i$ and $\mu_T = \bigcup \mu_i$ is another element of $I$. In other words every totally order subset of $I$ has a upper bound in $I$. By Zorn's lemma there exists a maximal element $(E, \sigma_E, \mu_E)$ in $I$. We claim that $E$ is an algebraic closure. Since by definition of $I$ the extension $E/F$ is algebraic, it suffices to show that $E$ is algebraically closed. \medskip\noindent To see this we argue by contradiction. Namely, suppose that $E$ is not algebraically closed. Then there exists an irreducible polynomial $P$ over $E$ of degree $> 1$, see Lemma \ref{lemma-algebraically-closed}. By Lemma \ref{lemma-finite-is-algebraic} we obtain a nontrivial finite extension $E' = E[x]/(P)$. Observe that $E'/F$ is algebraic by Lemma \ref{lemma-algebraic-permanence}. Thus the cardinality of $E'$ is $\leq \max(\aleph_0, |F|)$. By elementary set theory we can extend the given injection $E \subset S$ to an injection $E' \to S$. In other words, we may think of $E'$ as an element of our set $I$ contradicting the maximality of $E$. This contradiction completes the proof.
By Lemma \ref{lemma-galois-goes-up} given a subextension $L/M/K$ the extension $L/M$ is Galois. Of course $L/M$ is also finite (Lemma \ref{lemma-finite-goes-up}). Thus $|\text{Gal}(L/M)| = [L : M]$ by Lemma \ref{lemma-finite-Galois}. Conversely, if $H \subset G$ is a finite subgroup, then $[L : L^H] = |H|$ by Lemma \ref{lemma-galois-over-fixed-field}. It follows formally from these two observations that we obtain a bijective correspondence as in the theorem. \medskip\noindent If $H \subset G$ is normal, then $L^H$ is fixed by the action of $G$ and we obtain a canonical map $G/H \to \text{Aut}(L^H/K)$. This map has to be injective as $\text{Gal}(L/L^H) = H$. Hence $|G/H| = [L^H : K]$ and $L^H$ is Galois by Lemma \ref{lemma-finite-Galois}. \medskip\noindent Conversely, assume that $K \subset M \subset L$ with $M/K$ Galois. By Lemma \ref{lemma-lift-maps} we see that every element $\tau \in \text{Gal}(L/K)$ induces an element $\tau|_M \in \text{Gal}(M/K)$. This induces a homomorphism of Galois groups $\text{Gal}(L/K) \to \text{Gal}(M/K)$ whose kernel is $H$. Thus $H$ is a normal subgroup.
We will use the result of finite Galois theory (Theorem \ref{theorem-galois-theory}) without further mention. Let $S \subset L$ be a finite subset. There exists a tower $L/E/K$ such that $K(S) \subset E$ and such that $E/K$ is finite Galois, see Lemma \ref{lemma-normal-closure-inside-normal}. In other words, we see that $L/K$ is the union of its finite Galois subextensions. For such an $E$, by Lemma \ref{lemma-galois-infinite} the map $\text{Gal}(L/K) \to \text{Gal}(E/K)$ is surjective and continuous, i.e., the kernel is open because the topology on $\text{Gal}(E/K)$ is discrete. In particular we see that no element of $M \setminus K$ is fixed by $\text{Gal}(L/K)$ as $E^{\text{Gal}(E/K)} = K$. This proves that $L^G = K$. \medskip\noindent Lemma \ref{lemma-galois-goes-up} given a subextension $L/M/K$ the extension $L/M$ is Galois. It is immediate from the definition of the topology on $G$ that the subgroup $\text{Gal}(L/M)$ is closed. By the above applied to $L/M$ we see that $L^{\text{Gal}(L/M)} = M$ \medskip\noindent Conversely, let $H \subset G$ be a closed subgroup. We claim that $H = \text{Gal}(L/L^H)$. The inclusion $H \subset \text{Gal}(L/L^H)$ is clear. Suppose that $g \in \text{Gal}(L/L^H)$. Let $S \subset L$ be a finite subset. We will show that the open neighbourhood $U_S(g) = \{g' \in G \mid g'(s) = g(s)\}$ of $g$ meets $H$. This implies that $g \in H$ because $H$ is closed. Let $L/E/K$ be a finite Galois subextension containing $K(S)$ as in the first paragraph of the proof and consider the homomorphism $c : \text{Gal}(L/K) \to \text{Gal}(E/K)$. Then $L^H \cap E = E^{c(H)}$. Since $g$ fixes $L^H$ it fixes $E^{c(H)}$ and hence $c(g) \in c(H)$ by finite Galois theory. Pick $h \in H$ with $c(h) = c(g)$. Then $h \in U_S(g)$ as desired. \medskip\noindent At this point we have established the correspondence between closed subgroups and subextensions. \medskip\noindent Assume $H \subset G$ is open. Arguing as above we find that $H$ containes $\text{Gal}(E/K)$ for some large enough finite Galois subextension $E$ and we find that $L^H$ is contained in $E$ whence finite over $K$. Conversely, if $M$ is a finite subextension, then $M$ is generated by a finite subset $S$ and the corresponding subgroup is the open subset $U_S(e)$ where $e \in G$ is the neutral element. \medskip\noindent Assume that $K \subset M \subset L$ with $M/K$ Galois. By Lemma \ref{lemma-galois-infinite} there is a surjective continuous homomorphism of Galois groups $\text{Gal}(L/K) \to \text{Gal}(M/K)$ whose kernel is $\text{Gal}(L/M)$. Thus $\text{Gal}(L/M)$ is a normal closed subgroup. \medskip\noindent Finally, assume $N \subset G$ is normal and closed. For any $L/E/K$ as in the first paragraph of the proof, the image $c(N) \subset \text{Gal}(E/K)$ is a normal subgroup. Hence $L^N = \bigcup E^{c(N)}$ is a union of Galois extensions of $K$ (by finite Galois theory) whence Galois over $K$.
