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I am confused about the notion of "rational normal curve".

First of all, if a projective curve is rational and normal, then using the Zariski's Main Theorem, we can conclude that it is isomorphic to $\mathbf{P}^1$. Is there an easier way to see this? Or we really need to translate the argument in ZMT to get an elementary proof?

As for the rational normal curve, I am using the standard definition (See Wikipedia page https://en.wikipedia.org/wiki/Rational_normal_curve). It says: "The term "normal" refers to projective normality, not normal schemes." Projectively normality is necessary because Hartshorne II Exercise 3.18(b) provided such example: $(x, y) \to (x^4, x^3y, xy^3, y^4)$. Is it a sufficient condition? In other words, is every rational and projectively normal curve in $\mathbf{P}^n$ comes from an $d$-uple embedding: $\mathbf{P}^1 \hookrightarrow \mathbf{P}^d \subseteq \mathbf{P}^n$?

In this wikipedia page: https://en.wikipedia.org/wiki/Normal_scheme, there is a stronger statement: "linearly normal ... is the meaning of "normal" in the phrases rational normal curve and rational normal scroll." Is every linearly normal and rational curve a rational nomal curve? I try to find this result in standard textbook, but none of Hartshorne, Shafarevich or Harris explain the terminology. Maybe it is hided in some Exercise...

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After some search, I saw two results which might count as a proof. In this question Definitions of "linearly normal" variety, it is said that "$V$ as projective variety cannot be obtained by an isomorphic linear projection from a projective space of higher dimension, except in the trivial way of lying in a proper linear subspace." On the other hand, by Shafarevich I P158, "any rational map from $\mathbf{P}^m$ is obtained by composing the Veronese map with a projection". Combining the two facts, we see that a linear normal rational curve is a rational normal curve. Is there a standard reference for the first fact?

ZH L
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    I'm not sure I'm understanding the question fully. Any smooth rational curve is isomorphic to $\mathbb P^1$, simply because rational maps from a smooth curve can be uniquely extended to morphisms. And your Wikipedia page defines these as smooth, rational curves – Brandon Jan 18 '21 at 21:52
  • @Brandon Sorry, I have corrected a mistake in the statement. I wrote "projective normal" instead of "rational normal", which makes the question very confusing... I understand rational maps can be extended to morphisms, but why are the resulting morphisms necessary inverse to each other? – ZH L Jan 18 '21 at 22:05
  • Oh, good questions. So since the maps come from birational equivalences, they are inverse on a dense set. One good thing about varieties is that if two morphisms agree on a dense set, they must agree everywhere! I think you can prove this by showing the product map is contained in the diagonal. – Brandon Jan 18 '21 at 22:12
  • @Brandon Thanks! – ZH L Jan 18 '21 at 22:19

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Yes, so you can argue like this: First of all projective normality implies normality, but for a curve that amounts to smoothness, hence this curve is isomorphic to $\mathbb{P}^{1}$, but the only linear series on $\mathbb{P}^{1}$ are those $\mathcal{O}(n)$. But now projective normality says that the embedding is given by the complete linear series (equivalent to saying that it does not come from a projection) for those $n>0$, which is the definition of rational normal curve.

Y_q
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