# Does every space curve lie on a rational surface?

Let $C\subset \mathbb P^3$ be a smooth projective curve over $\mathbb C$. Is there a (singular) rational surface $S$ such that $C\subset S\subset \mathbb P^3$?

I'm also interested in the following higher-dimensional analogue:

Let $X\subset \mathbb P^n$ be a smooth projective variety. Is there a rational variety $Y$ of dimension $\dim X+1$ such that $X\subset Y\subset\mathbb P^n$?

Edit. There is a much simpler argument than the original answer below, and it also answers the second question. I will add that argument in a moment. However, I will leave the "rational simple connectedness" argument as well. The point is that the "rational simple connectedness" argument holds for any embedding of a curve $C$ in a rationally simply connected variety $P$, e.g., embedded in a general hypersurface of degree $m$ in $\mathbb{P}^n$ so long as $m^2\leq n$.
New answer to both questions. The following technique is usually called "retract rationality". The argument can be summarized as "rational varieties, such as the ambient $\mathbb{P}^n$, are retract rational". In fact, if you do not mind that $Y$ is very singular at the generic point of $X$, you can easily adapt the following argument to $X$ a subvariety of a unirational variety (of codimension $>1$). This actually leads to one of the proposed strategies for the open conjecture that there exist rationally connected varieties that are not unirational: find a rationally connected variety $P$ such that for some $X$ of codimension $>1$, there exists no rational $Y$ containing $X$. For $X$ a subvariety of $\mathbb{P}^n$, or any rational variety $P$, the following argument produces a rational subvariety $Y$ that contains $X$ and that is smooth at the generic point of $X$.
Denote by $\ell$ the dimension of $X$. If $\ell+1$ equals $n$, the result is trivially true by choosing $Y=\mathbb{P}^n$. Thus, assume that $n> \ell+1$. By Bertini, for a general linear projection $$\rho:\mathbb{P}^n \dashrightarrow \mathbb{P}^{\ell + 1},$$ the regular locus of $\rho$ intersects $X$ in a dense open subset of $X$, and the induced rational transformation $$\rho_X:X\dashrightarrow \mathbb{P}^{\ell + 1},$$ is birational to its image. Denote by $U\subset \mathbb{P}^{\ell+1}$ a dense, affine open subset such that $\rho_X^{-1}(U)\subset X$ is a dense open subset and such that $$\rho_{X,U}:\rho_X^{-1}(U)\to U,$$ is a closed immersion. Up to shrinking $U$ further, assume that the invertible sheaf $\mathcal{O}_{\mathbb{P}^n}(1)|_X$ is trivialized on $\rho_X^{-1}(U)$. Since $\rho_{X,U}$ is a closed immersion, the pullback $k$-algebra homomorphism is surjective, $$\rho_{X,U}^\#:\mathcal{O}_{\mathbb{P}^{\ell+1}}(U) \twoheadrightarrow \mathcal{O}_X(\rho_X^{-1}(U)).$$ Thus, the restrictions to $\rho_X^{-1}(U)$ of the $n+1$ defining global sections of $\mathcal{O}_{\mathbb{P}^n}(1)$ lift to $n+1$ global sections of $\mathcal{O}_{\mathbb{P}^{\ell+1}}$ on $U$. By a Bertini-type argument, for a general choice of these $n+1$ lifts, after replacing $U$ by a dense open that still intersects $X$, the $n+1$ global sections have empty base locus, and the associated rational transformation $$j:U\to \mathbb{P}^n,$$ is birational to its image, say $Y=\overline{j(U)}\subset \mathbb{P}^n$. Thus $Y$ is a rational subvariety of $\mathbb{P}^n$ of dimension $\ell+1 = \text{dim}(X)+1$. Since the $n+1$ global sections are chosen to extend the tautological sections on $\rho_X^{-1}(U)$, the composition $j\circ \rho$ is a well-defined rational transformation that agrees with the inclusion $i:X\to \mathbb{P}^n$. Thus, $X$ is contained in an $(\ell+1)$-dimensional rational subvariety $Y$ of $\mathbb{P}^n$.
