hcju/stacksqa huggingface dataset:
Let $\zeta \in K$ be a primitive $n$th root of $1$. Let $\sigma$ be a generator of $\text{Gal}(L/K)$. Consider $\sigma : L \to L$ as a $K$-linear operator. Note that $\sigma^n - 1 = 0$ as a linear operator. Applying linear independence of characters (Lemma \ref{lemma-independence-characters}), we see that there cannot be a polynomial over $K$ of degree $< n$ annihilating $\sigma$. Hence the minimal polynomial of $\sigma$ as a linear operator is $x^n - 1$.  Since $\zeta$ is a root of $x^n - 1$ by linear algebra there is a $0 \neq z \in L$ such that $\sigma(z) = \zeta z$. This $z$ satisfies $z^n \in K$ because $\sigma(z^n) = (\zeta z)^n = z^n$. Moreover, we see that $z, \sigma(z), \ldots, \sigma^{n - 1}(z) = z, \zeta z, \ldots \zeta^{n - 1} z$ are pairwise distinct which guarantees that $z$ generates $L$ over $K$. Hence $L = K[z]$ as required.
Observe that for $d | e$ the subfield $K(\alpha^d)$ has $[K(\alpha^d) : K] = d/e$ and $[L : K(\alpha^d)] = d$ and that both extensions $K(\alpha^d)/K$ and $L/K(\alpha^d)$ are extensions as in the lemma. \medskip\noindent We will use induction on the pair of integers $([L : L'], [L' : K])$ ordered lexicographically. Let $p$ be a prime number dividing $e$ and set $d = e/p$. If $K(\alpha^d)$ is contained in $L'$, then we win by induction, because then it suffices to prove the lemma for $L/L'/K(\alpha^d)$. If not, then $[L'(\alpha^d) : L'] = p$ and by induction hypothesis we have $L'(\alpha^d) = K(\alpha^i)$ for some $i | d$. If $i \not = 1$ we are done by induction. Thus we may assume that $[L : L'] = p$. \medskip\noindent If $e$ is not a power of $p$, then we can do this trick again with a second prime number and we win. Thus we may assume $e$ is a power of $p$. \medskip\noindent If the characteristic of $K$ is $p$ and $e$ is a $p$th power, then $L/K$ is purely inseparable. Hence $L/L'$ is purely inseparable of degree $p$ and hence $\alpha^p \in L'$. Thus $L' = K(\alpha^p)$ and this case is done. \medskip\noindent The final case is where $e$ is a power of $p$, the characteristic of $K$ is not $p$, $L/L'$ is a degree $p$ extension, and $L = L'(\alpha^{e/p})$. Claim: this can only happen if $e = p$ and $L' = K$. The claim finishes the proof. \medskip\noindent First, we prove the claim when $K$ contains a primitive $p$th root of unity $\zeta$. In this case the degree $p$ extension $K(\alpha^{e/p})/K$ is Galois with Galois group generated by the automorphism $\alpha^{e/p} \mapsto \zeta \alpha^{e/p}$. On the other hand, since $L$ is generated by $\alpha^{e/p}$ and $L'$ we see that the map $$ K(\alpha^{e/p}) \otimes_K L' \longrightarrow L $$ is an isomorphism of $K$-algebras (look at dimensions). Thus $L$ has an automorphism $\sigma$ of order $p$ over $K$ sending $\alpha^{e/p}$ to $\zeta \alpha^{e/p}$. Then $\sigma(\alpha) = \zeta' \alpha$ for some $e$th root of unity $\zeta'$ (as $\alpha^e$ is in $K$). Then on the one hand $(\zeta')^{e/p} = \zeta$ and on the other hand $\zeta'$ has to be a $p$th root of $1$ as $\sigma$ has order $p$. Thus $e/p = 1$ and the claim has been shown. \medskip\noindent Finally, suppose that $K$ does not contain a primitive $p$th root of $1$. Choose a primitive $p$th root $\zeta$ in some algebraic closure $\overline{L}$ of $L$. Consider the diagram $$ \xymatrix{ K(\zeta) \ar[r] & L(\zeta) \\ K \ar[u] \ar[r] & L \ar[u] } $$ By Lemma \ref{lemma-adjoint-pth-root-unity} the vertical extensions have degree prime to $p$. Hence $[L(\zeta) : K(\zeta)]$ is divisible by $e$. On the other hand, $L(\zeta)$ is generated by $\alpha$ over $K(\zeta)$ and hence $[L(\zeta) : K(\zeta)] \leq e$. Thus $[L(\zeta) : K(\zeta)] = e$. Similarly we have $[K(\alpha^{e/p}, \zeta) : K(\zeta)] = p$ and $[L(\zeta) : L'(\zeta)] = p$. Thus the fields $K(\zeta), L'(\zeta), L(\zeta)$ and the element $\alpha$ fall into the case discussed in the previous paragraph we conclude $e = p$ as desired.
