Is the stack of stable curves with no rational component algebraic? Let $g\geq 2$ be an integer and let $\overline{\mathcal{M}}_g$ be the (smooth proper Deligne-Mumford) algebraic  stack of stable curves of genus $g$.
Let $\mathcal{M}_g^{nr}$ be the substack of stable curves $C\to S$ such that, for every geometric point $\overline{s}$ of $S$, the fibres $C_{\overline{s}}$ does not admit a morphism from $\mathbb{P}^1_{\overline{s}}$. (In other words, none of the irreducible components of $C_{\overline{s}}$ are rational.)


Is $\mathcal{M}_g^{nr}$ an algebraic substack of $\overline{\mathcal{M}}_g$?


I feel like $\mathcal{M}_g^{nr}$ is   the complement of a some boundary divisor, but I'm not sure how to make this precise.
 A: I am adding a few details to my comment above.  For notational convenience, in addition to the usual notation $\overline{\mathcal{M}}_{g,n}$ for genus-$g$, $n$-pointed curves, also for every finite set $N$ with $n$ elements, use the notation $\overline{\mathcal{M}}_{g,N}$ when the $n$ marked sections are explicitly indexed by $N$.  Thus, there are boundary morphisms, $$\Delta_{(g',N'),(g'',N'')}:\overline{\mathcal{M}}_{g',N'\sqcup\{\ell'\}}\times \overline{\mathcal{M}}_{g'',N''\sqcup\{\ell''\}} \to \overline{\mathcal{M}}_{g'+g'',N'\sqcup N''},$$ $$\Delta_{g,N}:\overline{\mathcal{M}}_{g,N\sqcup\{\ell',\ell''\}}\to \overline{\mathcal{M}}_{g+1,N},$$ defined in the usual way.
Definition.  An assignment to every integer $g$ and finite set $N$ satisfying $2g-2+|N|>0$ of a reduced substack $Z_{g,N}\subset \overline{\mathcal{M}}_{g,N}$ is $\Delta$-compatible if it is compatible with permutations of $N$, and for every boundary morphism $\Delta_{(g',N'),(g'',N'')}$, resp. $\Delta_{g,N}$, the inverse image of $Z_{g'+g'',N'\sqcup N''}$, resp. of $Z_{g+1,N}$, equals the union of the pullbacks of $Z_{g',N'\sqcup\{\ell'\}}$ and $Z_{g'',N''\sqcup\{\ell''\}}$, resp. the pullback of $Z_{g,N\sqcup\{\ell',\ell''\}}$.  It is combinatorial if for every $(g,N)$ such that $Z_{g,N}$ does not equal $\overline{\mathcal{M}}_{g,N}$, then $Z_{g,N}$ is contained in the boundary divisor of $\overline{\mathcal{M}}_{g,N}$.
Proposition. Every combinatorial, $\Delta$-compatible assignment $(Z_{g,N})_{g,N}$ is a system of closed substacks $Z_{g,N}$ of $\overline{\mathcal{M}}_{g,N}$.
Proof.  By hypothesis, for every $(g,N)$ such that the boundary of $\overline{\mathcal{M}}_{g,N}$ is empty, then $Z_{g,N}$ is either empty or else equal to all of $\overline{\mathcal{M}}_{g,N}$, both of which are closed substacks of $\overline{\mathcal{M}}_{g,N}$.  Thus, by way of induction (on the integer $2g-2+|N|$, for example), assume that the boundary of $\overline{\mathcal{M}}_{g,N}$ is nonempty.  
If $Z_{g,N}$ equals all of $\overline{\mathcal{M}}_{g,N}$, then it is a closed substack of $\overline{\mathcal{M}}_{g,N}$.  Thus, assume that does not equal all of $\overline{\mathcal{M}}_{g,N}$.  Since the system is combiantorial, the substack $Z_{g,N}$ is contained in the boundary.  Since the boundary is a union of finitely many irreducible closed substacks of $\overline{\mathcal{M}}_{g,N}$, it suffices to check that the intersection with each of these irreducible closed substacks is closed.  
Each of these irreducible closed substacks is the image of a boundary morphism.  Since the system is $\Delta$-compatible, the inverse image of $Z_{g,N}$ under the boundary morphism is obtained from $Z_{h,P}$ with $2h-2+|P| < 2g-2+|N|$.  Thus, by the induction hypothesis, the inverse image of $Z_{g,N}$ under each boundary morphism is a closed substack of the domain of the boundary morphism.  Since the boundary morphism is finite, hence proper, the image of this closed substack of the domain is also a closed substack of $\overline{\mathcal{M}}_{g,N}$.  Thus, $Z_{g,N}$ is a closed substack of $\overline{\mathcal{M}}_{g,N}$.  By induction on $2g-2+|N|$, every $Z_{g,N}$ is a closed substack of $\overline{\mathcal{M}}_{g,N}$. QED
Notation.  For every nonnegative integer $h$, denote by $Z^{h}_{g,N} \subset \overline{\mathcal{M}}_{g,N}$ the reduced substack parameterizing those genus-$g$, $N$-marked curves such that there exists a subcurve that is the image of a nonconstant morphism from a proper, connected, reduced, at-worst-nodeal curve of arithmetic genus $\leq h$.
Corollary. The system $(Z^{h}_{g,N})$ is $\Delta$-compatible and combinatorial.  Thus, it is a system of closed substacks.
Proof.  For a curve that is a cofiber coproduct of connected subcurves, the coproduct curve contains an irreducible component of geometric genus $\leq h$ if and only if one of the connected subcurves contains an irreducible component of geometric genus $\leq h$.  Similarly, for a curve with a non-disconnecting node, the normalizations of the irreducible components all factor through the partial normalization of the non-disconnecting node.  Thus, the nodal curve has an irreducible component of geometric genus $\leq h$ if and only if the partial normalization has an irreducible component of geometric genus $\leq h$.  Thus, the system is $\Delta$-compatible.
If $g\leq h$, then $Z^h_{g,N}$ equals all of $\overline{\mathcal{M}}_{g,N}$.  If $h < g$, then every curve parameterized by the complement of the boundary is irreducible of genus $g>h$.  Thus, there is no irreducible component of this curve that has geometric genus $\leq h$.  So, in this case, $Z^h_{g,N}$ is contained in the boundary divisor.  In all cases, the substack $Z^h_{g,N}$ is combinatorial. QED
Finally, when $h$ equals $0$, the substack $Z^0_{g,n}$ is, by definition, the complement of the substack $\mathcal{M}^{\text{nr}}_{g,n}$.  As the complement of a closed substack, the substack $\mathcal{M}^{\text{nr}}_{g,n}$ is an open substack.  For every integer $g>2$, the closed substack $Z^0_{g,0}$ of $\overline{\mathcal{M}}_{g,0}$ contains no irreducible component of the boundary divisor of $\overline{\mathcal{M}}_{g,0}$.
