hcju/stacksqa huggingface dataset:
In Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-quasi-compact-existence-universally-closed} we have seen one of the implications. To prove the other, assume that $f$ is universally closed. Let $$ \xymatrix{ \Spec(K) \ar[r] \ar[d] & X \ar[d] \\ \Spec(A) \ar[r] & Y } $$ be a diagram as in Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-valuative-criterion}. Let $X_A = \Spec(A) \times_Y X$, so that we have $$ \xymatrix{ \Spec(K) \ar[r] \ar[rd] & X_A \ar[d] \\  & \Spec(A) } $$ By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-quasi-compact} we see that $X_A \to \Spec(A)$ is quasi-compact. Since $X_A \to X$ is representable, we see that $X_A$ is decent also, see Lemma \ref{lemma-representable-properties}. Moreover, as $f$ is universally closed, we see that $X_A \to \Spec(A)$ is universally closed. Hence we may and do replace $X$ by $X_A$ and $Y$ by $\Spec(A)$. \medskip\noindent Let $x' \in |X|$ be the equivalence class of $\Spec(K) \to X$. Let $y \in |Y| = |\Spec(A)|$ be the closed point. Set $y' = f(x')$; it is the generic point of $\Spec(A)$. Since $f$ is universally closed we see that $f(\overline{\{x'\}})$ contains $\overline{\{y'\}}$, and hence contains $y$. Let $x \in \overline{\{x'\}}$ be a point such that $f(x) = y$. Let $U$ be a scheme, and $\varphi : U \to X$ an \'etale morphism such that there exists a $u \in U$ with $\varphi(u) = x$. By Lemma \ref{lemma-specialization} and our assumption that $X$ is decent there exists a specialization $u' \leadsto u$ on $U$ with $\varphi(u') = x'$. This means that there exists a common field extension $K \subset K' \supset \kappa(u')$ such that $$ \xymatrix{ \Spec(K') \ar[r] \ar[d] & U \ar[d] \\ \Spec(K) \ar[r] \ar[rd] & X \ar[d] \\  & \Spec(A) } $$ is commutative. This gives the following commutative diagram of rings $$ \xymatrix{ K' & \mathcal{O}_{U, u} \ar[l] \\ K \ar[u] & \\  & A \ar[lu] \ar[uu] } $$ By Algebra, Lemma \ref{algebra-lemma-dominate} we can find a valuation ring $A' \subset K'$ dominating the image of $\mathcal{O}_{U, u}$ in $K'$. Since by construction $\mathcal{O}_{U, u}$ dominates $A$ we see that $A'$ dominates $A$ also. Hence we obtain a diagram resembling the second diagram of Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-valuative-criterion} and the proposition is proved.
By condition (1) and Morphisms, Lemma \ref{morphisms-lemma-unramified-permanence} we see that $j$ is a unramified. Then $\Delta_j : R \to R \times_{U \times_S U} R$ is an open immersion by Morphisms, Lemma \ref{morphisms-lemma-diagonal-unramified-morphism}. However, then condition (2) says $\Delta_j$ is bijective on $k$-valued points, hence $\Delta_j$ is an isomorphism, hence $j$ is a monomorphism. Then it easily follows from the commutative diagrams that $R(T) \subset U(T) \times U(T)$ is an equivalence relation for all schemes $T$ over $S$.
We recall that by Schemes, Lemma \ref{schemes-lemma-section-immersion} we have that $e$ is an immersion which is a closed immersion (resp.\ quasi-compact) if $G \to S$ is separated (resp.\ quasi-separated). For the converse, consider the diagram $$ \xymatrix{ G \ar[r]_-{\Delta_{G/S}} \ar[d] & G \times_S G \ar[d]^{(g, g') \mapsto m(i(g), g')} \\ S \ar[r]^e & G } $$ It is an exercise in the functorial point of view in algebraic geometry to show that this diagram is cartesian. In other words, we see that $\Delta_{G/S}$ is a base change of $e$. Hence if $e$ is a closed immersion (resp.\ quasi-compact) so is $\Delta_{G/S}$, see Schemes, Lemma \ref{schemes-lemma-base-change-immersion} (resp.\ Schemes, Lemma \ref{schemes-lemma-quasi-compact-preserved-base-change}).
