**Proof.**
By induction on the number of irreducible components of $X$. If $X$ is irreducible, then the lemma holds by Lemma 53.21.3 applied to $X$ viewed as a scheme over the field $k' = H^0(X, \mathcal{O}_ X)$. Assume $X$ is not irreducible. Before we continue, if $k$ is finite, then we replace $k$ by a purely transcendental extension $K$. This is allowed by Varieties, Lemmas 33.22.1, 33.44.2, 33.6.7, and 33.8.4, Cohomology of Schemes, Lemma 30.5.2, Lemma 53.4.4 and the elementary fact that $K$ is geometrically integral over $k$.

Assume that $\mathcal{L}$ is not globally generated to get a contradiction. Then we may choose a coherent ideal sheaf $\mathcal{I} \subset \mathcal{O}_ X$ such that $H^0(X, \mathcal{I}\mathcal{L}) = H^0(X, \mathcal{L})$ and such that $\mathcal{O}_ X/\mathcal{I}$ is nonzero with support of dimension $0$. For example, take $\mathcal{I}$ the ideal sheaf of any closed point in the common vanishing locus of the global sections of $\mathcal{L}$. We consider the short exact sequence

\[ 0 \to \mathcal{I}\mathcal{L} \to \mathcal{L} \to \mathcal{L}/\mathcal{I}\mathcal{L} \to 0 \]

Since the support of $\mathcal{L}/\mathcal{I}\mathcal{L}$ has dimension $0$ we see that $\mathcal{L}/\mathcal{I}\mathcal{L}$ is generated by global sections (Varieties, Lemma 33.33.3). From the short exact sequence, and the fact that $H^0(X, \mathcal{I}\mathcal{L}) = H^0(X, \mathcal{L})$ we get an injection $H^0(X, \mathcal{L}/\mathcal{I}\mathcal{L}) \to H^1(X, \mathcal{I}\mathcal{L})$.

Recall that the $k$-vector space $H^1(X, \mathcal{I}\mathcal{L})$ is dual to $\mathop{\mathrm{Hom}}\nolimits (\mathcal{I}\mathcal{L}, \omega _ X)$. Choose $\varphi : \mathcal{I}\mathcal{L} \to \omega _ X$. By Lemma 53.21.6 we have $H^1(X, \mathcal{L}) = 0$. Hence

\[ \dim _ k H^0(X, \mathcal{I}\mathcal{L}) = \dim _ k H^0(X, \mathcal{L}) = \deg (\mathcal{L}) + \chi (\mathcal{O}_ X) > \dim _ k H^1(X, \mathcal{O}_ X) = \dim _ k H^0(X, \omega _ X) \]

We conclude that $\varphi $ is not injective on global sections, in particular $\varphi $ is not injective. For every generic point $\eta \in X$ of an irreducible component of $X$ denote $V_\eta \subset \mathop{\mathrm{Hom}}\nolimits (\mathcal{I}\mathcal{L}, \omega _ X)$ the $k$-subvector space consisting of those $\varphi $ which are zero at $\eta $. Since every associated point of $\mathcal{I}\mathcal{L}$ is a generic point of $X$, the above shows that $\mathop{\mathrm{Hom}}\nolimits (\mathcal{I}\mathcal{L}, \omega _ X) = \bigcup V_\eta $. As $X$ has finitely many generic points and $k$ is infinite, we conclude $\mathop{\mathrm{Hom}}\nolimits (\mathcal{I}\mathcal{L}, \omega _ X) = V_\eta $ for some $\eta $. Let $\eta \in C \subset X$ be the corresponding irreducible component. Let $Y \subset X$ be the union of the other irreducible components of $X$. Then $Y$ is a nonempty reduced closed subscheme not equal to $X$. Let $\mathcal{J} \subset \mathcal{O}_ X$ be the ideal sheaf of $Y$. Please keep in mind that the support of $\mathcal{J}$ is $C$.