Let $\alpha, \beta \in E$ be algebraic over $k$. Then $k(\alpha, \beta)/k$ is a finite extension by Lemma \ref{lemma-algebraic-finitely-generated}. It follows that $k(\alpha + \beta) \subset k(\alpha, \beta)$ is a finite extension, which implies that $\alpha + \beta$ is algebraic by Lemma \ref{lemma-finite-is-algebraic}. Similarly for the difference, product and quotient of $\alpha$ and $\beta$.
Choose $\alpha \in F$. Then $\alpha$ is algebraic over $E$. The key observation is that $\alpha$ is algebraic over a finitely generated subextension of $k$. That is, there is a finite set $S \subset E$ such that $\alpha $ is algebraic over $k(S)$: this is clear because being algebraic means that a certain polynomial in $E[x]$ that $\alpha$ satisfies exists, and as $S$ we can take the coefficients of this polynomial. It follows that $\alpha$ is algebraic over $k(S)$. In particular, the extension $k(S, \alpha)/ k(S)$ is finite. Since $S$ is a finite set, and $k(S)/k$ is algebraic, Lemma \ref{lemma-algebraic-finitely-generated} shows that $k(S)/k$ is finite. Using multiplicativity (Lemma \ref{lemma-multiplicativity-degrees}) we find that $k(S,\alpha)/k$ is finite, so $\alpha$ is algebraic over $k$.
Let $\alpha \in R$ be nonzero. Then $1, \alpha, \alpha^2, \ldots$ are contained in $R$. By Lemma \ref{lemma-finite-is-algebraic} we find a nontrivial relation $a_0 + a_1 \alpha + \ldots + a_d \alpha^d = 0$. We may assume $a_0 \not = 0$ because if not we can divide the relation by $\alpha$ to decrease $d$. Then we see that $$ a_0 = \alpha (- a_1  - \ldots - a_d \alpha^{d - 1}) $$ which proves that the inverse of $\alpha$ is the element $a_0^{-1} (- a_1  - \ldots - a_d \alpha^{d - 1})$ of $R$.
If $E/F$ is finite, then $f : E \to E$ is an $F$-linear injective map (Lemma \ref{lemma-field-maps-injective}) of finite dimensional vector spaces, and hence bijective. In general we still see that $f$ is injective. Let $\alpha \in E$ and let $P \in F[x]$ be a polynomial such that $P(\alpha) = 0$. Let $E' \subset E$ be the subfield of $E$ generated by the roots $\alpha = \alpha_1, \ldots, \alpha_n$ of $P$ in $E$. Then $E'$ is finite over $F$ by Lemma \ref{lemma-algebraic-finitely-generated}. Since $f$ preserves the set of roots, we find that $f|_{E'} : E' \to E'$. Hence $f|_{E'}$ is an isomorphism by the first part of the proof and we conclude that $\alpha$ is in the image of $f$.
This is just a restatement of the argument in Lemma \ref{lemma-field-extension-generated-by-one-element}: the observation is that if $P$ is the minimal polynomial of $\alpha$, then the map $$ k[x]/(P) \to k(\alpha), \quad x \mapsto \alpha $$ is an isomorphism as in the aforementioned proof, and we have counted the degree of such an extension (see Example \ref{example-degree-simple-algebraic-extension}).
