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Let $X \subset {\bf P}^d$ be rational normal curve. $X$ is defined as the image of the map $v_d : {\bf P}^1 \to {\bf P}^d$ $$[a:b] \to [a^d:a^{d-1}b:\dots:ab^{d-1}:b^d].$$ The map $v_d$ induces a map $v_d^*:k[z_0,...,z_d] \to k[x_0,x_1]$ by setting $z_i \to x_0^{d-i}x_1^i$, therefore under the map $v_d^*$ every homogeneous polynomial of degree $m$ maps to a homogeneous polynomial of degree $md$ of $k[x_0,x_1]$.

Then in the book "Algebraic Geometry" by Joe Harris the following argument is written which i'm unable to understand:

Under the map $v_d$ the homogeneous polynomials of degree $m$ in the coordinate ring $[z_0:\dots:z_d]$ on ${\bf P}^d$ pull back to give all homogeneous polynomials of degree $md$ in the coordinate ring $[x_0:x_1]$ of ${\bf P}^1$. Therefore $$\left(\frac {k[z_0,\dots,z_d]}{I(X)}\right)_m\simeq k[x_0,x_1]_{md}.$$

Can someone explain me the above argument please?

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This is really an exercise in applying the division algorithm in $\mathbb Z$.

Pick a monomial of degree $md$ in $k[x_0,x_1]$, say $x_0^ax_1^b$ with $a+b=md$. Then, express $a=dk+r$ with $0\leq r<d$ and $b=dk'+r'$ with $0\leq r'<d$.

Then $0\leq r+r'<2d$ and $r+r'=a-dk+b-dk'=d(m-k-k')$ which causes $r+r'$ to be either $d$ or $0$.

Then, $z_0^k$ maps to $x_0^{dk}$, $z_d^{k'} $ maps to $x_1^{dk'}$ and $z_{r'}$ maps to $x_0^rx_1^{r'}$ under the map $v^*_d$. Putting this together, $z_0^kz_d^{k'}z_{r'}$ maps to $x_0^ax_1^b $. Finally the degree of the monomial $z_0^kz_d^{k'}z_{r'}$ is $k+k'+1=\frac{dk+dk'+d}{d}=\frac{dk+dk'+r+r'}{d}=\frac{a+b}{d}=m$ as required.

So the map $v^*_d$ is surjective on the claimed homogeneous components. Can you see why the kernel of this map is exactly $I(X)$?

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  • $\begingroup$ +Very nice:+1. (You have considered only the case $r+r'=d$. The case $r+r'=0$ is easier since then $k=k'=0$ and then $z_0^{k}\cdot z_d^{k'}$ already maps to $x_0^a\cdot x_1^b$) $\endgroup$ – Georges Elencwajg May 2 '18 at 13:39

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