hcju/stacksqa huggingface dataset:
Denote $g : Y \to \Spec(k)$ and $f : X \to \Spec(k)$ the structure morphisms. Then $g = f \circ i$. Denote $a, b, c$ the right adjoint of Duality for Schemes, Lemma \ref{duality-lemma-twisted-inverse-image} for $f, g, i$. Then $b = c \circ a$ by uniqueness of right adjoints and because $Rg_* = Rf_* \circ Ri_*$. In the proof of Lemma \ref{lemma-duality-dim-1} we set $\omega_X^\bullet = a(\mathcal{O}_{\Spec(k)})$ and $\omega_Y^\bullet = b(\mathcal{O}_{\Spec(k)})$. Hence $\omega_Y^\bullet = c(\omega_X^\bullet)$ which implies (1) by Duality for Schemes, Lemma \ref{duality-lemma-twisted-inverse-image-closed}. Since $\omega_X = H^{-1}(\omega_X^\bullet)$ and $\omega_Y = H^{-1}(\omega_Y^\bullet)$ we conclude that $\omega_Y = \SheafHom(\mathcal{O}_Y, \omega_X)$. This implies $i_*\omega_Y = \SheafHom_{\mathcal{O}_X}(i_*\mathcal{O}_Y, \omega_X)$ by Duality for Schemes, Lemma \ref{duality-lemma-sheaf-with-exact-support-ext}.
A reduced $1$-dimensional Noetherian scheme is Cohen-Macaulay, so (1) is true. Since $X$ is reduced, we see that $X = Y \cup Z$ scheme theoretically. With notation as in Morphisms, Lemma \ref{morphisms-lemma-scheme-theoretic-union} and by the statement of that lemma we have a short exact sequence $$ 0 \to \mathcal{O}_X \to \mathcal{O}_Y \oplus \mathcal{O}_Z \to \mathcal{O}_{Y \cap Z} \to 0 $$ Since $\mathcal{J} = \Ker(\mathcal{O}_X \to \mathcal{O}_Z)$, $\mathcal{J}' = \Ker(\mathcal{O}_Y \to \mathcal{O}_{Y \cap Z})$, $\mathcal{I} = \Ker(\mathcal{O}_X \to \mathcal{O}_Y)$, and $\mathcal{I}' = \Ker(\mathcal{O}_Z \to \mathcal{O}_{Y \cap Z})$ a diagram chase implies (2). Observe that $\mathcal{I} + \mathcal{J}$ is the ideal sheaf of $Y \cap Z$ and that $\mathcal{I} \cap \mathcal{J} = 0$. Hence we have the following exact sequences \begin{align*} 0 \to \mathcal{O}_X \to \mathcal{O}_Y \oplus \mathcal{O}_Z \to \mathcal{O}_{Y \cap Z} \to 0 \\ 0 \to \mathcal{J} \to \mathcal{O}_X \to \mathcal{O}_Z \to 0 \\ 0 \to \mathcal{I} \to \mathcal{O}_X \to \mathcal{O}_Y \to 0 \\ 0 \to \mathcal{J} \oplus \mathcal{I} \to \mathcal{O}_X \to \mathcal{O}_{Y \cap Z} \to 0 \\ 0 \to \mathcal{J}' \to \mathcal{O}_Y \to \mathcal{O}_{Y \cap Z} \to 0 \\ 0 \to \mathcal{I}' \to \mathcal{O}_Z \to \mathcal{O}_{Y \cap Z} \to 0 \end{align*} Since $X$ is Gorenstein $\omega_X$ is an invertible $\mathcal{O}_X$-module (Duality for Schemes, Lemma \ref{duality-lemma-gorenstein}). Since $Y \cap Z$ has dimension $0$ we have $\omega_X|_{Y \cap Z} \cong \mathcal{O}_{Y \cap Z}$. Thus if we prove (3) and (4), then we obtain the short exact sequences of the lemma by tensoring the above short exact sequence with the invertible module $\omega_X$. By symmetry it suffices to prove (3) and by (2) it suffices to prove $i_*(\omega_Y) = \mathcal{J}\omega_X$. \medskip\noindent We have $i_*\omega_Y = \SheafHom_{\mathcal{O}_X}(i_*\mathcal{O}_Y, \omega_X)$ by Lemma \ref{lemma-closed-immersion-dim-1-CM}. Again using that $\omega_X$ is invertible we finally conclude that it suffices to show $\SheafHom_{\mathcal{O}_X}(\mathcal{O}_X/\mathcal{I}, \mathcal{O}_X)$ maps isomorphically to $\mathcal{J}$ by evaluation at $1$. In other words, that $\mathcal{J}$ is the annihilator of $\mathcal{I}$. This follows from the above.