Consider the diagram $$ \xymatrix{ T \times_S G \ar[rrr]_{(t, g) \mapsto (t, m(\psi(t), g))} & & & T \times_S G \ar[r]_{\text{pr}} \ar[d] & G \ar[d] \\ & & & T \ar[r] & S } $$ The left top horizontal arrow is an isomorphism and the square is cartesian. Hence the lemma follows from Morphisms, Lemma \ref{morphisms-lemma-base-change-flat}.
By Morphisms, Lemma \ref{morphisms-lemma-base-change-differentials} we have $$ \Omega_{G \times_S G/G} = \text{pr}_0^*\Omega_{G/S} $$ where on the left hand side we view $G \times_S G$ as a scheme over $G$ using $\text{pr}_1$. Let $\tau : G \times_S G \to G \times_S G$ be the ``shearing map'' given by $(g, h) \mapsto (m(g, h), h)$ on points. This map is an automorphism of $G \times_S G$ viewed as a scheme over $G$ via the projection $\text{pr}_1$. Combining these two remarks we obtain an isomorphism $$ \tau^*\text{pr}_0^*\Omega_{G/S} \to \text{pr}_0^*\Omega_{G/S} $$ Since $\text{pr}_0 \circ \tau = m$ this can be rewritten as an isomorphism $$ m^*\Omega_{G/S} \to \text{pr}_0^*\Omega_{G/S} $$ Pulling back this isomorphism by $(e \circ f, \text{id}_G) : G \to G \times_S G$ and using that $m \circ (e \circ f, \text{id}_G) = \text{id}_G$ and $\text{pr}_0 \circ (e \circ f, \text{id}_G) = e \circ f$ we obtain an isomorphism $$ \Omega_{G/S} \to f^*e^*\Omega_{G/S} $$ as desired. By Morphisms, Lemma \ref{morphisms-lemma-differentials-relative-immersion-section} we have $\mathcal{C}_{S/G} \cong e^*\Omega_{G/S}$. If $S$ is the spectrum of a field, then any $\mathcal{O}_S$-module on $S$ is free and the final statement follows.
The multiplication map is isomorphic to the projection map $\text{pr}_0 : G \times_k G \to G$ because the diagram $$ \xymatrix{ G \times_k G \ar[d]^m \ar[rrr]_{(g, g') \mapsto (m(g, g'), g')} & & & G \times_k G \ar[d]^{(g, g') \mapsto g} \\ G \ar[rrr]^{\text{id}} & & & G } $$ is commutative with isomorphisms as horizontal arrows. The projection is open by Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open}.
For any field extension $k \subset K$ the morphism $G_K \to G$ is open (Morphisms, Lemma \ref{morphisms-lemma-scheme-over-field-universally-open}). Every point $\xi$ of $T \times_k U$ is the image of a morphism $(t, u) : \Spec(K) \to T \times_k U$ for some $K$. Then the image of $T_K \times_K U_K = (T \times_k U)_K \to G_K$ contains the translate $t \cdot U_K$ which is open. Combining these facts we see that the image of $T \times_k U \to G$ contains an open neighbourhood of the image of $\xi$. Since $\xi$ was arbitrary we win.
Say $S = \Spec(k)$ with $k$ a field, and let $G$ be a group scheme over $S$. By Lemma \ref{lemma-group-scheme-separated} we have to show that $e : S \to G$ is a closed immersion. By Morphisms, Lemma \ref{morphisms-lemma-algebraic-residue-field-extension-closed-point-fibre} the image of $e : S \to G$ is a closed point of $G$. It is clear that $\mathcal{O}_G \to e_*\mathcal{O}_S$ is surjective, since $e_*\mathcal{O}_S$ is a skyscraper sheaf supported at the neutral element of $G$ with value $k$. We conclude that $e$ is a closed immersion by Schemes, Lemma \ref{schemes-lemma-characterize-closed-immersions}.