Let $\sigma$ be a generator of $\text{Gal}(L/K)$. Consider $\sigma : L \to L$ as a $K$-linear operator. Observe that $\sigma^p - 1 = 0$ as a linear operator. Applying linear independence of characters (Lemma \ref{lemma-independence-characters}), there cannot be a polynomial of degree $< p$ annihilating $\sigma$. We conclude that the minimal polynomial of $\sigma$ is $x^p - 1 = (x - 1)^p$. This implies that there exists $w \in L$ such that $(\sigma - 1)^{p - 1}(w) = y$ is nonzero. Then $\sigma(y) = y$, i.e., $y \in K$. Thus $z = y^{-1}(\sigma - 1)^{p - 2}(w)$ satisfies $\sigma(z) = z + 1$. Since $z \not \in K$ we have $L = K[z]$. Moreover since $\sigma(z^p - z) = (z + 1)^p - (z + 1) = z^p - z$ we see that $z^p - z \in K$ and the proof is complete.
By multiplicativity of degrees of extensions (Lemma \ref{lemma-multiplicativity-degrees}) it suffices to prove this when $k'$ is generated by a single element $\alpha \in k'$ over $k$. Let $f \in k[T]$ be the minimal polynomial of $\alpha$ over $k$. Then $k'(x_1, \ldots, x_r)$ is generated by $\alpha, x_1, \ldots, x_r$ over $k$ and hence $k'(x_1, \ldots, x_r)$ is generated by $\alpha$ over $k(x_1, \ldots, x_r)$. Thus it suffices to show that $f$ is still irreducible as an element of $k(x_1, \ldots, x_r)[T]$. We only sketch the proof. It is clear that $f$ is irreducible as an element of $k[x_1, \ldots, x_r, T]$ for example because $f$ is monic as a polynomial in $T$ and any putative factorization in $k[x_1, \ldots, x_r, T]$ would lead to a factorization in $k[T]$ by setting $x_i$ equal to $0$. By Gauss' lemma we conclude.
Let $x_1, \ldots, x_r \in K$ be a transcendence basis for $K$ over $k$. Then $n = [K : k(x_1, \ldots, x_r)] < \infty$. Suppose that $k \subset k' \subset K$ with $k'/k$ finite. In this case $[k'(x_1, \ldots, x_r) : k(x_1, \ldots, x_r)] = [k' : k] < \infty$, see Lemma \ref{lemma-purely-transcendental-degree}. Hence $$ [k' : k] = [k'(x_1, \ldots, x_r) : k(x_1, \ldots, x_r)] < [K : k(x_1, \ldots, x_r)] = n. $$ In other words, the degrees of finite subextensions are bounded and the lemma follows.