We define the right hand side of the first formula as follows: $$ \chi(X, \omega_X^\bullet) = \sum\nolimits_{i \in \mathbf{Z}} (-1)^i\dim_k H^i(X, \omega_X^\bullet) $$ This is well defined because $\omega_X^\bullet$ is in $D^b_{\textit{Coh}}(\mathcal{O}_X)$, but also because $$ H^i(X, \omega_X^\bullet) = \Ext^i(\mathcal{O}_X, \omega_X^\bullet) = H^{-i}(X, \mathcal{O}_X) $$ which is always finite dimensional and nonzero only if $i = 0, -1$. This of course also proves the first formula. The second is a consequence of the first because $\omega_X^\bullet = \omega_X[1]$ in the CM case, see Lemma \ref{lemma-duality-dim-1-CM}.
Recall that Gorenstein schemes are Cohen-Macaulay (Duality for Schemes, Lemma \ref{duality-lemma-gorenstein-CM}) and hence $\omega_X$ is a dualizing module on $X$, see Lemma \ref{lemma-duality-dim-1-CM}. It follows more or less from the definition of the Gorenstein property that the dualizing sheaf is invertible, see Duality for Schemes, Section \ref{duality-section-gorenstein}. By (\ref{equation-rr}) applied to $\omega_X$ we have $$ \chi(X, \omega_X) =  \deg(c_1(\omega_X) \cap [X]_1) + \chi(X, \mathcal{O}_X) $$ Combined with Lemma \ref{lemma-euler} this gives $$ 2\chi(X, \mathcal{O}_X) = - \deg(c_1(\omega_X) \cap [X]_1) = - \deg(\omega_X) $$ the second equality by (\ref{equation-degree-c1}). Putting this back into (\ref{equation-rr}) for $\mathcal{E}$ gives the displayed formula of the lemma. The symmetry in dimensions is a consequence of duality for $X$, see Remark \ref{remark-rework-duality-locally-free}.
Since $\Gamma(X, \mathcal{O}_X) = k$ has no nontrivial idempotents, we see that $X$ is connected. This already shows that $X$ is equidimensional of dimension $1$ (any irreducible component of dimension $0$ would be a connected component). Let $\mathcal{I} \subset \mathcal{O}_X$ be the maximal coherent submodule supported in closed points. Then $\mathcal{I}$ exists (Divisors, Lemma \ref{divisors-lemma-remove-embedded-points}) and is globally generated (Varieties, Lemma \ref{varieties-lemma-chi-tensor-finite}). Since $1 \in \Gamma(X, \mathcal{O}_X)$ is not a section of $\mathcal{I}$ we conclude that $\mathcal{I} = 0$. Thus $X$ does not have embedded points (Divisors, Lemma \ref{divisors-lemma-remove-embedded-points}). Thus $X$ has $(S_1)$ by Divisors, Lemma \ref{divisors-lemma-S1-no-embedded}. Hence $X$ is Cohen-Macaulay.