For any point $g \in G$ there exists a field extension $k \subset K$ and a $K$-valued point $g' \in G(K)$ mapping to $g$. If we think of $g'$ as a $K$-rational point of the group scheme $G_K$, then we see that $\mathcal{O}_{G, g} \to \mathcal{O}_{G_K, g'}$ is a faithfully flat local ring map (as $G_K \to G$ is flat, and a local flat ring map is faithfully flat, see Algebra, Lemma \ref{algebra-lemma-local-flat-ff}). The result for $\mathcal{O}_{G_K, g'}$ implies the result for $\mathcal{O}_{G, g}$, see Algebra, Lemma \ref{algebra-lemma-injective-minimal-primes-in-image}. Hence in order to prove (1) it suffices to prove it for $k$-rational points $g$ of $G$. In this case translation by $g$ defines an automorphism $G \to G$ which maps $e$ to $g$. Hence $\mathcal{O}_{G, g} \cong \mathcal{O}_{G, e}$. In this way we see that (2) implies (1), since irreducible components passing through $e$ correspond one to one with minimal prime ideals of $\mathcal{O}_{G, e}$. \medskip\noindent In order to prove (2) and (3) it suffices to prove (2) when $k$ is algebraically closed. In this case, let $Z_1$, $Z_2$ be two irreducible components of $G$ passing through $e$. Since $k$ is algebraically closed the closed subscheme $Z_1 \times_k Z_2 \subset G \times_k G$ is irreducible too, see Varieties, Lemma \ref{varieties-lemma-bijection-irreducible-components}. Hence $m(Z_1 \times_k Z_2)$ is contained in an irreducible component of $G$. On the other hand it contains $Z_1$ and $Z_2$ since $m|_{e \times G} = \text{id}_G$ and $m|_{G \times e} = \text{id}_G$. We conclude $Z_1 = Z_2$ as desired.
Let $U = \psi(G') \subset G$. Let $Z = G \setminus \psi(G') = G \setminus U$ with the reduced induced closed subscheme structure. By Lemma \ref{lemma-group-scheme-over-field-translate-open} the image of $$ Z \times_k G' \longrightarrow Z \times_k U \longrightarrow G $$ is open (the first arrow is surjective). On the other hand, since $\psi$ is a homomorphism of group schemes, the image of $Z \times_k G' \to G$ is contained in $Z$ (because translation by $\psi(g')$ preserves $U$ for all points $g'$ of $G'$; small detail omitted). Hence $Z \subset G$ is an open subset (although not necessarily an open subscheme). Thus $U = \psi(G')$ is closed.
To show that $i$ is a closed immersion it suffices to show that $i(G')$ is a closed subset of $G$. Let $k \subset k'$ be a perfect extension of $k$. If $i(G'_{k'}) \subset G_{k'}$ is closed, then $i(G') \subset G$ is closed by Morphisms, Lemma \ref{morphisms-lemma-fpqc-quotient-topology} (as $G_{k'} \to G$ is flat, quasi-compact and surjective). Hence we may and do assume $k$ is perfect. We will use without further mention that products of reduced schemes over $k$ are reduced. We may replace $G'$ and $G$ by their reductions, see Lemma \ref{lemma-reduced-subgroup-scheme-perfect}. Let $\overline{G'} \subset G$ be the closure of $i(G')$ viewed as a reduced closed subscheme. By Varieties, Lemma \ref{varieties-lemma-closure-of-product} we conclude that $\overline{G'} \times_k \overline{G'}$ is the closure of the image of $G' \times_k G' \to G \times_k G$. Hence $$ m\Big(\overline{G'} \times_k \overline{G'}\Big) \subset \overline{G'} $$ as $m$ is continuous. It follows that $\overline{G'} \subset G$ is a (reduced) closed subgroup scheme. By Lemma \ref{lemma-open-subgroup-closed-over-field} we see that $i(G') \subset \overline{G'}$ is also closed which implies that $i(G') = \overline{G'}$ as desired.
