Smoothen a nodal curve

This is correct that you can always "deform" a nodal curve into a smooth one. A good reference for this fact is Corollary $7.11$ in these notes by Talpo and Vistoli.

UPD: As Qixiao points out Talpo and Vistoli prove the result only under the assumption that all singularities are rational. I'll explain in the UPD section below how to deduce the general result from their one. The proof crucially uses the fact that any nodal singularity is defined over a finite separable extension of a ground field. This allows us to use Galois descent to reduce the situation to the case of rational singularities.

The proof boils down to two (rather independent parts): classification of Ordinary Double Points and Deformation Theory of (isolated) singularities. The first part crucially relies on Artin Approximation Theorem which is beautifully explained in Chapter 3.6 of the book "Neron Models" by Bosch, Lutkebohmert and Raynaud. Classification of ordinary double points (and much more) is done in Chapter 3.2 Freitag and Kiehl's book "Etale Cohomology and Weil conjectures". Let me summarize the theorem below:

Definition 1: Let $k$ be an arbitrary field (not necessary algebraically closed or even perfect), then we say that a pair $(U,x)$ of a finite type $k$-scheme $U$ and a $k$-rational point $x\in U(k)$ defines an rational ordinary double point, if there is an isomorphism of $\bar{k}$-algebras $\mathcal O_{U\otimes_k \bar{k}, \bar x}\cong \bar{k}[[u,v]]/(uv)$.

Definition 2: Let $k$ be an arbitrary field, then we say that a pair $(U,x)$ of a finite type $k$-scheme and a closed point $x\in U$ defines an ordinary double point, if there is a point $\bar x\in U(\bar k)$ which lies over $x$ and such that $(U\otimes_k \bar k, \bar x)$ defines a rational ordinary double point.

Theorem 1: (Classification of rational ordinary double points). Let $(U, x)$ be a rational ordinary double point over a field $k$, then there is a pair $(W,y)$ and two etale residually trivial maps $f:(W,y) \to (U,x)$ and $g:(W,y) \to (\operatorname{Spec} k[[u,v]]/(uv), 0)$.

Theorem 2: (Classification of ordinary double points). Let $(U, x)$ be an ordinary double point over a field $k$, then the residue field $k(x)$ is finite separable over $k$ and there is a pair $(W,y)$ and two etale maps $f:(W,y) \to (U,x)$ and $g:(W,y) \to (\operatorname{Spec} k[[u,v]]/(uv), 0)$.

As for deformation theory, the main thing that we need to understand is tangent-obstruction theory for the deformation problem of local complete intersection, generically smooth finite type $k$-schemes. This is done in Chapters $3-5$ of Vistoli's notes. The main result of these Chapters is the following theorem:

Theorem 3: Let $X_0$ be a local complete intersection, generically smooth scheme of finite type over a field $k$, then the tangent space to the deformation problem of $X_0$ is given by $\operatorname{Ext^1}(\Omega^1_{X_0/k}, \mathcal O_{X_0})$ and the obstruction theory to this problem is given by $\operatorname{Ext^2}(\Omega^1_{X_0/k}, \mathcal O_{X_0})$.

Now, we are ready to explain the main idea behind the proof given in Talpo-Vistoli's notes. First of all, we deal with a 'standard model' of an affine nodal curve with $1$ (isolated) singularity. Namely, we mean a pair $(\operatorname{Spec} k[x,y]/(xy),0)$. Talpo and Vistoli compute explicitly its miniversal deformation and show that the generic fibre is 'smooth' (doesn't quite make sense, better to speak in terms of completed local rings). Then they relate a miniversal deformation of a general affine nodal curve with $1$ isolated singularity to this very specific case of a standard model (here they crucially use the classification result above and general theory of miniversal deformations developed in their notes before). This is Proposition $7.5$ in their notes.

Now we need to 'glue' these local deformations to a formal deformation of $X_0$. In order to do that we use a low degree terms exact sequence associated to a local-to-global spectral sequence for Ext-groups: $$ 0 \to H^1(X_0, \underline{Hom}(\Omega^1_{X_0/k}, \mathcal O_{X_0})) \to \operatorname{Ext}^1(\Omega^1_{X_0/k}, \mathcal O_{X_0}) \to H^0(X_0, \underline{Ext^1}(\Omega^1_{X_0/k}, \mathcal O_{X_0})) \to H^2(X_0, \underline{Hom}(\Omega^1_{X_0/k}, \mathcal O_{X_0})). $$