It follows from the definitions that (2) implies (1). Assume $P$ is as in (2). Write $P(T) = \sum\nolimits_{i = 0}^d a_i T^{d - i}$ and $a_i = b_i^p$. The polynomial $Q(T) = \sum\nolimits_{i = 0}^d b_i T^{d - i}$ has distinct roots in an algebraic closure as well, because the roots of $Q$ are the $p$th roots of the roots of $P$. If $\alpha$ is not a $p$th power, then $T^p - \alpha$ is an irreducible polynomial over $L$ (Lemma \ref{lemma-take-pth-root}). Moreover $Q$ and $T^p - \alpha$ have a root in common in an algebraic closure $\overline{L}$. Thus $Q$ and $T^p - \alpha$ are not relatively prime, which implies $T^p - \alpha | Q$ in $L[T]$. This contradicts the fact that the roots of $Q$ are pairwise distinct.
Assume (1). Consider $T = \lim_{i \in I} T_i$ as in (2). Let $(y, x_T)$ be an element of the fibre product $\colim_i G(T_i) \times_{G(T)} F(T)$. Then $y$ comes from $y_i \in G(T_i)$ for some $i$. Consider the functor $F_{y_i}$ on $(\Sch/T_i)_{fppf}$ as in Definition \ref{definition-locally-finite-presentation}. We see that $x_T \in F_{y_i}(T)$. Moreover $T = \lim_{i' \geq i} T_{i'}$ is a directed system of affine schemes over $T_i$. Hence (1) implies that $x_T$ the image of a unique element $x$ of $\colim_{i' \geq i} F_{y_i}(T_{i'})$. Thus $x$ is the unique element of $\colim F(T_i)$ which maps to the pair $(y, x_T)$. This proves that (2) holds. \medskip\noindent Assume (2). Let $T$ be a scheme and $y_T \in G(T)$. We have to show that $F_{y_T}$ is limit preserving. Let $T' = \lim_{i \in I} T'_i$ be an affine scheme over $T$ which is the directed limit of affine scheme $T'_i$ over $T$. Let $x_{T'} \in F_{y_T}$. Pick $i \in I$ which is possible as $I$ is a directed set. Denote $y_i \in F(T'_i)$ the image of $y_{T'}$. Then we see that $(y_i, x_{T'})$ is an element of the fibre product $\colim_i G(T'_i) \times_{G(T')} F(T')$. Hence by (2) we get a unique element $x$ of $\colim_i F(T'_i)$ mapping to $(y_i, x_{T'})$. It is clear that $x$ defines an element of $\colim_i F_y(T'_i)$ mapping to $x_{T'}$ and we win.
Let $T = \lim_{i \in I} T_i$ as in characterization (2) of Lemma \ref{lemma-characterize-relative-limit-preserving}. Consider the diagram of sets $$ \xymatrix{ \colim_i F(T_i) \ar[r] \ar[d]_a & F(T) \ar[d]^a \\ \colim_i G(T_i) \ar[r] \ar[d]_b & G(T) \ar[d]^b \\ \colim_i H(T_i) \ar[r] & H(T) } $$ By assumption the two squares are fibre product squares. Hence the outer rectangle is a fibre product diagram too which proves the lemma.
Let $T = \lim_{i \in I} T_i$ as in characterization (2) of Lemma \ref{lemma-characterize-relative-limit-preserving}. Consider the diagram of sets $$ \xymatrix{ \colim_i F(T_i) \ar[r] \ar[d]_a & F(T) \ar[d]^a \\ \colim_i G(T_i) \ar[r] \ar[d]_b & G(T) \ar[d]^b \\ \colim_i H(T_i) \ar[r] & H(T) } $$ By assumption the lower square and the outer rectangle are fibre products of sets. Hence the upper square is a fibre product square too which proves the lemma.
Let $T = \lim_{i \in I} T_i$ as in characterization (2) of Lemma \ref{lemma-characterize-relative-limit-preserving}. Then we have $$ \colim (F \times_G H)(T_i) = \colim F(T_i) \times_{\colim G(T_i)} \colim H(T_i) $$ as filtered colimits commute with finite products. Our goal is thus to show that $$ \xymatrix{ \colim F(T_i) \times_{\colim G(T_i)} \colim H(T_i) \ar[r] \ar[d] & F(T) \times_{G(T)} H(T) \ar[d] \\ \colim_i E(T_i) \ar[r] & E(T) } $$ is a fibre product diagram. This follows from the observation that given maps of sets $E' \to E$, $F \to G$, $H \to G$, and $G \to E$ we have $$ E' \times_E (F \times_G H) = (E' \times_E F) \times_{(E' \times_E G)} (E' \times_E H) $$ Some details omitted.