Let $\varphi : \mathcal{I}\mathcal{L} \to \omega _ X$ be arbitrary. Since $\mathcal{J}\mathcal{I}\mathcal{L}$ has no embedded associated points (as a submodule of $\mathcal{L}$) and as $\varphi $ is zero in the generic point $\eta $ of the support of $\mathcal{J}$, we find that $\varphi $ factors as

\[ \mathcal{I}\mathcal{L} \to \mathcal{I}\mathcal{L}/\mathcal{J}\mathcal{I}\mathcal{L} \to \omega _ X \]

We can view $\mathcal{I}\mathcal{L}/\mathcal{J}\mathcal{I}\mathcal{L}$ as the pushforward of a coherent sheaf on $Y$ which by abuse of notation we indicate with the same symbol. Since $\omega _ Y = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Y, \omega _ X)$ by Lemma 53.4.5 we find a factorization

\[ \mathcal{I}\mathcal{L} \to \mathcal{I}\mathcal{L}/ \mathcal{J}\mathcal{I}\mathcal{L} \xrightarrow {\varphi '} \omega _ Y \to \omega _ X \]

of $\varphi $. Let $\mathcal{I}' \subset \mathcal{O}_ Y$ be the image of $\mathcal{I} \subset \mathcal{O}_ X$. There is a surjective map $\mathcal{I}\mathcal{L}/\mathcal{J}\mathcal{I}\mathcal{L} \to \mathcal{I}'\mathcal{L}|_ Y$ whose kernel is supported in closed points. Since $\omega _ Y$ is a Cohen-Macaulay module on $Y$, the map $\varphi '$ factors through a map $\varphi '' : \mathcal{I}'\mathcal{L}|_ Y \to \omega _ Y$. Thus we have commutative diagrams

\[ \vcenter { \xymatrix{ 0 \ar[r] & \mathcal{I}\mathcal{L} \ar[r] \ar[d] & \mathcal{L} \ar[r] \ar[d] & \mathcal{L}/\mathcal{I}\mathcal{L} \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathcal{I}'\mathcal{L}|_ Y \ar[r] & \mathcal{L}|_ Y \ar[r] & \mathcal{L}|_ Y/\mathcal{I}'\mathcal{L}|_ Y \ar[r] & 0 } } \quad \text{and}\quad \vcenter { \xymatrix{ \mathcal{I}\mathcal{L} \ar[r]_\varphi \ar[d] & \omega _ X \\ \mathcal{I}'\mathcal{L}|_ Y \ar[r]^{\varphi ''} & \omega _ Y \ar[u] } } \]

Now we can finish the proof as follows: Since for every $\varphi $ we have a $\varphi ''$ and since $\omega _ X \in \textit{Coh}(\mathcal{O}_ X)$ represents the functor $\mathcal{F} \mapsto \mathop{\mathrm{Hom}}\nolimits _ k(H^1(X, \mathcal{F}), k)$, we find that $H^1(X, \mathcal{I}\mathcal{L}) \to H^1(Y, \mathcal{I}'\mathcal{L}|_ Y)$ is injective. Since the boundary $H^0(X, \mathcal{L}/\mathcal{I}\mathcal{L}) \to H^1(X, \mathcal{I}\mathcal{L})$ is injective, we conclude the composition

\[ H^0(X, \mathcal{L}/\mathcal{I}\mathcal{L}) \to H^0(X, \mathcal{L}|_ Y/\mathcal{I}'\mathcal{L}|_ Y) \to H^1(X, \mathcal{I}'\mathcal{L}|_ Y) \]

is injective. Since $\mathcal{L}/\mathcal{I}\mathcal{L} \to \mathcal{L}|_ Y/\mathcal{I}'\mathcal{L}|_ Y$ is a surjective map of coherent modules whose supports have dimension $0$, we see that the first map $H^0(X, \mathcal{L}/\mathcal{I}\mathcal{L}) \to H^0(X, \mathcal{L}|_ Y/\mathcal{I}'\mathcal{L}|_ Y)$ is surjective (and hence bijective). But by induction we have that $\mathcal{L}|_ Y$ is globally generated (if $Y$ is disconnected this still works of course) and hence the boundary map

\[ H^0(X, \mathcal{L}|_ Y/\mathcal{I}'\mathcal{L}|_ Y) \to H^1(X, \mathcal{I}'\mathcal{L}|_ Y) \]

cannot be injective. This contradiction finishes the proof.
$\square$

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