Suppose that $k \subset K$ is a field extension. If $G_K$ is quasi-compact, then $G$ is too as $G_K \to G$ is surjective. By Lemma \ref{lemma-group-scheme-field-geometrically-irreducible} we see that $G_K$ is irreducible. Hence it suffices to prove the lemma after replacing $k$ by some extension. Choose $K$ to be an algebraically closed field extension of very large cardinality. Then by Varieties, Lemma \ref{varieties-lemma-make-Jacobson}, we see that $G_K$ is a Jacobson scheme all of whose closed points have residue field equal to $K$. In other words we may assume $G$ is a Jacobson scheme all of whose closed points have residue field $k$. \medskip\noindent Let $U \subset G$ be a nonempty affine open. Let $g \in G(k)$. Then $gU \cap U \not = \emptyset$. Hence we see that $g$ is in the image of the morphism $$ U \times_{\Spec(k)} U \longrightarrow G, \quad (u_1, u_2) \longmapsto u_1u_2^{-1} $$ Since the image of this morphism is open (Lemma \ref{lemma-group-scheme-over-field-open-multiplication}) we see that the image is all of $G$ (because $G$ is Jacobson and closed points are $k$-rational). Since $U$ is affine, so is $U \times_{\Spec(k)} U$. Hence $G$ is the image of a quasi-compact scheme, hence quasi-compact.
By Varieties, Lemma \ref{varieties-lemma-geometrically-connected-criterion} we see that $G$ is geometrically connected. If we show that $G_K$ is irreducible for some field extension $k \subset K$, then the lemma follows. Hence we may apply Varieties, Lemma \ref{varieties-lemma-make-Jacobson} to reduce to the case where $k$ is algebraically closed, $G$ is a Jacobson scheme, and all the closed points are $k$-rational. \medskip\noindent Let $Z \subset G$ be the unique irreducible component of $G$ passing through the neutral element, see Lemma \ref{lemma-group-scheme-field-geometrically-irreducible}. Endowing $Z$ with the reduced induced closed subscheme structure, we see that $Z \times_k Z$ is reduced and irreducible (Varieties, Lemmas \ref{varieties-lemma-geometrically-reduced-any-base-change} and \ref{varieties-lemma-bijection-irreducible-components}). We conclude that $m|_{Z \times_k Z} : Z \times_k Z \to G$ factors through $Z$. Hence $Z$ becomes a closed subgroup scheme of $G$. \medskip\noindent To get a contradiction, assume there exists another irreducible component $Z' \subset G$. Then $Z \cap Z' = \emptyset$ by Lemma \ref{lemma-group-scheme-field-geometrically-irreducible}. By Lemma \ref{lemma-irreducible-group-scheme-over-field-qc} we see that $Z$ is quasi-compact. Thus we may choose a quasi-compact open $U \subset G$ with $Z \subset U$ and $U \cap Z' = \emptyset$. The image $W$ of $Z \times_k U \to G$ is open in $G$ by Lemma \ref{lemma-group-scheme-over-field-translate-open}. On the other hand, $W$ is quasi-compact as the image of a quasi-compact space. We claim that $W$ is closed. If the claim is true, then $W \subset G \setminus Z'$ is a proper open and closed subset of $G$, which contradicts the assumption that $G$ is connected. \medskip\noindent Proof of the claim. Since $W$ is quasi-compact, we see that points in the closure of $W$ are specializations of points of $W$ (Morphisms, Lemma \ref{morphisms-lemma-reach-points-scheme-theoretic-image}). Thus we have to show that any irreducible component $Z'' \subset G$ of $G$ which meets $W$ is contained in $W$. As $G$ is Jacobson and closed points are rational, $Z'' \cap W$ has a rational point $g \in Z''(k) \cap W(k)$ and hence $Z'' = Zg$. But $W = m(Z \times_k W)$ by construction, so $Z'' \cap W \not = \emptyset$ implies $Z'' \subset W$.