(The parenthetical statement follows from Cohomology of Schemes, Lemma \ref{coherent-lemma-coherent-support-dimension-0}.) Since $H^0(X, \mathcal{O}_X) = k$ is dual to $H^1(X, \omega_X)$ (see Section \ref{section-Riemann-Roch}) we see that $\dim H^1(X, \omega_X) = 1$. The sheaf $\omega_X$ represents the functor $\mathcal{F} \mapsto \Hom_k(H^1(X, \mathcal{F}), k)$ on the category of coherent $\mathcal{O}_X$-modules (Duality for Schemes, Lemma \ref{duality-lemma-dualizing-module-proper-over-A}). Consider an exact sequence as in the statement of the lemma and assume that $H^1(X, \mathcal{F}) \not = 0$. Since $H^1(X, \mathcal{Q}) = 0$ we see that $H^1(X, \omega_X) \to H^1(X, \mathcal{F})$ is an isomorphism. By the universal property of $\omega_X$ stated above, we conclude there is a map $\mathcal{F} \to \omega_X$ whose action on $H^1$ is the inverse of this isomorphism. The composition $\omega_X \to \mathcal{F} \to \omega_X$ is the identity (by the universal property) and the lemma is proved.
Let $\mathcal{F}' \subset \mathcal{F}$ be the submodule generated by global sections and the image of $\omega_X \to \mathcal{F}$. Since $\dim_k H^0(X, \mathcal{Q}) \geq 2$ and $\dim_k H^1(X, \omega_X) = \dim_k H^0(X, \mathcal{O}_X) = 1$, we see that $\mathcal{F}' \to \mathcal{Q}$ is not zero and $\omega_X \to \mathcal{F}'$ is not an isomorphism. Hence $H^1(X, \mathcal{F}') = 0$ by Lemma \ref{lemma-vanishing} and our assumption on $\mathcal{F}$. Consider the short exact sequence $$ 0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{Q}/\Im(\mathcal{F}' \to \mathcal{Q}) \to 0 $$ If the quotient on the right is nonzero, then we obtain a contradiction because then $H^0(X, \mathcal{F})$ is bigger than $H^0(X, \mathcal{F}')$.
Consider the submodule $\mathcal{F}'$ generated by the global sections. By Lemma \ref{lemma-globally-generated} we see that $\mathcal{F}' \to \mathcal{Q}$ is surjective, in particular $\mathcal{F}' \not = 0$. Since $X$ is a curve, we see that $\mathcal{F}' \subset \mathcal{F}$ is an inclusion of rank $1$ sheaves, hence $\mathcal{Q}' = \mathcal{F}/\mathcal{F}'$ is supported in finitely many points. To get a contradiction, assume that $\mathcal{Q}'$ is nonzero. Then we see that $H^1(X, \mathcal{F}') \not = 0$. Then we get a nonzero map $\mathcal{F}' \to \omega_X$ by the universal property (Duality for Schemes, Lemma \ref{duality-lemma-dualizing-module-proper-over-A}). The image of the composition $\mathcal{F}' \to \omega_X \to \mathcal{F}$ is generated by global sections, hence is inside of $\mathcal{F}'$. Thus we get a nonzero self map $\mathcal{F}' \to \mathcal{F}'$. Since $\mathcal{F}'$ is torsion free of rank $1$ on a proper curve this has to be an automorphism (details omitted). But then this implies that $\mathcal{F}'$ is contained in $\omega_X \subset \mathcal{F}$ contradicting the surjectivity of $\mathcal{F}' \to \mathcal{Q}$.