Original answer to first question. The answer to the first question is positive by "rational simple connectedness". Let $B$ be a copy of $\mathbb{P}^1$. Let $f:C\to B$ be a finite morphism that is étale over a dense open subset $B^o\subset B$. Let $F$ be a zero-dimensional, reduced $k$-scheme of length $d$ whose connected components are an ordered $d$-tuple of $k$-rational points. Denote by $$I^o = \text{Isom}_{B^o}(B^o \times F,C^o)\to B^o$$ the relative Isom scheme. For $e\geq 1$, denote by $\overline{\mathcal{M}}_{0,d}(\mathbb{P}^3,e)$ the moduli stack of $d$-pointed, genus $0$ stable maps to $\mathbb{P}^3$ of degree $e$. Denote the evaluation $1$-morphism by $$\text{ev}:\overline{\mathcal{M}}_{0,d}(\mathbb{P}^3,e)\to (\mathbb{P}^3)^d.$$ There is a natural action of the symmetric group $\mathfrak{S}_d$ on both source and target, and the $1$-morphism is equivariant for these actions. There is also a natural action of $\mathfrak{S}_d$ on $I^o$, and this action makes $I^o$ into a $\mathfrak{S}_d$-torsor over $B^o$. The inclusion $i:C\hookrightarrow \mathbb{P}^3$ determines a morphism $$i_F: \text{Isom}_{B^o}(B^o\times F,C^o) \to (\mathbb{P}^3)^d.$$ This morphism is also $\mathfrak{S}_d$-equivariant. Form the $2$-fibered product, $$\overline{\mathcal{M}}_{0,d}(\mathbb{P}^3,e)_i = \overline{\mathcal{M}}_{0,d}(\mathbb{P}^3,e)\times_{\text{ev}, i_F} I^o,$$ together with its projection morphism, $$\text{pr}_{I^o}:\overline{\mathcal{M}}_{0,d}(\mathbb{P}^3,e)_i \to I^o.$$ This morphism is $\mathfrak{S}_d$-equivariant. Thus, forming the quotients of domain and target gives a $1$-morphism, $$\text{pr}_{B^o}:\overline{\mathcal{M}}_{0,f^o}(\mathbb{P}^3,e) \to B^o.$$ The geometric generic fiber of this morphism equals the geometric generic fiber of $\text{pr}_{I^o}$, namely, the stack parameterizing $d$-pointed, degree $e$ stable maps to $\mathbb{P}^3$ whose $d$ marked points map to the $d$ points of the geometric generic fiber of $f^o$.
By "rational simple connectedness" (or more elementary arguments), for $e\gg 0$, the geometric generic fiber of $\text{pr}_{B^o}$ is rationally connected. By the Rationally Connected Fibration Theorem, together with the Weak Approximation Theorem of Hassett-Tschinkel, there exists a rational section of $\text{pr}_{B^o}$, $$\sigma^o:B^o\to \overline{\mathcal{M}}_{0,f^o}(\mathbb{P}^3,e),$$ mapping the geometric generic point of $B^o$ to a point parameterizing a stable map with smooth domain mapping isomorphically to its image in $\mathbb{P}^3$. In particular, up to shrinking $B^o$, assume that $\sigma^o$ is a regular $1$-morphism with image in the non-stacky locus. The pullback via $\sigma^o$ of the universal stable map is a quasi-projective surface $S^o$ together with a proper, smooth morphism $\pi^o:S^o\to B^o$ with connected, genus $0$ geometric fibers, a morphism $u^o:S^o\to \mathbb{P}^3$, and a factorization $j^o:C^o\to S^o$ of $f^o$ such that $u^o\circ j^o$ equals $i^o$.
Altogether, $S^o$ is a quasi-projective rational surface with a morphism $u^o$ to $\mathbb{P}^3$ such that $i^o:C^o\to \mathbb{P}^3$ factors through $u^o$.