By taking an affine open covering we reduce to the case of an affine $X$. Say $X = \Spec(B)$. Write $A = \colim A_i$ with $A_i = \prod_{t \in T_i} k$ and $T_i$ finite. Then $T_i = |\Spec(A_i)|$ with the discrete topology and the transition morphisms $A_i \to A_{i'}$ are given by set maps $T_{i'} \to T_i$. Thus $|T| = \lim T_i$ as a topological space, see Limits, Lemma \ref{limits-lemma-topology-limit}. Similarly we have \begin{align*} |T \times_k X| & = |\Spec(A \otimes_k B)| \\ & = |\Spec(\colim A_i \otimes_k B)| \\ & = \lim |\Spec(A_i \otimes_k B)| \\ & = \lim |\Spec(\prod\nolimits_{t \in T_i} B)| \\ & = \lim T_i \times |X| \\ & = (\lim T_i) \times |X| \\ & = |T| \times |X| \end{align*} by the lemma above and the fact that limits commute with limits.
Let $e \in U(k)$ be a quasi-compact open neighbourhood of the identity element. By replacing $U$ by $U \cap i(U)$ we may assume that $U$ is invariant under the inverse map. As $G$ is separated this is still a quasi-compact set. Set $$ G' = \bigcup\nolimits_{n \geq 1} m_n(U \times_k \ldots \times_k U) $$ where $m_n : G \times_k \ldots \times_k G \to G$ is the $n$-slot multiplication map $(g_1, \ldots, g_n) \mapsto m(m(\ldots (m(g_1, g_2), g_3), \ldots ), g_n)$. Each of these maps are open (see Lemma \ref{lemma-group-scheme-over-field-open-multiplication}) hence $G'$ is an open subgroup scheme. By Lemma \ref{lemma-open-subgroup-closed-over-field} it is also a closed subgroup scheme.
Let us first prove that $\dim_g(G) = \dim_{g'}(G)$ for any pair of points $g, g' \in G$. By Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-after-base-change} we may extend the ground field at will. Hence we may assume that both $g$ and $g'$ are defined over $k$. Hence there exists an automorphism of $G$ mapping $g$ to $g'$, whence the equality. By Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-at-a-point} we have $\dim_g(G) = \dim(\mathcal{O}_{G, g}) + \text{trdeg}_k(\kappa(g))$. On the other hand, the dimension of $G$ (or any open subset of $G$) is the supremum of the dimensions of the local rings of $G$, see Properties, Lemma \ref{properties-lemma-codimension-local-ring}. Clearly this is maximal for closed points $g$ in which case $\text{trdeg}_k(\kappa(g)) = 0$ (by the Hilbert Nullstellensatz, see Morphisms, Section \ref{morphisms-section-points-finite-type}). Hence the lemma follows.
We first argue by induction on $n$ that we may assume all $g_i$ are on the same connected component of $G$. Namely, if not, then we can find a decomposition $G = W_1 \amalg W_2$ with $W_i$ open in $G$ and (after possibly renumbering) $g_1, \ldots, g_r \in W_1$ and $g_{r + 1}, \ldots, g_n \in W_2$ for some $0 < r < n$. By induction we can find affine opens $U_1$ and $U_2$ of $G$ with $g_1, \ldots, g_r \in U_1$ and $g_{r + 1}, \ldots, g_n \in U_2$. Then $$ g_1, \ldots, g_n \in (U_1 \cap W_1) \cup (U_2 \cap W_2) $$ is a solution to the problem. Thus we may assume $g_1, \ldots, g_n$ are all on the same connected component of $G$. Translating by $g_1^{-1}$ we may assume $g_1, \ldots, g_n \in G^0$ where $G^0 \subset G$ is as in Proposition \ref{proposition-connected-component}. Choose an affine open neighbourhood $U$ of $e$, in particular $U \cap G^0$ is nonempty. Since $G^0$ is irreducible we see that $$ G^0 \cap (Ug_1^{-1} \cap \ldots \cap Ug_n^{-1}) $$ is nonempty. Since $G \to \Spec(k)$ is locally of finite type, also $G^0 \to \Spec(k)$ is locally of finite type, hence any nonempty open has a $k$-rational point. Thus we can pick $g \in G^0(k)$ with $g \in Ug_i^{-1}$ for all $i$. Then $g_i \in g^{-1}U$ for all $i$ and $g^{-1}U$ is the affine open we were looking for.
