(Co)tangent Map of the d^th Veronese Embedding This question is regarding Vakil 19.1 D, self-study.
What does the (co)tangent map of the $d^{th}$ Veronese embedding over an algebraically closed field $k$
$$\mathbb P^n_k \to \mathbb P^N_k$$
look like on closed points? Specifically, I would like to see the tangent map is injective (or the cotangent map is surjective). If we choose a closed point in $\mathbb P^N_k$ and look at it inside an affine patch, we get a residue field that looks like (say in $U_0$)
$$k[y_1,...,y_N]/(y_1-a_1,...,y_N-a_N)$$
where we think of the $y_i$ as degree $d$ monomials in the $x_i$ divided by $x_0^d$. This maps to $k[x_1,...,x_n]/(x_1-a_1,...,x_n-a_n)$ by (I think) sending each $y_i$ to the corresponding monomial in the $x_i$. The cotangent map appears to be
$$(y_1-a_1,...,y_N-a_N)/(y_1-a_1,...,y_N-a_N)^2 \to (x_1-a_1,...,x_n-a_n)/(x_1-a_1,...,x_n-a_n)^2$$
sending the $y_i$ to the same monomials as above, and keeping the constants fixed. How is this onto? What maps to $x_i-a_i$ for a fixed $i$, for example?
 A: As in your previous question I continue to think it is more intuitive to look at the tangent map. The Veronese embedding can be written in a coordinate-free way as
$$\mathbb{P}(V) \ni [v] \mapsto [v^n] \in \mathbb{P}(S^n(V))$$
where $V$ is a finite-dimensional vector space and $S^n(V)$ is its $n^{th}$ symmetric power. The Zariski tangent space at a point $[v] \in \mathbb{P}(V)$ can naturally be identified with the quotient $V/v$ of $V$ by the subspace spanned by $v$. Writing a tangent vector at $[v]$ in the form $[v + \epsilon w]$ where $\epsilon^2 = 0$, the differential of the Veronese embedding is
$$[v + \epsilon] \mapsto [(v + \epsilon w)^n] = \left[ v^n + n \epsilon wv^{n-1} \right] \in \mathbb{P}(S^n(V))$$
so to show injectivity it suffices to show that if $nw v^{n-1} = 0$ in the tangent space $S^n(V)/v^n$ to $v^n$, then $w = 0$ in $V/v$. If the characteristic of $k$ does not divide $n$, so that we can divide by $n$, this follows from the fact that the symmetric algebra $S(V) = \sum_{n \ge 0} S^n(V)$, which is a polynomial algebra, is an integral domain, so if $wv^{n-1} = c v^n$ then $v^{n-1}$ can be canceled from both sides to give $w = cv$.
If the characteristic of $k$ does divide $n$ then this definition of the Veronese embedding differs from the one using monomials. I assume that whatever source this came from is implicitly working in characteristic $0$. It may be possible to rescue this version of the Veronese embedding using divided powers.
Edit: Here is some detail on the identification of the Zariski tangent space at $v$ with $V/v$, as requested. Working slightly informally (every manipulation I'm doing with infinitesimals can be justified by defining the Zariski tangent space using $k[\epsilon]/\epsilon^2$ but I have not done so here), consider the quotient map
$$V \setminus \{ 0 \} \ni v \mapsto [v] \in \mathbb{P}(V).$$
Since $V$ is a vector space the Zariski tangent space at every point can be canonically identified with $V$. The differential of the quotient map at $v \in V$ sends a tangent vector $v + \epsilon w$ to $[v + \epsilon w]$. The kernel of the differential consists of $\epsilon w$ such that $[v + \epsilon w] = [v]$, meaning $v + \epsilon w$ is a scalar multiple of $v$, meaning $w$ is a scalar multiple of $v$. And the differential is surjective (I don't know what the cleanest way to see this is but it follows from working in an affine chart of $\mathbb{P}(V)$). So it identifies the tangent space at $[v]$ with $V/v$ as desired.
Edit #2: As Georges says in the comments, this identification of the Zariski tangent space is not quite right; it should be $\text{Hom}(v, V/v)$ where $v$ is shorthand for the line spanned by $v$. I neglected to take into account the effect of scaling $v$ properly. However, if we pick a specific $v \in V$ lifting $[v] \in \mathbb{P}(V)$ we can ignore this which is implicitly what I've done above so the calculations above are still fine (as far as I can tell).