Assume $k$ is algebraically closed. Let $x \in X$ be a closed point. Let $C_i \subset X$ be the irreducible components and for each $i$ let $x_i \in C_i$ be the generic point. By Varieties, Lemma \ref{varieties-lemma-very-ample-vanish-at-point} we can choose a section $s \in H^0(X, \mathcal{L})$ such that $s$ vanishes at $x$ but not at $x_i$ for all $i$. The corresponding module map $s : \mathcal{O}_X \to \mathcal{L}$ is injective with cokernel $\mathcal{Q}$ supported in finitely many points and with $H^0(X, \mathcal{Q}) \geq 2$. Consider the corresponding exact sequence $$ 0 \to \omega_X \to \omega_X \otimes \mathcal{L} \to \omega_X \otimes \mathcal{Q} \to 0 $$ By Lemma \ref{lemma-globally-generated} we see that the module generated by global sections surjects onto $\omega_X \otimes \mathcal{Q}$. Since $x$ was arbitrary this proves the lemma. Some details omitted. \medskip\noindent We will reduce the case where $k$ is not algebraically closed, to the algebraically closed field case. We suggest the reader skip the rest of the proof. Choose an algebraic closure $\overline{k}$ of $k$ and consider the base change $X_{\overline{k}}$. Let us check that $X_{\overline{k}} \to \Spec(\overline{k})$ is an example of Situation \ref{situation-Cohen-Macaulay-curve}. By flat base change (Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}) we see that $H^0(X_{\overline{k}}, \mathcal{O}) = \overline{k}$. The scheme $X_{\overline{k}}$ is proper over $\overline{k}$ (Morphisms, Lemma \ref{morphisms-lemma-base-change-proper}) and equidimensional of dimension $1$ (Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-after-base-change}). The pullback of $\omega_X$ to $X_{\overline{k}}$ is the dualizing module of $X_{\overline{k}}$ by Lemma \ref{lemma-sanity-check-duality}. The pullback of $\mathcal{L}$ to $X_{\overline{k}}$ is very ample (Morphisms, Lemma \ref{morphisms-lemma-very-ample-base-change}). The degree of the pullback of $\mathcal{L}$ to $X_{\overline{k}}$ is equal to the degree of $\mathcal{L}$ on $X$ (Varieties, Lemma \ref{varieties-lemma-degree-base-change}). Finally, we see that $\omega_X \otimes \mathcal{L}$ is globally generated if and only if its base change is so (Varieties, Lemma \ref{varieties-lemma-globally-generated-base-change}). In this way we see that the result follows from the result in the case of an algebraically closed ground field.
Let $s$ be a regular global section of $\mathcal{L}$. Let $i : Z = Z(s) \to X$ be the zero scheme of $s$, see Divisors, Section \ref{divisors-section-effective-Cartier-invertible}. By condition (3) we see that $Z \not = \emptyset$ (small detail omitted). Consider the short exact sequence $$ 0 \to \mathcal{O}_X \xrightarrow{s} \mathcal{L} \to i_*(\mathcal{L}|_Z) \to 0 $$ Tensoring with $\mathcal{L}$ we obtain $$ 0 \to \mathcal{L} \to \mathcal{L}^{\otimes 2} \to i_*(\mathcal{L}^{\otimes 2}|_Z) \to 0 $$ Observe that $Z$ has dimension $0$ (Divisors, Lemma \ref{divisors-lemma-effective-Cartier-makes-dimension-drop}) and hence is the spectrum of an Artinian ring (Varieties, Lemma \ref{varieties-lemma-algebraic-scheme-dim-0}) hence $\mathcal{L}|_Z \cong \mathcal{O}_Z$ (Algebra, Lemma \ref{algebra-lemma-locally-free-semi-local-free}). The short exact sequence also shows that $H^1(X, \mathcal{L}^{\otimes 2}) = 0$ (for example using Varieties, Lemma \ref{varieties-lemma-chi-tensor-finite} to see vanishing in the spot on the right). Using induction on $n \geq 1$ and the sequence $$ 0 \to \mathcal{L}^{\otimes n} \xrightarrow{s} \mathcal{L}^{\otimes n + 1} \to i_*(\mathcal{L}^{\otimes n + 1}|_Z) \to 0 $$ we see that $H^1(X, \mathcal{L}^{\otimes n}) = 0$ for $n > 0$ and that there exists a global section $t_{n + 1}$ of $\mathcal{L}^{\otimes n + 1}$ which gives a trivialization of $\mathcal{L}^{\otimes n + 1}|_Z \cong \mathcal{O}_Z$. \medskip\noindent Consider the multiplication map $$ \mu_n : H^0(X, \mathcal{L}) \otimes_k H^0(X, \mathcal{L}^{\otimes n}) \oplus H^0(X, \mathcal{L}^{\otimes 2}) \otimes_k H^0(X, \mathcal{L}^{\otimes n - 1}) \longrightarrow H^0(X, \mathcal{L}^{\otimes n + 1}) $$ We claim this is surjective for $n \geq 3$. To see this we consider the short exact sequence $$ 0 \to \mathcal{L}^{\otimes n} \xrightarrow{s} \mathcal{L}^{\otimes n + 1} \to i_*(\mathcal{L}^{\otimes n + 1}|_Z) \to 0 $$ The sections of $\mathcal{L}^{\otimes n + 1}$ coming from the left in this sequence are in the image of $\mu_n$. On the other hand, since $H^0(\mathcal{L}^{\otimes 2}) \to H^0(\mathcal{L}^{\otimes 2}|_Z)$ is surjective (see above) and since $t_{n - 1}$ maps to a trivialization of $\mathcal{L}^{\otimes n - 1}|_Z$ we see that $\mu_n(H^0(X, \mathcal{L}^{\otimes 2}) \otimes t_{n - 1})$ gives a subspace of $H^0(X, \mathcal{L}^{\otimes n + 1})$ surjecting onto the global sections of $\mathcal{L}^{\otimes n + 1}|_Z$. This proves the claim. \medskip\noindent From the claim in the previous paragraph we conclude that the graded $k$-algebra $$ S = \bigoplus\nolimits_{n \geq 0} H^0(X, \mathcal{L}^{\otimes n}) $$ is generated in degrees $0, 1, 2, 3$ over $k$. Recall that $X = \text{Proj}(S)$, see Morphisms, Lemma \ref{morphisms-lemma-proper-ample-is-proj}. Thus $S^{(6)} = \bigoplus_{n} S_{6n}$ is generated in degree $1$. This means that $\mathcal{L}^{\otimes 6}$ is very ample as desired.
Choose basis $s_0, \ldots, s_n$ of $H^0(X, \mathcal{L}^{\otimes 2})$ over $k$. By property (1) we see that $\mathcal{L}^{\otimes 2}$ is globally generated and we get a morphism $$ \varphi_{\mathcal{L}^{\otimes 2}, (s_0, \ldots, s_n)} : X \longrightarrow \mathbf{P}^n_k $$ See Constructions, Section \ref{constructions-section-projective-space}. The lemma asserts that this morphism is a closed immersion. To check this we may replace $k$ by its algebraic closure, see Descent, Lemma \ref{descent-lemma-descending-property-closed-immersion}. Thus we may assume $k$ is algebraically closed. \medskip\noindent Assume $k$ is algebraically closed. For each generic point $\eta_i \in X$ let $V_i \subset H^0(X, \mathcal{L})$ be the $k$-subvector space of sections vanishing at $\eta_i$. Since $\mathcal{L}$ is globally generated, we see that $V_i \not = H^0(X, \mathcal{L})$. Since $X$ has only a finite number of irreducible components and $k$ is infinite, we can find $s \in H^0(X, \mathcal{L})$ nonvanishing at $\eta_i$ for all $i$. Then $s$ is a regular section of $\mathcal{L}$ (because $X$ is Cohen-Macaulay by Lemma \ref{lemma-automatic} and hence $\mathcal{L}$ has no embedded associated points). \medskip\noindent In particular, all of the statements given in the proof of Lemma \ref{lemma-criterion-very-ample} hold with this $s$. Moreover, as $\mathcal{L}$ is globally generated, we can find a global section $t \in H^0(X, \mathcal{L})$ such that $t|_Z$ is nonvanishing (argue as above using the finite number of points of $Z$). Then in the proof of Lemma \ref{lemma-criterion-very-ample} we can use $t$ to see that additionally the multiplication map $$ \mu_n : H^0(X, \mathcal{L}) \otimes_k H^0(X, \mathcal{L}^{\otimes 2}) \longrightarrow H^0(X, \mathcal{L}^{\otimes 3}) $$ is surjective. Thus $$ S = \bigoplus\nolimits_{n \geq 0} H^0(X, \mathcal{L}^{\otimes n}) $$ is generated in degrees $0, 1, 2$ over $k$. Arguing as in the proof of Lemma \ref{lemma-criterion-very-ample} we find that $S^{(2)} = \bigoplus_{n} S_{2n}$ is generated in degree $1$. This means that $\mathcal{L}^{\otimes 2}$ is very ample as desired. Some details omitted.
