d-uple embedding When one restricts the $d$-uple embedding $\mathbb{P}^n \hookrightarrow \mathbb{P}^N$ to $\mathbb{P}^{n-1} \hookrightarrow \mathbb{P}^n$, does this yield the $d$-uple embedding $\mathbb{P}^{n-1} \hookrightarrow \mathbb{P}^N$?
The degree of $\mathbb{P}^n$ in $\mathbb{P}^N$ is $d^n$. Is the degree of $\mathbb{P}^{n-1}$ under the above embedding equal to $d^{n-1}$?
 A: The d-uple embedding can be described in a coordinate-free way as the map $\mathbb{P}(V^*)$ to $\mathbb{P}(Sym^d(V^*))$ given by sending a quotient ($V \rightarrow L$) (considered as a point of  $\mathbb{P}(V^*)$) to the quotient ($Sym^dV \rightarrow Sym^dL$) (considered as a point of  $\mathbb{P}(Sym^dV^*)$). If $W$ is any quotient of $V$, with say $\phi:V \rightarrow W \rightarrow 0$, then we have an inclusion $i:\mathbb{P}(W^*) \rightarrow \mathbb{P}(V^*)$ sending the quotient $W \rightarrow L$ to the composite quotient with $\phi$. Similarly, one has $i^d:\mathbb{P}(Sym^d(W^*))\rightarrow \mathbb{P}(Sym^d(V^*))$ by sending a quotient $P=(Sym^dW^*\rightarrow N)$ to the composite quotient $Q$ with $Sym(\phi^d)$ - i.e. $Q=(Sym^dV^*\rightarrow Sym^dW^* \rightarrow N)$.
It is clear from these definitions that the diagram of inclusions commutes:
$\mathbb{P}(V^*)\rightarrow\mathbb{P}(Sym^dV^*)$
$\uparrow$______________$\uparrow$
$\mathbb{P}(W^*)\rightarrow \mathbb{P}(Sym^dW^*)$
where the up arrows are $i$ and $i^d$ (sorry for the crappiness of my commutative diagram; I'm bad at Tex). This proves that the d-uple embedding restricts to the d-uple embedding in a very functorial way.
In coordinates, it's also easy. For example, the 3-uple embedding of $\mathbb{P}^2$ into $\mathbb{P}^8$ is given by $(x:y:z) \rightarrow (x^3:x^2y:x^2z:xy^2:xz^2:y^3:y^2z:yz^2:z^3)$. It sends the hyperplane $x=0$ to the set of all coordinates $(0:0:0:0:0:y^3:y^2z:yz^2:z^3)$ which, when you ignore the 0's, is exactly the 3-uple embedding of $\mathbb{P}^1$ with coordinates $(y:z)$ in $\mathbb{P}^3$
EDITED TO INCLUDE: The Hilbert polynomial of a closed subvariety $Y$ of $\mathbb{P}^N$ is given for $m>>0$ by $P(m)=dim_kH^0(Y,O_Y(1)^{\otimes m})$ where $O_Y(1)$ is defined to be the restriction of $O_{\mathbb{P}^N}(1)$ to $Y$. In the case of the d-uple embedding, where $Y=\mathbb{P}^n$, we have $O_Y(1)=O_{\mathbb{P}^n}(d)$. It follows that $P(m)$ is equal to the dimension of the space of monomials of degree $md$ in $n+1$ variables. This is $n+md\choose md$. It may be rewritten as $\frac{d^nm^n}{n!} +$ things of lower order in $m$.
Now considering the $Z=\mathbb{P}^{n-1}$ under the composite embedding first to $\mathbb{P}^n$ as a hyperplane, and next as a d-uple embedding into $\mathbb{P}^N$, we have $O_Z(1)=O_{\mathbb{P}^{n-1}}(d)$. The same argument shows that the Hilbert polynomial of $Z$ under this embedding can be written as $\frac{d^{n-1}m^{n-1}}{(n-1)!}$ + lower order.
Since the degree of a closed subscheme is defined as the factorial of the dimension times the lead coefficient of the Hilbert polynomial of the embedding, we see that indeed the composite embedding has degree $d^{n-1}$. 
