d_fold Veronese Embedding

Question

I was doing some exercise in the first chapter of Ulrich and I am stuck on this one:

Let $n,d>0$ be integers and $N=\binom{n+d}n-1$. Let $M_0,...M_N \in k[X_0,...X_n]$ be all monomials of degree d in $X_0,...,X_n$. Consider the morphism

$\nu_d:\mathbb P^n \to \mathbb P^N$, $(x_0,...,x_n) \mapsto ((M_0)(x_0,...,x_n):...:(M_n)(x_0,...,x_n))$.

$(a)$ Define a $k$-algebra homomorphism $\theta : k[Y_0,...,Y_n] \to k[X_0,...,X_n]$ by $Y_i \mapsto M_i$ and let $I=ker \theta$. Show that I is a homogeneous prime ideal.

Let $V_+=\{x\in \mathbb P^n \mid F(x)=0 \forall F \in \overline I\}$

$(b)$ show that $\nu_d$ induces an isomorphism $\mathbb P^n \cong V_+(I)$ of prevarietes. Is $V_+(I)$ a linear subspace of $\mathbb P^N$?

$(c)$ Let $f \in k[X_0,...,X_n]$ be homogeneous of degree $d$. Show that $\nu_d(V_+(f))$ is the intersection of $V_+(I)$ and a linear subspace of $\mathbb P^N$.

I have already solved part $(a)$ but I am stuck in $(b)$ (how to define th morphism of prevarieties) and $(c)$ (I know $\nu_d (\mathbb P^n)= ker\theta$).

Any comment and guidance will be highly appreciated.

Answer 1

Disclaimer: I'm not adding anything substantially different from the other answer, just fleshing out the details in (b). I agree that you should try to get (c) by yourself. Just write down what the relevant objects are: $\nu_d(V_+(f))$ and the intersection of the embedding and a hyperplane.

Define the standard open cover $\{U_i\}_{i=0,\dots,m}$ of $P^m$ as $U_i=\{[x_0,\dots,x_m]:x_i\neq 0\}$. As an exercise show that $U_i\cong A^m$, the $m$-dimensional affine space.

(b) One can define the morphism locally on the source, this is, take an open cover of the domain and define a map from each of the open sets in the cover into the target. Then, you have a well defined morphism if the local maps you defined agree in the overlaps of the open sets. We say that the maps glue when they agree in the overlaps.

You already defined $\nu_d:P^n\to P^N$ without having to do any sort of gluing. However, as explained in the other answer, to define an inverse we have to do some gluing (perhaps someone else has another, more elegant, answer?).

Choose an ordering of the degree $d$ monomials in the $n+1$ variables and let's label the coordinates of $P^N$ using this convention. For example, let the first coordinate be $X_{(d,0,\dots,0)}$ and $X_{(d-1,1,0\dots,0)}$ the second and so on... Similarly, label the standard open sets of $P^N$ using this, so that the standard open sets of $P^N$ are $U_{(d,0,\dots,0)}$ and so on.

1) We only need $n+1$ standard open sets to cover $V_+(I)$. Indeed, the open sets corresponding to the vectors $(d,0,\dots,0),\, (0,d,0,\dots,0),\dots$ cover it. Denote these open sets by $W_0,\dots,W_n\subset P^N$, e.g., $W_0=U_{(d,0,\dots,0)}$.

2) Define the inverse morphisms locally: for example, let $f_0:W_0\to P^n$ be defined via $$ [X_{(i_0,\dots,i_n)}]\mapsto \left[ \frac{X_{(d,0,\dots,0)}}{X_{(d,0,\dots,0)}}, \frac{X_{(d-1,1,\dots,0)}}{X_{(d,0,\dots,0)}}, \frac{X_{(d-1,0,1,\dots,0)}}{X_{(d,0,\dots,0)}}, \dots \right]. $$ Similarly, $f_1:W_1=U_{(0,d,\dots,0)}\to P^n$ is defined as $$ [X_{(i_0,\dots,i_n)}]\mapsto \left[ \frac{X_{(1,d-1,0\dots,0)}}{X_{(0,d,\dots,0)}}, \frac{X_{(0,d,\dots,0)}}{X_{(0,d,\dots,0)}}, \frac{X_{(0,d-1,1,\dots,0)}}{X_{(0,d,\dots,0)}}, \dots \right]. $$ (This is the same as in the other answer.)

3) Check that all these $f_i$ for $i=0,\dots,n$ glue! This means that, for example, you need to check that $f_0|=f_1|$ in $W_0\cap W_1$.