Assume $F$ is limit preserving. It suffices to show that $F^+$ is limit preserving, since $F^\# = (F^+)^+$, see Sites, Theorem \ref{sites-theorem-plus}. Let $T$ be an affine scheme over $S$, and let $T = \lim T_i$ be written as the directed limit of an inverse system of affine $S$ schemes. Recall that $F^+(T)$ is the colimit of $\check H^0(\mathcal{V}, F)$ where the limit is over all coverings of $T$ in $(\Sch/S)_{fppf}$. Any fppf covering of an affine scheme can be refined by a standard fppf covering, see Topologies, Lemma \ref{topologies-lemma-fppf-affine}. Hence we can write $$ F^+(T) = \colim_{\mathcal{V}\text{ standard covering }T} \check H^0(\mathcal{V}, F). $$ Any $\mathcal{V} = \{T_k \to T\}_{k = 1, \ldots, n}$ in the colimit may be written as $V_i \times_{T_i} T$ for some $i$ and some standard fppf covering $\mathcal{V}_i = \{T_{i, k} \to T_i\}_{k = 1, \ldots, n}$ of $T_i$. Denote $\mathcal{V}_{i'} = \{T_{i', k} \to T_{i'}\}_{k = 1, \ldots, n}$ the base change for $i' \geq i$. Then we see that \begin{align*} \colim_{i' \geq i} \check H^0(\mathcal{V}_i, F) & = \colim_{i' \geq i} \text{Equalizer}( \xymatrix{ \prod F(T_{i', k}) \ar@<1ex>[r] \ar@<-1ex>[r] & \prod F(T_{i', k} \times_{T_{i'}} T_{i', l}) } \\ & = \text{Equalizer}( \xymatrix{ \colim_{i' \geq i} \prod F(T_{i', k}) \ar@<1ex>[r] \ar@<-1ex>[r] & \colim_{k' \geq k} \prod F(T_{i', k} \times_{T_{i'}} T_{i', l}) } \\ & = \text{Equalizer}( \xymatrix{ \prod F(T_k) \ar@<1ex>[r] \ar@<-1ex>[r] & \prod F(T_k \times_T T_l) } \\ & = \check H^0(\mathcal{V}, F) \end{align*} Here the second equality holds because filtered colimits are exact. The third equality holds because $F$ is limit preserving and because $\lim_{i' \geq i} T_{i', k} = T_k$ and $\lim_{i' \geq i} T_{i', k} \times_{T_{i'}} T_{i', l} = T_k \times_T T_l$ by Limits, Lemma \ref{limits-lemma-scheme-over-limit}. If we use this for all coverings at the same time we obtain \begin{align*} F^+(T) & = \colim_{\mathcal{V}\text{ standard covering }T} \check H^0(\mathcal{V}, F) \\ & = \colim_{i \in I} \colim_{\mathcal{V}_i\text{ standard covering }T_i} \check H^0(T \times_{T_i}\mathcal{V}_i, F) \\ & = \colim_{i \in I} F^+(T_i) \end{align*} The switch of the order of the colimits is allowed by Categories, Lemma \ref{categories-lemma-colimits-commute}.