If $k$ is perfect then this follows from Lemma \ref{lemma-group-scheme-characteristic-zero-smooth} (characteristic zero) and Lemma \ref{lemma-reduced-group-scheme-prefect-field-characteristic-p-smooth} (positive characteristic). We can reduce the general case to this case by descent for smoothness (Descent, Lemma \ref{descent-lemma-descending-property-smooth}) and going to the perfect closure using Lemma \ref{lemma-abelian-variety-change-field}.
Let $k$ be a field. Let $A$ be an abelian variety over $k$. By Lemma \ref{lemma-abelian-variety-change-field} we may replace $k$ by its algebraic closure. Consider the morphism $$ h : A \times_k A \longrightarrow A \times_k A,\quad (x, y) \longmapsto (x, xyx^{-1}y^{-1}) $$ This is a morphism over $A$ via the first projection on either side. Let $e \in A(k)$ be the unit. Then we see that $h|_{e \times A}$ is constant with value $(e, e)$. By More on Morphisms, Lemma \ref{more-morphisms-lemma-flat-proper-family-cannot-collapse-fibre} there exists an open neighbourhood $U \subset A$ of $e$ such that $h|_{U \times A}$ factors through some $Z \subset U \times A$ finite over $U$. This means that for $x \in U(k)$ the morphism $A \to A$, $y \mapsto xyx^{-1}y^{-1}$ takes finitely many values. Of course this means it is constant with value $e$. Thus $(x, y) \mapsto xyx^{-1}y^{-1}$ is constant with value $e$ on $U \times A$ which implies that the group law on $A$ is abelian.
Apply the theorem of the cube (More on Morphisms, Theorem \ref{more-morphisms-theorem-of-the-cube}) to the difference $$ \mathcal{M} = m_{1, 2, 3}^*\mathcal{L} \otimes m_1^*\mathcal{L} \otimes m_2^*\mathcal{L} \otimes m_3^*\mathcal{L} \otimes m_{1, 2}^*\mathcal{L}^{\otimes -1} \otimes m_{1, 3}^*\mathcal{L}^{\otimes -1} \otimes m_{2, 3}^*\mathcal{L}^{\otimes -1} $$ This works because the restriction of $\mathcal{M}$ to $A \times A \times e = A \times A$ is equal to $$ n_{1, 2}^*\mathcal{L} \otimes n_1^*\mathcal{L} \otimes n_2^*\mathcal{L} \otimes n_{1, 2}^*\mathcal{L}^{\otimes -1} \otimes n_1^*\mathcal{L}^{\otimes -1} \otimes n_2^*\mathcal{L}^{\otimes -1} \cong \mathcal{O}_{A \times_k A} $$ where $n_{i_1, \ldots, i_t} : A \times_k A \to A$ is the morphism $(x_1, x_2) \mapsto \sum x_{i_j}$. Similarly for $A \times e \times A$ and $e \times A \times A$.
Consider the morphism $A \to A \times_k A \times_k A$, $x \mapsto (x, x, -x)$ where $-x = [-1](x)$. Pulling back the relation of Lemma \ref{lemma-apply-cube} we obtain $$ \mathcal{L} \otimes \mathcal{L} \otimes \mathcal{L} \otimes [-1]^*\mathcal{L} \cong [2]^*\mathcal{L} $$ which proves the result for $n = 2$. By induction assume the result holds for $1, 2, \ldots, n$. Then consider the morphism $A \to A \times_k A \times_k A$, $x \mapsto (x, x, [n - 1]x)$. Pulling back the relation of Lemma \ref{lemma-apply-cube} we obtain $$ [n + 1]^*\mathcal{L} \otimes \mathcal{L} \otimes \mathcal{L} \otimes [n - 1]^*\mathcal{L} \cong [2]^*\mathcal{L} \otimes [n]^*\mathcal{L} \otimes [n]^*\mathcal{L} $$ and the result follows by elementary arithmetic.
