Let $Y$ be a closed subscheme of $X=\mathbb P{^n}$ with
Hilbert Polynomial given by $h(m) =\binom{m+r}{r}$
Prove that $Y$ is isomorphic to $\mathbb P^r$ embedded linearly inside $X$.
Can someone help me solve this? I know for $\mathbb P^r$ that is indeed the Hilbert Polynomial.
Edit : The context of this problem is a step towards the construction of Hilbert Schemes as done in FGA explained (chapter 2).
Looking at the Hilbert polynomial I can say that the closed subscheme has degree 1 and dimension r. Let $I$ be the ideal of $Y$, I would like to show that it is generated by a single linear polynomial(?). I tried to prove that it can't be generated by polynomials of degree greater than 1. To do this I was trying to use $h_X+h_Y= h_{X{\cup}Y}+h_{X{\cap}Y}$ but I wasn't able to show it.
Best Answer
If you knew that $Y\cong \Bbb P^r$, then you'd be much closer to a solution: a closed immersion $\Bbb P^r\to\Bbb P^n$ comes from a line bundle $\mathcal{L}$ on $\Bbb P^r$ with a choice of $n+1$ global sections which generate the line bundle at every point. Such a line bundle must be $\mathcal{O}(d)$ for some $d>0$, and I claim that the degree of the image is exactly $d$: pick a hyperplane $H$ in $\Bbb P^n$ not containing $Y$ so that $\deg H\cap Y=\deg Y$ by the generalized Bezout theorem, while the degree is also $(r-1)!$ times the leading coefficient of Hilbert polynomial of $H\cap Y$. The Hilbert polynomial of $H\cap Y$ is $\binom{m+r}{r}-\binom{m+d+r}{r}=\frac{d}{(r-1)!}m^{r-1}+\cdots$ because $H$ restricts to a hypersurface on $Y$ cut out by a polynomial of degree $d$, so we must have $\deg Y=d$. As $\deg Y=1$, this shows that $Y$ must be embedded by $\mathcal{O}(1)$, which is a linear embedding.
To show $Y\cong \Bbb P^r$, we'll first show that we have a finite map $Y\to \Bbb P^r$ and then calculate that the degree of this map is 1, showing it's an isomorphism. Let $\Bbb P^{n-r-1}\subset\Bbb P^n$ be a linear $(n-r-1)$-plane which is disjoint from $Y$, and let $\Bbb P^r\subset \Bbb P^n$ be a linear $r$-plane disjoint from $\Bbb P^{n-r-1}$. Let $\pi: \Bbb P^n\setminus \Bbb P^{n-r-1}\to\Bbb P^r$ denote the projection from $\Bbb P^{n-r-1}$ to $\Bbb P^r$. I claim that $\pi$ restricts to a finite map on $Y$. First, the fiber over $p\in\Bbb P^r$ is the intersection of $Y$ with the linear $(n-r)$-plane spanned by $p$ and $\Bbb P^{n-r-1}$: this must be a dimension-zero closed subscheme of $\Bbb P^{n-r}$ for dimension reasons, as $Y$ does not intersect $\Bbb P^{n-r-1}$ while any two closed subschemes of $\Bbb P^a$ with dimensions adding to $a$ must intersect. Thus the fiber is a finite set of points, so $\pi|_Y$ is quasi-finite. But $\pi|_Y$ is also projective, and quasi-finite + projective implies finite, so $\pi|_Y$ is actually finite.
This implies that $(\pi|_Y)_*\mathcal{O}_Y$ is a coherent sheaf of $\mathcal{O}_{\Bbb P^r}$-algebras and $Y\cong \mathop{\mathbf{Spec}} (\pi|_Y)_*\mathcal{O}_Y$. If we can show that $(\pi|_Y)_*\mathcal{O}_Y$ is of rank 1, then $(\pi|_Y)_*\mathcal{O}_Y$ is a line bundle which will imply that $Y\cong \Bbb P^r$. As the rank of $(\pi|_Y)_*\mathcal{O}_Y$ at a point $p\in\Bbb P^r$ is the degree of the intersection of $Y$ with the $(n-r)$-plane spanned by $\Bbb P^{n-r-1}$ and $p$, we need to do a little intersection theory to show that this is 1. The following argument will do the trick:
Claim. If $Y\subset \Bbb P^n$ is a closed subscheme with Hilbert polynomial $h(m)=\binom{m+r}{r}$ and $L$ is a linear subspace of dimension $n-r$ so that $Y$ and $L$ intersect properly, then $Y\cap L$ is a single point (ie a point with Hilbert polynomial $1$).
Proof. $L$ has an ideal which is generated by $r$ linear forms $h_1,\cdots,h_r$, and $Y\cap L=Y\cap V(h_1)\cap\cdots\cap V(h_r)$, so we can prove this by induction. $Y\cap V(h_1)$ has Hilbert polynomial $\binom{r+m}{r}-\binom{r+m-1}{r}=\binom{r-1+m}{r-1}$, and by induction we see that the Hilbert polynomial of $Y\cap V(h_1)\cap\cdots\cap V(h_i)$ is $\binom{r-i+m}{r-i}$. Therefore when $i=r$ this Hilbert polynomial is just $1$ and the claim is proven.