Let $T$ be an affine scheme over $S$. Let $I$ be a directed set, and let $T_i$ be an inverse system of affine schemes over $S$ such that $T = \lim T_i$. We have to show that the canonical map $\colim F(T_i) \to F(T)$ is bijective. \medskip\noindent Choose some $0 \in I$ and choose a standard fppf covering $\{V_{0, k} \to T_{0}\}_{k = 1, \ldots, m}$ which refines the pullback $\{U_j \times_S T_0 \to T_0\}$ of the given fppf covering of $S$. For each $i \geq 0$ we set $V_{i, k} = T_i \times_{T_0} V_{0, k}$, and we set $V_k = T \times_{T_0} V_{0, k}$. Note that $V_k = \lim_{i \geq 0} V_{i, k}$, see Limits, Lemma \ref{limits-lemma-scheme-over-limit}. \medskip\noindent Suppose that $x, x' \in \colim F(T_i)$ map to the same element of $F(T)$. Say $x, x'$ are given by elements $x_i, x'_i \in F(T_i)$ for some $i \in I$ (we may choose the same $i$ for both as $I$ is directed). By assumption (2) and the fact that $x_i, x'_i$ map to the same element of $F(T)$ this implies that $$ x_i|_{V_{i', k}} = x'_i|_{V_{i', k}} $$ for some suitably large $i' \in I$. We can choose the same $i'$ for each $k$ as $k \in \{1, \ldots, m\}$ ranges over a finite set. Since $\{V_{i', k} \to T_{i'}\}$ is an fppf covering and $F$ is a sheaf this implies that $x_i|_{T_{i'}} = x'_i|_{T_{i'}}$ as desired. This proves that the map $\colim F(T_i) \to F(T)$ is injective. \medskip\noindent To show surjectivity we argue in a similar fashion. Let $x \in F(T)$. By assumption (2) for each $k$ we can choose a $i$ such that $x|_{V_k}$ comes from an element $x_{i, k} \in F(V_{i, k})$. As before we may choose a single $i$ which works for all $k$. By the injectivity proved above we see that $$ x_{i, k}|_{V_{i', k} \times_{T_{i'}} V_{i', l}} = x_{i, l}|_{V_{i', k} \times_{T_{i'}} V_{i', l}} $$ for some large enough $i'$. Hence by the sheaf condition of $F$ the elements $x_{i, k}|_{V_{i', k}}$ glue to an element $x_{i'} \in F(T_{i'})$ as desired.
Suppose that $T$ is a scheme and $y \in G^\#(T)$. We have to show the functor $F^\#_y : (\Sch/T)_{fppf}^{opp} \to \textit{Sets}$ constructed from $F^\# \to G^\#$ and $y$ as in Definition \ref{definition-locally-finite-presentation} is limit preserving. By Equation (\ref{equation-fibre-map-functors}) we see that $F^\#_y$ is a sheaf. Choose an fppf covering $\{V_j \to T\}_{j \in J}$ such that $y|_{V_j}$ comes from an element $y_j \in F(V_j)$. Note that the restriction of $F^\#$ to $(\Sch/V_j)_{fppf}$ is just $F^\#_{y_j}$. If we can show that $F^\#_{y_j}$ is limit preserving then Lemma \ref{lemma-sheaf-finite-presentation} guarantees that $F^\#_y$ is limit preserving and we win. This reduces us to the case $y \in G(T)$. \medskip\noindent Let $y \in G(T)$. In this case we claim that $F^\#_y = (F_y)^\#$. This follows from Equation (\ref{equation-fibre-map-functors}). Thus this case follows from Lemma \ref{lemma-sheafify-finite-presentation}.
