Hilbert polynomial of iterated Veronese embedding Let $X=\mathbb{V}(x^2-yz)\subset\mathbb{P}^2$ and consider the Veronese embedding $Y=\mathcal{v}_2(X)\subset\mathbb{P}^5$.
Find the Hilbert polynomial, and thus the degree, of $Y$.
I know how we can read the degree of a projective variety off from the Hilbert polynomial, but I'm struggling to calculate the polynomial in this case.
I'm pretty sure that we can parametrise $$Y=\big\{[\lambda^2\mu^2,\mu^4,\lambda^4,\lambda\mu^3,\lambda^3\mu,\lambda^2\mu^2]\mid\lambda,\mu\neq0\big\}\cup\big\{[0:1:0:0:0:0],[0:0:1:0:0:0]\big\}.$$
Further, $X=\mathcal{v}_2(\mathbb{P}^1)$ and so $Y=(\mathcal{v}_2\circ\mathcal{v}_2)(\mathbb{P}^1)$, but I've never come across iterated applications of the Veronese embedding before. It looks like $Y\cong\mathcal{v}_4(\mathbb{P}^1)$ by ignoring the first/last coordinate in my parametrisation.
But if I carry on naively, I argue as follows (trying to repeat an argument that seems to be commonly used for finding the degree of the Veronese embedding):

We can restrict a degree $d$ polynomial on $\mathbb{P}^5$ to $Y$ and write it in terms of $\lambda,\mu$ and it will be a degree $4d$ polynomial.
   Thus $$h_Y(d)=\binom{2+4d}{2}=\frac{(2+4d)(1+4d)}{2}=8d^2+6d+1.$$
   Then the leading term is $8d^2=(16/2!)d^2$, which tells us that $\dim Y=2$ and $\deg Y=16$.

This doesn't seem like a correct answer though. I know that the degree depends on the embedding, but this just seems like an unusually large number to get...
Edit: Using the answer given to this question I'm pretty sure that the degree of $Y$ is equal to the degree of $\mathcal{v}_4(\mathbb{P}^1)=4^1=4$, which does disagree with my above argument.
Edit 2: As another question, is it generally true that $(\mathcal{v}_d\circ\mathcal{v}_e)(X)\cong\mathcal{v}_{d\cdot e}(X)$? If, so, is this isomorphism always 'nice', in the sense that it induces an isomorphism of homogeneous coordinate rings?
 A: Note that the $d-$tule embedding is an isomorphic copy of $\mathbb P^n$ where the linear forms $O_X(1)$ corresponds to the $d$-forms of $\mathbb P^n$.($O_{\mathbb P^n}(d)$). (look up [Hartshorne II, Ex. 5.13]). This should help you to see your second question.
To answer your original question, I will just compute the Hilbert polynomial of $v_d: \mathbb P^n \to \mathbb P^{N}$. Let $v_d(\mathbb P^n) = X$. Let $S(X)$ be the homogeneous coordinate ring. Then it follows [H, II, Ex. 5.9] that for $m \gg 0$, $S(X)_m $ is isomorphic to the group of homogeneous $m$ forms on $X$ ($\mathcal O_X(m)$), but that just equals the $md$ forms on $\mathbb P^n$. 
Therefore, $h_X(m) = \dim_k (S(X))_m = \binom{md+n}{n}$ for $m \gg 0$. So the Hilbert polynomial is $\binom{zd+n}{n}$.
So the answer in your case would be $4z+1$.
A: This is not a complete answer, but it is too long to write a single comment.
Let $V$ be a vector space of dimension $n+1$. Let $S^d V$ denote the space of homogeneous polynomials of degree $d$ on $V^*$ (you can think $S^d V$ as polynomials whose variables are the elements of a basis of $V$).
Then we have
$$
\mathbb{P} V \to \mathbb{P} S^d V \to \mathbb{P} S^e(S^d V)
$$
where the first map is $v_d$ and the second map is $v_e$.
The Veronese reembedding $v_e(v_d(\mathbb{P} V))$ is degenerate in $\mathbb{P} S^e(S^d V)$ and its span is a linear space of the same dimension of $S^{ed} V$ (and in fact if we play with the representation theory of $GL(V)$, we can see that this subspace is the unique copy of $S^{ed}V$ in $S^e(S^d V)$ as $GL(V)$-representation).
In particular $v_e(v_d(\mathbb{P} V)) = v_{ed}(\mathbb{P}V) \subseteq S^{ed}V \subseteq S^e(S^d V)$.
The same argument applies to subvarieties $X \subseteq \mathbb{P} V$.
