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I know that this is not a research question, but I am stuck from quite a while and I don't know what else to do.

Given a semistable curve $C$ over $k$, it is asserted that we may construct its stable model by taking $C'=\operatorname{Proj}(\oplus_{j\geq 1} H^0(\omega^{4j}))$. I just do not see how this works (I have read both The Projectivity of Moduli Space of Curves by Knudsen and The Geometry of Algebraic Curves by Arbarello-Cornalba-Griffiths, but they all gloss over the details I don't understand). For example: why is $C'$ even a nodal curve?

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  • $\begingroup$ "It is asserted", where? This is false. If you take any plane curve $C\subset \mathbb{P}^2$ of degree $d\geq 4$, then $\omega _C=\mathcal{O}_C(d-3)$ is very ample, so $C'=C$. $\endgroup$
    – abx
    Dec 24, 2016 at 10:18
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    $\begingroup$ @abx: this proves that every nodal plane curve of degree $\geq4$ is stable. No contradiction there, I think. $\endgroup$ Dec 24, 2016 at 10:29
  • $\begingroup$ Oops, my bad. I overlooked "semistable". $\endgroup$
    – abx
    Dec 24, 2016 at 10:40
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    $\begingroup$ You can prove this by induction on the number of unstable components of $C$, i.e., genus $0$ components $E$ with at most $2$ nodes. Let $f:C\to \widehat{C}$ be the contraction of $E$. If $E$ has $2$ nodes, then $\omega_C$ equals $f^*\omega_{\widehat{C}}$, so the canonical rings are equal. If $E$ has $1$ node, then $\omega_C$ equals $f^*\omega_{\widehat{C}}(\underline{E})$. Since $\omega_C$ has negative degree on $E$, every global section of $\omega_C^{\otimes n}$ comes from $f^*(\omega_{\widehat{C}}^{\otimes n})$. Again the canonical rings are equal. $\endgroup$ Dec 24, 2016 at 12:01

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I am just posting my comment above as a solution. For every integer $g\geq 2,$ for every field $k,$ for every proper, at-worst-nodal, geometrically connected $k$-curve $C$ of arithmetic genus $g,$ there exists a $k$-morphism $u:C\to C'$ with $C'$ a proper, at-worst-nodal, geometrically connected $k$-curve $C$ of arithmetic genus $g$ such that $\omega_{C'/k}$ is ample, such that there exists a dense open $C'_o\subset C'$ over which the restriction, $u:u^{-1}(C'_o)\to C'$, is an isomorphism, and such that for every integer $r\geq 0,$ the pullback map $$H^0(C',\omega_{C'/k}^{\otimes r}) \to H^0(C,\omega_{C/k}^{\otimes r})$$ is an isomorphism of $k$-vector spaces. It follows that there is an isomorphism of graded $k$-algebras, $$\bigoplus_{r\geq 0} H^0(C',\omega_{C'/k}^{\otimes r})\to \bigoplus_{r\geq 0} H^0(C,\omega_{C/k}^{\otimes r}).$$ Since $\omega_{C'/k}$ is ample, it follows that $C'$ is isomorphic to $\text{Proj}$ of this common graded $k$-algebra.

Because $C'$ is uniquely recovered as $\text{Proj}$ of the common algebra (if there exists $u$ as above), any such $C'$ will satisfy the cocycle condition for an étale descent. Thus, without loss of generality, assume that $k$ is separably closed. In particular, every irreducible component of $C$ is geometrically integral.

The claim is proved by induction on the number of unstable irreducible components $E$ of $C$. The result is tautological when $\omega_{C/k}$ is ample. Thus, assume that $\omega_{C/k}$ is not ample, i.e., for some irreducible compnent $C_i$ of $C$, $\omega_{C/k}|_{C_i}$ is not ample. For an irreducible component $C_i$ that intersects $\overline{C\setminus C_i}$ in nodes $\{p_1,\dots,p_\delta\}$, then $\omega_{C/k}|_{C_i}$ is isomorphic to $\omega_{C_i/k}(\underline{p_1}+\dots + \underline{p_\delta})$. This is ample except for an irreducible component $E$ that is a smooth curve of genus $0$ and $\delta$ equals $1$ or $2$. In that case, let $f:C\to \widehat{C}$ be the contraction of $E$. If $\delta$ equals $2$, then $f^*\omega_{\widehat{C}/k}\to \omega_{C/k}$ is an isomorphism. Thus, the pullback map $$H^0(\widehat{C},\omega_{\widehat{C}/k}^{\otimes r})\to H^0(C,\omega_{C/k}^{\otimes r})$$ is an isomorphism for every $r\geq 0$. Similarly, if $\delta$ equals $1$, then $\omega_{C/k}$ is isomorphic to $f^*\omega_{\widehat{C}/k}(\underline{E})$. This has degree $-1$ on $E$. Thus, for every $r\geq 0$, the pullback map $$H^0(\widehat{C},\omega_{\widehat{C}/k}^{\otimes r}) \to H^0(C,\omega_{C/k}^{\otimes r})$$ is an isomorphism. Thus the result holds by induction on the number of unstable components.

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