Choose a scheme $V$ and a surjective \'etale morphism $g : V \to Y$. Next, choose a scheme $U$ and a surjective \'etale morphism $h : U \to V \times_Y X$. It suffices to show for $T = \lim T_i$ as in the lemma that the map $$ \colim U(T_i) \longrightarrow U(T) \times_{V(T)} \colim V(T_i) $$ is surjective, because then $U \to V$ will be locally of finite presentation by Limits, Lemma \ref{limits-lemma-surjection-is-enough} (modulo a set theoretic remark exactly as in the proof of Proposition \ref{proposition-characterize-locally-finite-presentation}). Thus we take $a : T \to U$ and $b_i : T_i \to V$ which determine the same morphism $T \to V$. Picture $$ \xymatrix{ T \ar[d]_a \ar[rr]_{p_i} & & T_i \ar[d]^{b_i} \ar@{..>}[ld] \\ U \ar[r]^-h & X \times_Y V \ar[d] \ar[r] & V \ar[d]^g \\ & X \ar[r]^f & Y } $$ By the assumption of the lemma after increasing $i$ we can find a morphism $c_i : T_i \to X$ such that $h \circ a = (b_i, c_i) \circ p_i : T_i \to V \times_Y X$ and such that $f \circ c_i = g \circ b_i$. Since $h$ is an \'etale morphism of algebraic spaces (and hence locally of finite presentation), we have the surjectivity of $$ \colim U(T_i) \longrightarrow U(T) \times_{(X \times_Y V)(T)} \colim (X \times_Y V)(T_i) $$ by Proposition \ref{proposition-characterize-locally-finite-presentation}. Hence after increasing $i$ again we can find the desired morphism $a_i : T_i \to U$ with $a = a_i \circ p_i$ and $b_i = (U \to V) \circ a_i$.
Part (2) is a formal consequence of the existence of the limit $X = \lim X_i$ as an algebraic space over $S$. Choose an element $0 \in I$ (this is possible as a directed set is nonempty). Choose a scheme $U_0$ and a surjective \'etale morphism $U_0 \to X_0$. Set $R_0 = U_0 \times_{X_0} U_0$ so that $X_0 = U_0/R_0$. For $i \geq 0$ set $U_i = X_i \times_{X_0} U_0$ and $R_i = X_i \times_{X_0} R_0 = U_i \times_{X_i} U_i$. By Limits, Lemma \ref{limits-lemma-directed-inverse-system-has-limit} we see that $U = \lim_{i \geq 0} U_i$ and $R = \lim_{i \geq 0} R_i$ are schemes. Moreover, the two morphisms $s, t : R \to U$ are the base change of the two projections $R_0 \to U_0$ by the morphism $U \to U_0$, in particular \'etale. The morphism $R \to U \times_S U$ defines an equivalence relation as directed a limit of equivalence relations is an equivalence relation. Hence the morphism $R \to U \times_S U$ is an \'etale equivalence relation. We claim that the natural map \begin{equation} \label{equation-isomorphism-sheaves} U/R \longrightarrow \lim X_i \end{equation} is an isomorphism of fppf sheaves on the category of schemes over $S$. The claim implies $X = \lim X_i$ is an algebraic space by Spaces, Theorem \ref{spaces-theorem-presentation}. \medskip\noindent Let $Z$ be a scheme and let $a : Z \to \lim X_i$ be a morphism. Then $a = (a_i)$ where $a_i : Z \to X_i$. Set $W_0 = Z \times_{a_0, X_0} U_0$. Note that $W_0 = Z \times_{a_i, X_i} U_i$ for all $i \geq 0$ by our choice of $U_i \to X_i$ above. Hence we obtain a morphism $W_0 \to \lim_{i \geq 0} U_i = U$. Since $W_0 \to Z$ is surjective and \'etale, we conclude that (\ref{equation-isomorphism-sheaves}) is a surjective map of sheaves. Finally, suppose that $Z$ is a scheme and that $a, b : Z \to U/R$ are two morphisms which are equalized by (\ref{equation-isomorphism-sheaves}). We have to show that $a = b$. After replacing $Z$ by the members of an fppf covering we may assume there exist morphisms $a', b' : Z \to U$ which give rise to $a$ and $b$. The condition that $a, b$ are equalized by (\ref{equation-isomorphism-sheaves}) means that for each $i \geq 0$ the compositions $a_i', b_i' : Z \to U \to U_i$ are equal as morphisms into $U_i/R_i = X_i$. Hence $(a_i', b_i') : Z \to U_i \times_S U_i$ factors through $R_i$, say by some morphism $c_i : Z \to R_i$. Since $R = \lim_{i \geq 0} R_i$ we see that $c = \lim c_i : Z \to R$ is a morphism which shows that $a, b$ are equal as morphisms of $Z$ into $U/R$. \medskip\noindent Part (1) follows as we have seen above that $U_i \times_{X_i} X = U$ and $U \to U_i$ is affine by construction.