We identify $\operatorname{Ext}^1(\Omega^1_{X_0/k}, \mathcal O_{X_0})$ with the tangent space to deformations of $X_0$, $H^0(X_0, \underline{Ext^1}(\Omega^1_{X_0/k}, \mathcal O_{X_0}))$ with local deformations (a covering of a curve by affines and deformation of each affine part s.t. these deformations are isomorphic on overlaps) and also we need to identify a map between these groups with a natural map that sends a deformarion to an associated local deformation. But since $X_0$ is a curve we know that the $H^2$ term is automatically zero. So, given a covering of our nodal projective curve $X_0$ by affines and a bunch of deformations of these affines charts that are isomorphic on intersections we can always glue them to a deformation of the whole curve $X_0$. Apply this to the situation where each affine chart $U_i$ has exactly one distinct singular point (then all intersections are smooth and affine, so any deformation of such schemes are trivial). So, we can indeed glue these local "smoothifications" to a global one. I should say explicitly that this means that we can glue to a formal deformation $\mathcal X \to \operatorname{Spf} k[[t]]$ (Note that $\mathcal X$ is only a formal scheme, not a usual scheme. So we must prove that it is algebraizable in order to speak about its generic fiber as a scheme).

The last part is to algebraize $\mathcal X$. In general it may be a difficult problem. But here we are in a one-dimensional situation, so it is easily seen that there are no obstruction to deform an ample line bundle on $X_0$ to an ample line bundle $\mathcal X$ and then by Grothendieck's Existence we can algebraize it. So, finally we have a flat projective curve $X\to \operatorname{Spec} k[[t]]$ s.t. the special fibre is isomorphic to our initial curve $X_0$ and it is easy to see (following the construction) that the generic fibre should be smooth (the easiest way is to control all completed local rings. But one should be careful since we want to check that the generic fibre is smooth and not just regular. For details see Corollary 7.11 in the notes).

UPD: Let me show how to get of rid of the assumption that all singularities are rational.

Let $X_0$ be a reduced projective curve over $k$ with only ordinary double points. We want to find a flat projective relative curve $X\to \operatorname{Spec} k[[t]]$ such that the generic fibre is smooth and the special fibre is isomorphic to $X_0$.

According to theorem 2, we can choose a finite separable extension $k'/k$ such that all singularities of $X_0\otimes_k k'$ are rational. If necessarily, extend $k'$ even further to assume that all irreducible components of $X_0\otimes_{k} k'$ of genus zero are actually isomorphic to $\mathbf P^1$ (any form of $\mathbf P^1$ splits after a finite separable extension). This allows us to choose a $\operatorname{Gal}(k'/k)$-invariant divisor $D'$ on $X_0':=X_0\otimes_k k'$ s.t. a pair $(X_0',D')$ is a semistable pointed curve (in a sense of Knudsen). Consider a deformation problem $Def_{k''[[t]]}(X_0',D')$.

Step 1: The functor $Def_{k''[[t]]}(X_0',D')$ is pro-representable by $k'[[t]][[T_0,\dots,T_n]]=:A'$ for some n.

In order to check that this functor is pro-representable it is sufficient to show that this deformation problem admits a tangent-obstruction theory and that the Automorphism scheme $\underline{\operatorname{Aut}}_{X_0',D'}$ is unramified (This is due to Schlessinger's criterion).

The existence of tangent-obstruction theory is standard and, for example, is explained in Chapter 5 here. As for the Automorphism-scheme, I don't know a good reference for this fact. But a proof is very close to the proof of the same fact in the case of not-pointed semi-stable curves (Theorem $1.11$ in a celebrated Deligne-Mumford paper).

The last thing to verify is that a ring that pro-represents $Def_{k''[[t]]}(X_0',D')$ is actually isomorphic to power series ring $k'[[t]][[T_0, \dots,T_n]]$. Again, we know that it is enough to show that this deformation problem is unobstructed. By Theorem $5.4$ the obstruction space is given by $$ \operatorname{Ext^2}(\Omega^1_{X_0'/k'}, \mathcal I_D)=\operatorname{Ext^2}(\Omega^1_{X_0'/k'}(D), \mathcal O_{X_0}). $$ We want to show that this space is zero. Using local-to-global spectral sequence we reduce the question to showing that $$ H^0(X_0,\underline{Ext^2}(\Omega^1_{X_0'/k'}(D), \mathcal O_{X_0})))=H^1(X_0,\underline{Ext^1}(\Omega^1_{X_0'/k'}(D), \mathcal O_{X_0})))=H^2(X_0,\underline{Hom}(\Omega^1_{X_0'/k'}(D), \mathcal O_{X_0})))=0. $$ OK, a local calculation (exercise) shows that $\underline{Ext^2}(\Omega^1_{X_0'/k'}(D), \mathcal O_{X_0})=0$. Since $X_0$ is a curve we also conclude that $H^2(X_0,\underline{Hom}(\Omega^1_{X_0'/k'}(D), \mathcal O_{X_0})))$ vanishes. Finally, note that $\underline{Ext^1}(\Omega^1_{X_0'/k'}(D), \mathcal O_{X_0}))$ is supported at a closed subscheme of singular points. So, this sheaf is actually a pushforward from some closed $0$-dimensional subscheme, this implies that $H^1(X_0,\underline{Ext^1}(\Omega^1_{X_0'/k'}(D), \mathcal O_{X_0})))=0$ as well. So, our problem is indeed unobstructed.