The dimension of $X_{k'}$ is $1$ for example by Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-after-base-change}. The morphism $X_{k'} \to \Spec(k')$ is proper by Morphisms, Lemma \ref{morphisms-lemma-base-change-proper}. The equality $H^0(X_{k'}, \mathcal{O}_{X_{k'}}) = k'$ follows from Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}. The equality of the genus follows from the same lemma.
This is a special case of Divisors, Lemma \ref{divisors-lemma-equation-codim-1-in-projective-space} (see also Varieties, Lemma \ref{varieties-lemma-equation-codim-1-in-projective-space}). The parenthetical statement follows from the fact that a $1$ dimensional Noetherian scheme is Cohen-Macaulay if and only if it has no embedded points, see Divisors, Lemma \ref{divisors-lemma-noetherian-dim-1-CM-no-embedded-points}.
If $F$ is reducible, say $F = F' F''$ then let $Z'$ be the closed subscheme of $\mathbf{P}^2_k$ defined by $F'$. It is clear that $Z' \subset Z$ and that $Z' \not = Z$. Since $Z'$ has dimension $1$ as well, we conclude that either $Z$ is not reduced, or that $Z$ is not irreducible. Conversely, write $Z = \sum a_i D_i$ where $D_i$ are the irreducible components of $Z$, see Divisors, Lemmas \ref{divisors-lemma-codim-1-part} and \ref{divisors-lemma-codimension-1-is-effective-Cartier}. Let $F_i \in k[T_0, T_1, T_2]$ be the homogeneous polynomial generating the ideal of $D_i$. Then it is clear that $F$ and $\prod F_i^{a_i}$ cut out the same closed subscheme of $\mathbf{P}^2_k$. Hence $F = \lambda \prod F_i^{a_i}$ for some $\lambda \in k^*$ because both generate the ideal of $Z$. Thus we see that if $F$ is irreducible, then $Z$ is a prime divisor, i.e., a curve.
Let $S = k[T_0, T_1, T_2]$. There is an exact sequence of graded modules $$ 0 \to S(-d) \xrightarrow{F} S \to S/(F) \to 0 $$ Denote $i : Z \to \mathbf{P}^2_k$ the given closed immersion. Applying the exact functor $\widetilde{\ }$ (Constructions, Lemma \ref{constructions-lemma-proj-sheaves}) we obtain $$ 0 \to \mathcal{O}_{\mathbf{P}^2_k}(-d) \to \mathcal{O}_{\mathbf{P}^2_k} \to i_*\mathcal{O}_Z \to 0 $$ because $F$ generates the ideal of $Z$. Note that the cohomology groups of $\mathcal{O}_{\mathbf{P}^2_k}(-d)$ and $\mathcal{O}_{\mathbf{P}^2_k}$ are given in Cohomology of Schemes, Lemma \ref{coherent-lemma-cohomology-projective-space-over-ring}. On the other hand, we have $H^q(Z, \mathcal{O}_Z) = H^q(\mathbf{P}^2_k, i_*\mathcal{O}_Z)$ by Cohomology of Schemes, Lemma \ref{coherent-lemma-relative-affine-cohomology}. Applying the long exact cohomology sequence we first obtain that $$ k = H^0(\mathbf{P}^2_k, \mathcal{O}_{\mathbf{P}^2_k}) \longrightarrow H^0(Z, \mathcal{O}_Z) $$ is an isomorphism and next that the boundary map $$ H^1(Z, \mathcal{O}_Z) \longrightarrow H^2(\mathbf{P}^2_k, \mathcal{O}_{\mathbf{P}^2_k}(-d)) \cong k[T_0, T_1, T_2]_{d - 3} $$ is an isomorphism. Since it is easy to see that the dimension of this is $(d - 1)(d - 2)/2$ the proof is finished.