Consider the set $I$ of pairs $(E, \varphi)$ where $F \subset E \subset M$ is a subextension and $\varphi : E \to \overline{F}$ is a morphism of $F$-extensions. We partially order the set $I$ by declaring $(E, \varphi) \leq (E', \varphi')$ if and only if $E \subset E'$ and $\varphi'|_E = \varphi$. If $T = \{(E_t,  \varphi_t)\} \subset I$ is a totally ordered subset, then $\bigcup \varphi_t : \bigcup E_t \to \overline{F}$ is an element of $I$. Thus every totally ordered subset of $I$ has an upper bound. By Zorn's lemma there exists a maximal element $(E, \varphi)$ in $I$. We claim that $E = M$, which will finish the proof. If not, then pick $\alpha \in M$, $\alpha \not \in E$. The $\alpha$ is algebraic over $E$, see Lemma \ref{lemma-algebraic-goes-up}. Let $P$ be the minimal polynomial of $\alpha$ over $E$. Let $P^\varphi$ be the image of $P$ by $\varphi$ in $\overline{F}[x]$. Since $\overline{F}$ is algebraically closed there is a root $\beta$ of $P^\varphi$ in $\overline{F}$. Then we can extend $\varphi$ to $\varphi' : E(\alpha) = E[x]/(P) \to \overline{F}$ by mapping $x$ to $\beta$. This contradicts the maximality of $(E, \varphi)$ as desired.
Let $F$ be a field. If $M$ and $\overline{F}$ are algebraic closures of $F$, then there exists a morphism of $F$-extensions $\varphi : M \to \overline{F}$ by Lemma \ref{lemma-map-into-algebraic-closure}. Now the image $\varphi(M)$ is algebraically closed. On the other hand, the extension $\varphi(M) \subset \overline{F}$ is algebraic by Lemma \ref{lemma-algebraic-goes-up}. Thus $\varphi(M) = \overline{F}$.
We will use Lemma \ref{lemma-irreducible-polynomials} without further mention. Let $P$ be the minimal polynomial of $\alpha$ over $F$. Let $Q$ be the minimal polynomial of $\alpha$ over $E$. Then $Q$ divides $P$ in the polynomial ring $E[x]$, say $P = QR$. Then $P' = Q'R + QR'$. Thus if $Q' = 0$, then $Q$ divides $P$ and $P'$ hence $P' = 0$ by the lemma. This proves (1). Part (2) follows immediately from (1) and the definitions.
Suppose that $\alpha \in \overline{F}$ is a root of both $P$ and $P'$. Then $P = (x - \alpha)Q$ for some polynomial $Q$. Taking derivatives we obtain $P' = Q + (x - \alpha)Q'$. Thus $\alpha$ is a root of $Q$. Hence we see that if $P$ and $P'$ have a common root, then $P$ does not have pairwise distinct roots. Conversely, if $P$ has a repeated root, i.e., $(x - \alpha)^2$ divides $P$, then $\alpha$ is a root of both $P$ and $P'$. Combined with Lemma \ref{lemma-relatively-prime-polynomials} this proves the lemma.
If $\alpha_i$ is separable over $K_{i - 1}$ then $\deg_s(P_i) = \deg(P_i) = [K_i : K_{i - 1}]$ (last equality by Lemma \ref{lemma-degree-minimal-polynomial}). By multiplicativity (Lemma \ref{lemma-multiplicativity-degrees}) we have $$ [K : F] = \prod [K_i : K_{i - 1}] = \prod \deg(P_i) = \prod \deg_s(P_i) = |\Mor_F(K, \overline{F})| $$ where the last equality is Lemma \ref{lemma-count-embeddings-explicitly}. By the exact same argument we get the strict inequality $|\Mor_F(K, \overline{F})| < [K : F]$ if one of the $\alpha_i$ is not separable over $K_{i - 1}$. \medskip\noindent Finally, assume again that each $\alpha_i$ is separable over $K_{i - 1}$. We will show $K/F$ is separable. Let $\gamma = \gamma_1 \in K$ be arbitrary. Then we can find additional elements $\gamma_2, \ldots, \gamma_m$ such that $K = F(\gamma_1, \ldots, \gamma_m)$ (for example we could take $\gamma_2 = \alpha_1, \ldots, \gamma_{n + 1} = \alpha_n$). Then we see by the last part of the lemma (already proven above) that if $\gamma$ is not separable over $F$ we would have the strict inequality $|\Mor_F(K, \overline{F})| < [K : F]$ contradicting the very first part of the lemma (already prove above as well).
