Let $f:X \to Y$ be a proper surjective morphism of projective surfaces such that there exists a curve $C \subset X$ for which $f|\_{X\backslash C}$ is an isomorphism and $f(C)$ is a set of points. Suppose $X$ is a closed subscheme of $\mathbb{P}^n$ for some integer $n$ and $C$ contracts to a rational singularity i.e., $f(C)$ are rational singularities on $Y$. Does there exist a closed immersion of $Y$ into $\mathbb{P}^n$ for the same integer $n$?



Contractibility of curves and embedding into projective space
Let $X\subset \mathbb P^3$ be an arbitrary smooth projective surface of degree $d>2$ and assume that $X$ contains a line $\ell\simeq \mathbb P^1$ If $d=4$, assume in addition that $X$ is general among such surfaces. This implies that its Picard number is $2$. A simple adjunction computation shows that $\ell^2=2-d<0$ ($l^2$ computed on $X$). This implies that $\ell\subset X$ is contractible, so we have a morphism $f:X\to Y$ as requested. 


**Claim.** $Y$ cannot be embedded into $\mathbb P^3$. If $d>4$, then $Y$ cannot be embedded into any smooth $3$-fold. 


**Proof:**


* If $d=3$ (case *abx*): Since $X$ is rational, so is $Y$, so if it could be embedded to $\mathbb P^3$, its degree would have to be $1,2$ or $3$. Its easy to see using adjunction and that fact that $f$ is the blow up a smooth point that none of these is possible.
* If $d=4$ (case *Olivier*): The Kodaira dimension of $X$ is $0$, so if $Y$ were embeddable, it would have to have degree $4$. In this case adjunction does not help, because $0=K\_X=f^\*K\_Y$. However, if $H'\subset Y$ is the hyperplane section of $Y\subset \mathbb P^3$, then $f^\*H'\cdot\ell =0$ and since the Picard number of $X$ is $2$, we can compute that in this case $f^\*H'=2mH+m\ell$ for some $m\in \mathbb N\_+$. But then $H'^2=18m^2\neq 4$, so this leads to a contradiction and hence $Y$ cannot be embedded into $\mathbb P^3$.
* If $d>4$ (case *new*):This is actually the easiest. If $Y$ embeds into any smooth $3$-fold, then it is Gorenstein and $K\_Y$ is a Cartier divisor. Then we have $K\_X=f^\*K\_Y+ a\ell$ for some $a\in \mathbb Z$. Using adjunction for $\ell$ and our earlier computation tells us that $$2-d=\ell ^2=\frac {-2}{a+1}$$ which means that $$2=(d-2)\cdot(a+1),$$ that is, $d-2$ divides $2$ so $d>4$ is out of the question. This completes the proof.