This part will require you to have some understanding of the generators of the ideal $I$. For example, let $[X_{(i_0,\dots,i_n)}]\in W_0\cap W_1$. Check that $$ \left( 1, \frac{X_{(d-1,1,\dots,0)}}{X_{(d,0,\dots,0)}}, \frac{X_{(d-1,0,1,\dots,0)}}{X_{(d,0,\dots,0)}}, \dots \right) = \frac{X_{(d-1,1,\dots,0)}}{X_{(d,0,\dots,0)}}\cdot \left( \frac{X_{(1,d-1,0\dots,0)}}{X_{(0,d,\dots,0)}}, 1, \frac{X_{(0,d-1,1,\dots,0)}}{X_{(0,d,\dots,0)}}, \dots \right). $$ All you need to notice is that this equality holds if the point in question lies in $V_+(I)$. The horrible notation we've been using is your friend here.

Therefore all the $f_i$ glue to form a morphism $f:V_+(I)\to P^n$. I'll let you verify that this is the inverse morphism of $\nu_d$.

The second part of (b) has already been answered in this website: Image of the Veronese Embedding

Answer 2

First of all \begin{equation*} N=\binom{n+d}{d}-1. \end{equation*} a) You prove that $\nu_d$ is a continuous injective map (use the canonical affine open covering of $\mathbb{P}^N$); so its image is an irreducible closed subset of $\mathbb{P}^N$, equivalentely $\ker\nu_d^{*}=I$ is a homogeneous prime ideal of $\mathbb{K}[Y_0,\dots,Y_N]$ (warning to dimension of codomain).

b) From (a), $\nu_d$ is a bijection with $\mathbb{P}^n$ and $V_+(I)$; using again the canonical affine open covering of $\mathbb{P}^{N-1}$, you construct the explicit inverse map, and you see that it is a regular map as well.

Hint: On $U_0$ you can define the local inverse morphism $\left(1:\frac{z_1z_0^{d-1}}{z_0^d}:\dots:\frac{z_n^d}{z_0^d}\right)\mapsto\left(1:\frac{z_1}{z_0}:\dots:\frac{z_n}{z_0}\right)$.

c) Let $f$ be a homogeneous of degree $d$, then $f$ is the image via $\nu_d^{*}$ of a degree $1$ polynomial $g$ in $\mathbb{K}[Y_0,\dots,Y_N]$. It's easy to prove that \begin{equation*} \nu_d(V_+(f))=V_+(I)\cap V_+(g) \end{equation*} where the last set is a hyperplane in $\mathbb{P}^N$.

Update.

1. By canonical open affine covering of $\mathbb{P}^m$ I mean the covering given by following open affine sets: \begin{equation*} \forall i\in\{0,\dots,m\},\,U_i=\{[z^0:\dots:z^m]\in\mathbb{P}^m\mid z^i\neq0\}. \end{equation*}
2. $\nu_d^{*}$ is the morphism of rings associated to $\nu_d$.

Answer 3

Now I can understand the answer, but for this let me define $\nu_2^{-1}$, where $\nu_2$ is $2$-fold Veronese embedding given by:

$\nu_2: \mathbb P^2 \to \mathbb P^5$ $\ \ \ \ $,$\ $ $\nu_2([x_0:x_1;x_2])=[x_0^2:x_0x_1:x_0x_2:x_1^2:x_1x_2:x_2^2]$

Now I must define inverse map $\nu_2$ on $U_0,U_1,U_2$ (am I correct?).

$\nu_2^{-1}\mid_{\nu2(\mathbb P^2)}: U_i \to \mathbb P^2$ $\ \ \ $, $i=0,1,2$

$\nu_2^{-1} \mid_{U_0} [x_0:x_1:x_2:x_3:x_4:x_5]=[{x_0\over x_0}: {x_1\over x_0}:{x_2\over x_0}], x_0 \neq0 $

$\nu_2^{-1} \mid_{U_1} [x_0:x_1:x_2:x_3:x_4:x_5]=[{x_0\over x_1}: {x_1\over x_1}:{x_2\over x_1}], x_1 \neq0 $

$\nu_2^{-1} \mid_{U_2} [x_0:x_1:x_2:x_3:x_4:x_5]=[{x_0\over x_2}: {x_1\over x_2}:{x_2\over x_2}], x_2\neq0 $

Are these inverse maps correct? and we should have $\nu_2\circ (\nu_2^{-1} \mid_{U_i}) = id_{v_2(\mathbb P^2)}$ and $\ \ \ \ $ $(\nu_2^{-1} \circ\nu_2 (\mathbb P^2))\mid_{U_i} = id_{v_2^{-1}(\mathbb P^2) \mid_{U_i}}$. Please let me know if I did a mistake.