This is a corollary of Lemma \ref{lemma-separably-generated-separable}. Namely, since $K/F$ is finite we can find finitely many elements $\alpha_1, \ldots, \alpha_n \in K$ generating $K$ over $F$ (for example we can choose the $\alpha_i$ to be a basis of $K$ over $F$). If $K/F$ is separable, then each $\alpha_i$ is separable over $F(\alpha_1, \ldots, \alpha_{i - 1})$ by Lemma \ref{lemma-separable-goes-up} and we get equality by Lemma \ref{lemma-separably-generated-separable}. On the other hand, if we have equality, then no matter how we choose $\alpha_1, \ldots, \alpha_n$ we get that $\alpha_1$ is separable over $F$ by Lemma \ref{lemma-separably-generated-separable}. Since we can start the sequence with an arbitrary element of $K$ it follows that $K$ is separable over $F$.
Choose $\alpha \in F$. Then $\alpha$ is separable algebraic over $E$. Let $P = x^d + \sum_{i < d} a_i x^i$ be the minimal polynomial of $\alpha$ over $E$. Each $a_i$ is separable algebraic over $k$. Consider the tower of fields $$ k \subset k(a_0) \subset k(a_0, a_1) \subset \ldots \subset k(a_0, \ldots, a_{d - 1}) \subset k(a_0, \ldots, a_{d - 1}, \alpha) $$ Because $a_i$ is separable algebraic over $k$ it is separable algebraic over $k(a_0, \ldots, a_{i - 1})$ by Lemma \ref{lemma-separable-goes-up}. Finally, $\alpha$ is separable algebraic over $k(a_0, \ldots, a_{d - 1})$ because it is a root of $P$ which is irreducible (as it is irreducible over the possibly bigger field $E$) and separable (as it is separable over $E$). Thus $k(a_0, \ldots, a_{d - 1}, \alpha)$ is separable over $k$ by Lemma \ref{lemma-separably-generated-separable} and we conclude that $\alpha$ is separable over $k$ as desired.
Choose a basis $\alpha_1, \ldots, \alpha_n$ of $K$ as a vector space over $F$. By Lemma \ref{lemma-separable-equality} (and a tiny omitted argument) the set $\Hom_F(K, L)$ has $n$ elements, say $\sigma_1, \ldots, \sigma_n$. In particular, the two sides have the same dimension $n$ as vector spaces over $L$. Thus if the map is not an isomorphism, then it has a kernel. In other words, there would exist $\mu_j \in L$, $j = 1, \ldots, n$ not all zero, with $\sum \alpha_j \otimes \mu_j$ in the kernel. In other words, $\sum \sigma_i(\alpha_j)\mu_j = 0$ for all $i$. This would mean the $n \times n$ matrix with entries $\sigma_i(\alpha_j)$ is not invertible. Thus we can find $\lambda_1, \ldots, \lambda_n \in L$ not all zero, such that $\sum \lambda_i\sigma_i(\alpha_j) = 0$ for all $j$. Now any element $\alpha \in K$ can be written as $\alpha = \sum \beta_j \alpha_j$ with $\beta_j \in F$ and we would get $$ \sum \lambda_i\sigma_i(\alpha) = \sum \lambda_i\sigma_i(\sum \beta_j \alpha_j) = \sum \beta_j \sum \lambda_i\sigma_i(\alpha_j) = 0 $$ which contradicts Lemma \ref{lemma-independence-embeddings}.
By induction on the degree of $E/F$. If the degree of the extension is $1$ then the result is clear (with $n = 0$). If not, then choose $\alpha \in E$, $\alpha \not \in F$. Say $\alpha^{p^r} \in F$ for some $r > 0$. Pick $r$ minimal and replace $\alpha$ by $\alpha^{p^{r - 1}}$. Then $\alpha \not \in F$, but $\alpha^p \in F$. Then $t = \alpha^p$ is not a $p$th power in $F$ (because that would imply $\alpha \in F$, see Lemma \ref{lemma-nr-roots-unchanged} or its proof). Thus $F \subset F(\alpha)$ is a subextension of degree $p$ (Lemma \ref{lemma-take-pth-root}). By induction we find $\alpha_1, \ldots, \alpha_n \in E$ generating $E/F(\alpha)$ satisfying the conclusions of the lemma. The sequence $\alpha, \alpha_1, \ldots, \alpha_n$ does the job for the extension $E/F$.
If the characteristic is zero we set $E_{sep} = E$. Assume the characteristic is $p > 0$. Let $E_{sep}$ be the set of elements of $E$ which are separable over $F$. This is a subextension by Lemma \ref{lemma-separable-elements} and of course $E_{sep}$ is separable over $F$. Given an $\alpha$ in $E$ there exists a $p$-power $q$ such that $\alpha^q$ is separable over $F$. Namely, $q$ is that power of $p$ such that the minimal polynomial of $\alpha$ is of the form $P(x^q)$ with $P$ separable algebraic, see Lemma \ref{lemma-irreducible-polynomials}. Hence $E/E_{sep}$ is purely inseparable. Uniqueness is clear.