On the stable model 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?
 A: 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.