By Lemma \ref{lemma-abelian-variety-projective} (and More on Morphisms, Lemma \ref{more-morphisms-lemma-projective}) we see that $A$ has an ample invertible module $\mathcal{L}$. Since $[-1] : A \to A$ is an automorphism, we see that $[-1]^*\mathcal{L}$ is an ample invertible $\mathcal{O}_X$-module as well. Thus $\mathcal{N} = \mathcal{L} \otimes [-1]^*\mathcal{L}$ is ample, see Properties, Lemma \ref{properties-lemma-ample-tensor-globally-generated}. Since $\mathcal{N} \cong [-1]^*\mathcal{N}$ we see that $[d]^*\mathcal{N} \cong \mathcal{N}^{\otimes d^2}$ by Lemma \ref{lemma-pullbacks-by-n}. \medskip\noindent To get a contradiction $C \subset X$ be a proper curve contained in a fibre of $[d]$. Then $\mathcal{N}^{\otimes d^2}|_C \cong \mathcal{O}_C$ is an ample invertible $\mathcal{O}_C$-module of degree $0$ which contradicts Varieties, Lemma \ref{varieties-lemma-ample-curve} for example. (You can also use Varieties, Lemma \ref{varieties-lemma-ample-positive}.) Thus every fibre of $[d]$ has dimension $0$ and hence $[d]$ is finite for example by Cohomology of Schemes, Lemma \ref{coherent-lemma-characterize-finite}. Moreover, since $A$ is smooth over $k$ by Lemma \ref{lemma-abelian-variety-smooth} we see that $[d] : A \to A$ is flat by Algebra, Lemma \ref{algebra-lemma-CM-over-regular-flat} (we also use that schemes smooth over fields are regular and that regular rings are Cohen-Macaulay, see Varieties, Lemma \ref{varieties-lemma-smooth-regular} and Algebra, Lemma \ref{algebra-lemma-regular-ring-CM}). Thus $[d]$ is finite flat hence finite locally free by Morphisms, Lemma \ref{morphisms-lemma-finite-flat}. \medskip\noindent Finally, we come to the formula for the degree. By Varieties, Lemma \ref{varieties-lemma-degree-finite-morphism-in-terms-degrees} we see that $$ \deg_{\mathcal{N}^{\otimes d^2}}(A) = \deg([d]) \deg_\mathcal{N}(A) $$ Since the degree of $A$ with respect to $\mathcal{N}^{\otimes d^2}$, respectively $\mathcal{N}$ is the coefficient of $n^{\dim(A)}$ in the polynomial $$ n \longmapsto \chi(A, \mathcal{N}^{\otimes nd^2}),\quad \text{respectively}\quad n \longmapsto \chi(A, \mathcal{N}^{\otimes n}) $$ we see that $\deg([d]) = d^{2 \dim(A)}$.
Observe that $[d](x + y) = [d](x) + [d](y)$. Since translation by a point is an automorphism of $A$, we see that the set of points where $[d] : A \to A$ is \'etale is either empty or equal to $A$ (some details omitted). Thus it suffices to check whether $[d]$ is \'etale at the unit $e \in A(k)$. Since we know that $[d]$ is finite locally free (Lemma \ref{lemma-degree-multiplication-by-d}) to see that it is \'etale at $e$ is equivalent to proving that $\text{d}[d] : T_{A/k, e} \to T_{A/k, e}$ is injective. See Varieties, Lemma \ref{varieties-lemma-injective-tangent-spaces-unramified} and Morphisms, Lemma \ref{morphisms-lemma-flat-unramified-etale}. By Lemma \ref{lemma-group-scheme-addition-tangent-vectors} we see that $\text{d}[d]$ is given by multiplication by $d$ on $T_{A/k, e}$.