Step 2: Algebraization and descent.

Ok, so we've proven that $A'$ pro-represents our deformation functor. Thus there is a universal formal family $\mathcal X' \to \operatorname{Spf} A'$ together with $m$-sections $\mathcal X'(A')$ extending the divisor $D'$ (where $m=\deg D'$).

Firstly, note that we can algebraize $\mathcal X'$ to a flat projective scheme $X'$ over $\operatorname{Spec} A'$ by Grothendieck's existence (+deformation theory of vector bundles) since $X_0'$ is a curve (so no obstructions to deform line bundles). And likewise we can lift all sections of $\mathcal X'$ to section $\sigma_i$ of $X'$. Moreover, note that a locus in $\operatorname{Spec} A'$ over which $X'\to \operatorname{Spec} A'$ has only ordinary double points singularities is open, so $(X' \to \operatorname{Spec} A', \sigma_i)$ is a semi-stable pointed curve.

We claim that there is a descent of this scheme to a projective semistable curve $X\to \operatorname{Spec} k[[t]][[T_0,...,T_n]]$. Let us denote by $A$ the ring $k[[t]][[T_0,...,T_n]]$.

In order to descent $X'$ we need to find an $A'$-ample line bundle $L'$ on $X'$ such that the pair $(X',L')$ has a descent datum w.r.t a morphism $A\to A'$. The desired sheaf is the dualizing sheaf $\omega_{X'/A'}(\sum \sigma_i)$ twisted by sections $\sigma_i$. Two questions naturally arise: Why is it $A'$-ample and why does a pair $(X', \omega_{X'/A'}(\sum \sigma_i))$ has a descent datum?

As always, the first question is answered in Knudsen's paper, see Corollary 1.10.

As for the second question, let us understand explicitly what it means to have a descent datum. Observe that $(A')^{\operatorname{Gal}(k'/k)}=A$ and $A\otimes_k k'\cong A'$. So $A'\otimes_A A'\cong A\times \operatorname{Gal}(k'/k)$. And a descent datum is "just" an action of the Galois group $\operatorname{Gal}(k'/k)$ on $X'$ s.t. for any $s\in \operatorname{Gal}(k'/k)$ we have an isomorphism $s^*((X', \omega_{X'/A'}(\sum \sigma_i)))\cong (X', \omega_{X'/A'}(\sum \sigma_i))$ + cocycle condition.

Note that $X'$ has a natural action of $\operatorname{Gal}(k'/k)$ because the Galois group $\operatorname{Gal}(k'/k)$ acts on a functor $\operatorname{Def}_{k'[[t]]}(X_0', D')$ (because $X_0'$ and $D'$ are both defined over $k$). So it acts on an object that pro-represents this functor as well as on its algebraization.

To see that $\omega_{X'/A'}(\sum \sigma_i))$ is preserved by this action just note that the divisor $D'$ was chosen to be $\operatorname{Gal}(k'/k)$-invariant and the formation of dualizing sheaf commutes with etale base change.

Step 3: Use the known case of rational singularities to finish the proof.

Great, so now we have a flat projective curve $X \to \operatorname{Spec} A$ s.t. the special fibre is isomorphic to $X_0$. Moreover, from the known case of rational singularities we know that $X' \to \operatorname{Spec} A'$ is smooth over a generic fibre. How do we know this?

Just choose an arbitrary deformation $Y' \to \operatorname{Spec} k'[[t]]$ s.t. the generic fibre is smooth (it exists because $X_0 '$ has only rational ordinary double points). This defines a formal deformation $\mathcal Y' \to \operatorname{Spf} k'[[t]]$. By the universal property it comes as a pullback of $\mathcal X' \to \operatorname{Spf}A'$ through some morphism $\operatorname{Spf} k'[[t]] \to \operatorname{Spf} A'$. This corresponds to a morphism $\operatorname{Spec} k'[[t]] \to \operatorname{Spec} A'$ and since algebraization commutes with base change we see that some fiber of $X' \to \operatorname{Spec} A'$ must be smooth (as $Y' \to \operatorname{Spec} k'[[t]]$ is smooth over its generic fibre). Smoothness is an open condition, therefore we see that the generic fibre of $X' \to \operatorname{Spec} A'$ must be smooth as well.