Observe that $\lim X_i$ and $\lim Y_i$ exist by Lemma \ref{lemma-directed-inverse-system-has-limit}. Pick $0 \in I$ and choose an affine scheme $V_0$ and an \'etale morphism $V_0 \to Y_0$. Then the morphisms $V_i = Y_i \times_{Y_0} V_0 \to U_i = X_i \times_{Y_0} V_0$ are closed immersions of affine schemes. Hence the morphism $V = Y \times_{Y_0} V_0 \to U = X \times_{Y_0} V_0$ is a closed immersion because $V = \lim V_i$, $U = \lim U_i$ and because a limit of closed immersions of affine schemes is a closed immersion: a filtered colimit of surjective ring maps is surjective. Since the \'etale morphisms $V \to Y$ form an \'etale covering of $Y$ as we vary our choice of $V_0 \to Y_0$ we see that the lemma is true.
Observe that $\lim X_i$ exists by Lemma \ref{lemma-directed-inverse-system-has-limit}. Pick $0 \in I$ and choose an affine scheme $V_0$ and an \'etale morphism $U_0 \to X_0$. Then the affine schemes $U_i = X_i \times_{X_0} U_0$ are reduced. Hence $U = X \times_{X_0} U_0$ is a reduced affine scheme as a limit of reduced affine schemes: a filtered colimit of reduced rings is reduced. Since the \'etale morphisms $U \to X$ form an \'etale covering of $X$ as we vary our choice of $U_0 \to X_0$ we see that the lemma is true.
It is clear that (3) implies (2). We will assume (2) and prove (3). The proof is rather formal and we encourage the reader to find their own proof. \medskip\noindent Let us first prove that (3) holds when $T_i$ is in addition assumed separated for all $i$. Choose $i \in I$ and choose a surjective \'etale morphism $U_i \to T_i$ where $U_i$ is affine. Using Lemma \ref{lemma-space-over-limit} we see that with $U = U_i \times_{T_i} T$ and $U_{i'} = U_i \times_{T_i} T_{i'}$ we have $U = \lim_{i' \geq i} U_{i'}$. Of course $U$ and $U_{i'}$ are affine (see Lemma \ref{lemma-directed-inverse-system-has-limit}). Since $T_i$ is separated, the fibre product $V_i = U_i \times_{T_i} U_i$ is an affine scheme as well and we obtain affine schemes $V = V_i \times_{T_i} T$ and $V_{i'} = V_i \times_{T_i} T_{i'}$ with $V = \lim_{i' \geq i} V_{i'}$. Observe that $U \to T$ and $U_i \to T_i$ are surjective \'etale and that $V = U \times_T U$ and $V_{i'} = U_{i'} \times_{T_{i'}} U_{i'}$. Note that $\Mor(T, X)$ is the equalizer of the two maps $\Mor(U, X) \to \Mor(V, X)$; this is true for example because $X$ as a sheaf on $(\Sch/S)_{fppf}$ is the coequalizer of the two maps $h_V \to h_u$. Similarly $\Mor(T_{i'}, X)$ is the equalizer of the two maps $\Mor(U_{i'}, X) \to \Mor(V_{i'}, X)$. And of course the same thing is true with $X$ replaced with $Y$. Condition (2) says that the diagrams of in (3) are fibre products in the case of $U = \lim U_i$ and $V = \lim V_i$. It follows formally that the same thing is true for $T = \lim T_i$. \medskip\noindent In the general case, choose an affine scheme $U$, an $i \in I$, and a surjective \'etale morphism $U \to T_i$. Repeating the argument of the previous paragraph we still achieve the proof: the schemes $V_{i'}$, $V$ are no longer affine, but they are still quasi-compact and separated and the result of the preceding paragraph applies.