Namely, we have $\dim_k H^0(X, \mathcal{L}) \geq 0 + 1 - 0 = 1$ by Riemann-Roch (Lemma \ref{lemma-rr}), hence $\mathcal{L}$ has a nonzero section, hence $\mathcal{L} \cong \mathcal{O}_X$ by Varieties, Lemma \ref{varieties-lemma-check-invertible-sheaf-trivial}.
By definition of degree and genus we have $$ \dim_k H^0(X, \mathcal{L}) - \dim_k H^1(X, \mathcal{L}) = d + 1 $$ Let $s$ be a nonzero section of $\mathcal{L}$. Then the zero scheme of $s$ is an effective Cartier divisor $D \subset X$, we have $\mathcal{L} = \mathcal{O}_X(D)$ and we have a short exact sequence $$ 0 \to \mathcal{O}_X \to \mathcal{L} \to \mathcal{L}|_D \to 0 $$ see Divisors, Lemma \ref{divisors-lemma-characterize-OD} and Remark \ref{divisors-remark-ses-regular-section}. Since $H^1(X, \mathcal{O}_X) = 0$ by assumption, we see that $H^0(X, \mathcal{L}) \to H^0(X, \mathcal{L}|_D)$ is surjective. As $\mathcal{L}|_D$ is generated by global sections (because $\dim(D) = 0$, see Varieties, Lemma \ref{varieties-lemma-chi-tensor-finite}) we conclude that the invertible module $\mathcal{L}$ is generated by global sections. In fact, since $D$ is an Artinian scheme we have $\mathcal{L}|_D \cong \mathcal{O}_D$\footnote{In our case this follows from Divisors, Lemma \ref{divisors-lemma-finite-trivialize-invertible-upstairs} as $D \to \Spec(k)$ is finite.} and hence we can find a section $t$ of $\mathcal{L}$ whose restriction of $D$ generates $\mathcal{L}|_D$. The short exact sequence also shows that $H^1(X, \mathcal{L}) = 0$. \medskip\noindent For $n \geq 1$ consider the multiplication map $$ \mu_n : H^0(X, \mathcal{L}) \otimes_k H^0(X, \mathcal{L}^{\otimes n}) \longrightarrow H^0(X, \mathcal{L}^{\otimes n + 1}) $$ We claim this is surjective. To see this we consider the short exact sequence $$ 0 \to \mathcal{L}^{\otimes n} \xrightarrow{s} \mathcal{L}^{\otimes n + 1} \to \mathcal{L}^{\otimes n + 1}|_D \to 0 $$ The sections of $\mathcal{L}^{\otimes n + 1}$ coming from the left in this sequence are in the image of $\mu_n$. On the other hand, since $H^0(\mathcal{L}) \to H^0(\mathcal{L}|_D)$ is surjective and since $t^n$ maps to a trivialization of $\mathcal{L}^{\otimes n}|_D$ we see that $\mu_n(H^0(X, \mathcal{L}) \otimes t^n)$ gives a subspace of $H^0(X, \mathcal{L}^{\otimes n + 1})$ surjecting onto the global sections of $\mathcal{L}^{\otimes n + 1}|_D$. This proves the claim. \medskip\noindent Observe that $\mathcal{L}$ is ample by Varieties, Lemma \ref{varieties-lemma-ample-curve}. Hence Morphisms, Lemma \ref{morphisms-lemma-proper-ample-is-proj} gives an isomorphism $$ X \longrightarrow \text{Proj}\left( \bigoplus\nolimits_{n \geq 0} H^0(X, \mathcal{L}^{\otimes n})\right) $$ Since the maps $\mu_n$ are surjective for all $n \geq 1$ we see that the graded algebra on the right hand side is a quotient of the symmetric algebra on $H^0(X, \mathcal{L})$. Choosing a $k$-basis $s_0, \ldots, s_d$ of $H^0(X, \mathcal{L})$ we see that it is a quotient of a polynomial algebra in $d + 1$ variables. Since quotients of graded rings correspond to closed immersions of $\text{Proj}$ (Constructions, Lemma \ref{constructions-lemma-surjective-graded-rings-generated-degree-1-map-proj}) we find a closed immersion $X \to \mathbf{P}^d_k$. We omit the verification that this morphism is the morphism of Constructions, Lemma \ref{constructions-lemma-projective-space} associated to the sections $s_0, \ldots, s_d$ of $\mathcal{L}$.
Consider the invertible sheaf $\mathcal{L} = \omega_X^{\otimes -1}$ where $\omega_X$ is as in Lemma \ref{lemma-duality-dim-1}. Then $\deg(\omega_X) = -2$ by Lemma \ref{lemma-genus-gorenstein} and hence $\deg(\mathcal{L}) = 2$. By Lemma \ref{lemma-genus-zero-positive-degree} we conclude that choosing a basis $s_0, s_1, s_2$ of the $k$-vector space of global sections of $\mathcal{L}$ we obtain a closed immersion $$ \varphi_{(\mathcal{L}, (s_0, s_1, s_2))} : X \longrightarrow \mathbf{P}^2_k $$ Thus $X$ is a plane curve of some degree $d$. Let $F \in k[T_0, T_1, T_2]_d$ be its equation (Lemma \ref{lemma-equation-plane-curve}). Because the genus of $X$ is $0$ we see that $d$ is $1$ or $2$ (Lemma \ref{lemma-genus-plane-curve}). Observe that $F$ restricts to the zero section on $\varphi(X)$ and hence $F(s_0, s_1, s_2)$ is the zero section of $\mathcal{L}^{\otimes 2}$. Because $s_0, s_1, s_2$ are linearly independent we see that $F$ cannot be linear, i.e., $d = \deg(F) \geq 2$. Thus $d = 2$ and the proof is complete.
Since $X$ and $Y$ are smooth, the sheaves $\Omega_{X/k}$ and $\Omega_{Y/k}$ are invertible modules, see Morphisms, Lemma \ref{morphisms-lemma-smooth-omega-finite-locally-free}. Using the exact sequence $$ f^*\Omega_{Y/k} \longrightarrow \Omega_{X/k} \longrightarrow \Omega_{X/Y} \longrightarrow 0 $$ of Morphisms, Lemma \ref{morphisms-lemma-triangle-differentials} we see that (1) and (2) are equivalent and equivalent to the condition that $f^*\Omega_{Y/k} \to \Omega_{X/k}$ is nonzero in the generic point. The equivalence of (2) and (3) follows from Morphisms, Lemma \ref{morphisms-lemma-unramified-omega-zero}. The equivalence between (3) and (4) follows from Morphisms, Lemma \ref{morphisms-lemma-flat-unramified-etale} and the fact that flatness is automatic (Lemma \ref{lemma-flat}). To see the equivalence of (5) and (4) use Algebra, Lemma \ref{algebra-lemma-smooth-at-generic-point}. Some details omitted.