Since smoothness descends along etale morphisms we conclude that $X \to \operatorname{Spec} A$ is smooth over the generic point. Recall again that locus in $\operatorname{Spec} A$ over which $X$ is smooth is open. So the complement is given by some non-trivial ideal $I$ in $k[[t]][[T_0,\dots,T_n]]=k[[t, T_0,\dots,T_n]]$ and we want to check that some pullback of $X$ is smooth over a generic fibre for some morphism $k[[t]] \to k[[t,T_0,\dots,T_n]]$.

By what being said we reduced the question to purely algebraic statement: Given a non-trivial ideal $I$ in $k[[t,T_0,\dots,T_n]]$ there always exists a local $k$-morphism $\phi:k[[t, T_0,\dots,T_n]] \to k[[t]]$ such that $\phi(I)\neq 0$ (This means that the associated morphism $\operatorname{Spec} k[[t]] \to \operatorname{Spec} A$ sends a closed to a closed point and a generic point into the smooth locus $\operatorname{Spec} A -V(I)$. That's exactly what we are looking for).

Let us prove this fact by induction. First of all, choose any non-zero element $f\in I$ and forget about $I$. We try to construct a $k$-homomorphism $\phi:k[[t, T_0,\dots,T_n]] \to k[[t]]$ s.t. $\phi(f)\neq 0$. For simplicity of notations reformulate the question for an abstract power-series ring $k[[x_1,\dots, x_n]]$ and we construct a $k$-morphism to $k[[t]]$.

Observe that in order to construct $\phi$ it suffices to construct $\psi:k[[x_1, \dots, x_n]] \to k[[y_1, \dots, y_{n-1}]]$ s.t. $\psi(f)\neq 0$ (assuming that $n\geq 2$). So, we will actually construct that morphism instead of $\phi$ (and then we will get the desired $\phi$ as a composition of $\psi$'s).

OK, being said this let us do some real work. Note that any element $y\in k[[x_1, \dots, x_n]]$ with a non-trivial linear form and zero constant term can be extended to a regular system of parameters $(y=y_0, y_1, \dots, y_{n-1})$. This implies that $k[[x_1,\dots, x_n]]=k[[y_0, y_1,\dots, y_{n-1}]]$. Hence, if we find an element $y$ with a non-trivial linear form, zero constant term and such that $f\notin (y)$, we can consider a natural projection $$ \psi:k[[x_1,\dots, x_n]]=k[[y_0, y_1,\dots, y_{n-1}]] \to k[[y_1,\dots, y_{n-1}]] $$ and we are done since $\psi(f)\neq 0$.

So, the only thing we are left to show is that there always exists $y$ such that $y$ has a non-trivial linear form and $f$ doesn't lie in an ideal generated by $y$.

Suppose the contrary, this means that a power series $f$ is divisible by any power series with non-trivial linear form. This is impossible because there infinitely many distinct irreducible power series with non-trivial linear form over any field (for example, $x_1+x_2^m$ for all $m$. Here is the place where we use that $n\geq 2$!). But $k[[x_1,\dots, x_n]]$ is UFD, so it is impossible that an element is divisible by infinitely many distinct (up to units) irreducible polynomials. Hence, the desired $y$ always exists.

Yes. One reference I know is Brian Conrad's appendix to Matt Baker's paper, "Specialization of linear systems from curves to graphs". See, in particular, Theorem B.2 therein.

In fact, there are several problems concerning your question and Cheng's answer.

First, Brian Conrad's theorem "only" proves the existence of a proper flat regular family with dual graph of the reduction isomorphic to the dual graph of your curve. This is sufficient if all the irreducible components of your curve are genus 0. But it does not consider the issue of lifting the irreducible components of genus $>0$ (which I am not sure you can do).

Second, there is a small problem when your field $k$ is finite, since Brian Conrad's theorem is written only for infinite fields. Although one could adapt the proof to the finite field case when the valence of your graph is not too high (i.e. it is at most $|k|+1$), you can find a proof of the result by using Mumford curves (with a small restriction on the valence of one node) in that paper, corollary 7.8 (sorry for the autocite).

What it is proved is that, if you have a "totally degenerate semistable curve" $ C$ over a field $k$ (in the sense that all irreducible components are isomorphic to the projective line and all singularities are $k$-rational), all irreducible components have al least two singular points, and (this is the minor restricting condition) there is at least one component with some $k$-rational point non-singular, then there exists a Mumford curve $X$ over $k((t))$ such that the special fiber of the semistable model over $k[[t]]$ is isomorphic to $C$.