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This is Vakil 18.6 M, self-study.

We are to show that an integral degree $d$ curve $C$ in $\mathbb P^N_k$ (with $N \geq d$ and $k$ WLOG algebraically closed) is contained in a "linear $\mathbb P^d_k \subset \mathbb P^N_k$."

So, far I see that if you take a hyperplane $H$ in $\mathbb P^N_k$ intersecting $C$ but not containing it, then by Bezout's Theorem (18.6 K in Vakil),

$$\deg C \cap H = \deg C \deg H = d$$

where the degree indicates the leading coefficient of the Hilbert polynomial times $n!$, where $n$ is the dimension of the space. The problem is that I do not see what this tells us geometrically. I can see the leading coefficient of the Hilbert polynomial of $C \cap H$ must have been $d$ to begin with, but this seems purely formal.

I have studied both answers here, but we have not covered general position (the content of the first answer), and I do not follow the conclusion of the second answer, in addition to the fact that it does not seem to use Bezout's Theorem, at least as we have covered it. I suspect neither answer is what Vakil is looking for, and that I am simply missing some geometric realization about degree that would quickly convert what I have done thus far into an answer.

By "a linear $\mathbb P^d_k$" I assume Vakil means a linear space of dimension $d$ in $\mathbb P^N_k$, meaning a closed subscheme cut out by $N - d$ homogeneous linearly independent linear polynomials.

Johnny Apple
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  • If $X_1$ and $X_2$ are two (disjoint) zero-dimensional closed subschemes of $\Bbb P^N$, can you compute the degree of their union in terms of each of their degrees? – KReiser Dec 30 '20 at 05:01
  • Yes; it's the sum of the degrees since they do not intersect. I only know this on a technical level because I have proven a more general statement before. – Johnny Apple Dec 30 '20 at 05:14
  • Okay, so what's the maximum number of points in a zero-dimensional closed subscheme of $\Bbb P^N$ of degree $d$? – KReiser Dec 30 '20 at 05:19
  • That would be $d$. I think you are trying to get me to see that $|C \cap H| \leq d$, and while I now understand why this is so better (I knew it only in a vague sense prior to your comment), I still do not see how this pushes us to the answer. – Johnny Apple Dec 30 '20 at 05:26

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Continued from the comments - we established there that if $C\not\subset H$, then $|C\cap H|\leq d$. If you can only ever get at most $d$ points with a proper intersection, what happens when you pick a hyperplane through $d+1$ points on $C$? Full details under the spoiler.

Pick $d+1$ distinct points on $C$. If $d<n$, then there is a hyperplane $H$ through these $d+1$ points. If $C\not\subset H$, then $H\cap C$ is a zero-dimensional subscheme of degree $\geq d+1$, contradicting Bezout's theorem. So $C\subset H$. Repeat until $d=n$.

KReiser
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