Proof.
By Lemma 53.19.16 we find that X is Gorenstein, i.e., \omega _ X is an invertible \mathcal{O}_ X-module.
If the genus of X is zero, then \deg (\omega _ X) < 0, hence if X has more than one irreducible component, we get a contradiction with Lemma 53.22.4. In the irreducible case we see that X is isomorphic to an irreducible plane conic and \omega _ X^{\otimes -1} is very ample by Lemma 53.10.3.
If the genus of X is 1, then \omega _ X has a global section and \deg (\omega _ X|_ C) = 0 for all irreducible components. Namely, \deg (\omega _ X|_ C) \geq 0 for all irreducible components C by Lemma 53.22.4, the sum of these numbers is 0 by Lemma 53.8.3, and we can apply Varieties, Lemma 33.44.6. Then \omega _ X \cong \mathcal{O}_ X by Varieties, Lemma 33.44.13.
Assume the genus g of X is greater than or equal to 2. If X is irreducible, then we are done by Lemma 53.21.3. Assume X reducible. By Lemma 53.22.4 the inequalities of Lemma 53.21.7 hold for every Y \subset X as in the statement, except for Y = X. Analyzing the proof of Lemma 53.21.7 we see that (in the reducible case) the only inequality used for Y = X are
\deg (\omega _ X^{\otimes m}) > -2 \chi (\mathcal{O}_ X) \quad \text{and}\quad \deg (\omega _ X^{\otimes m}) + \chi (\mathcal{O}_ X) > \dim _ k H^1(X, \mathcal{O}_ X)
Since these both hold under the assumption g \geq 2 and m \geq 2 we win.
\square
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