A rational normal curve $C \subset \mathbb{P}\_k^d$ (assume $k= \mathbb{C}$) can be defined usually up to projective equivalence in two equivalent ways:


* smooth irreducible nondegenerate curve $C \subset \mathbb{P}^d$ of minimal degree $\text{deg}(C)=d$
* projectively equivalent to the image of the Veronese map


$$ \nu\_d: \mathbb{P}^1 \to \mathbb{P}^d, \ \ \
[X\_0:X\_1] \mapsto [X\_0^d: X\_0^{d-1}X\_1:..., X\_1^d] $$


The second definition can be reprased as that there exist a $\mathbb{C}$-basis of the vector space of homogeneous polynomials $p\_0,p\_1,..., p\_d$ of degree $d$ in two variables $X\_0, X\_1$ such that $C$ is the image under


$$ P: \mathbb{P}^1 \to \mathbb{P}^d, \ \ \
[X\_0:X\_1] \mapsto [p\_0(X): p\_1(X):..., p\_d(X)] $$


In Harris' book *[Algebraic Geometry](http://www.springer.com/it/book/9780387977164)*, on page 14 there is an alternative presumably equivalent construction of a rational normal curve:  

Start by choosing $d$ codimension two linear
spaces $\Lambda\_i \cong \mathbb{P}^{d-2} \subset \mathbb{P}^{d}$. The family $\{H\_i(\lambda)\}$ of hyperplanes in $\mathbb{P}^{d}$ containing $\Lambda\_i$ is
then parameterized by $\lambda \in \mathbb{P}^1$; choose such parameterizations, subject to the condition
that for each $\lambda$ the planes $H\_1(\lambda), ... , H\_d(\lambda)$ are independent, i.e., intersect in a
point $ p(\lambda)$. Claim: It is then the case that the locus of these points $p(\lambda)$ as $\lambda$ varies in $\mathbb{P}^1$ is a rational normal curve.


*Question:* Why this construction gives a rational normal curve $\bigcup\_{\lambda \in \mathbb{P}^{1}} H\_1(\lambda) \cap ... \cap H\_d(\lambda)$?


I found [here](https://math.stackexchange.com/questions/4170453/steiner-construction-of-rational-normal-curve) a rather promising approach discussing identical problem: in the linked discussion is showed how one can construct explicitly the point $p(\lambda)= [p\_0(\lambda): p\_1(\lambda):...: p\_d(\lambda)]$ in terms of certain homogenous polynomials $p\_i(\lambda)$ of degree $d$ in coordinates $(\lambda)=\lambda\_0, \lambda\_1$.


The idea is quite simple. Consider the $(d+1) \times (d+1)$ matrix $A(\lambda)$ where each $j$-th row of $A(\lambda)$ is encoded by the $j$-th family of hyperplanes as follows. The parametrized family $H\_j(\lambda)$ of hyperplanes which contain $\Lambda\_i \cong \mathbb{P}^{d-2}$ is given as vanishing loci of $\lambda\_0S\_j +\lambda\_1T\_j$, where
the two linear forms $S\_j, T\_j$ in variables $Y\_0,..., Y\_d$ determine $\Lambda\_j$. The $j$-th row of $A(\lambda)$ is encoded in


$$a\_0(\lambda)Y\_0+ a\_1(\lambda)Y\_1 +...+ a\_d(\lambda)Y\_d= \lambda\_0S\_j +\lambda\_1T\_j $$


The $(d+1)$-row is filled with zeroes. The adjugate matrix $P(\lambda)$ having $d$-minors of $A$ as entries satisfies $A \cdot P=P \cdot A= \text{det}(A) \cdot E\_{d+1} $ and $\text{det}(A) $ is zero. The $(d+1)$-th column of $P(\lambda)$ is the unique point $p(\lambda)=[p\_0(\lambda): p\_1(\lambda):...: p\_d(\lambda)]$ with homogeneous $p\_i(\lambda)$ of degree $d$.


One expects finally to obtain the rational normal curve as image of


$$[X\_0:X\_1] \mapsto [p\_0(X): p\_1(X):..., p\_d(X)] $$


but it isn't clear why the $p\_i(\lambda)$ are linearly *independent*. If they aren't, we obtain a degenerate curve in the image, that's not what we want.



Equivalent characterizations of rational normal curve
Let $\mathbb{P}^d = \mathbb{P}(V)$. A hyperplane in $\mathbb{P}(V)$ corresponds to an epimorphism $V \to k$, a pencil of hyperplanes to an epimorphism
$$
V \otimes \mathcal{O}\_{\mathbb{P}^1} \to \mathcal{O}\_{\mathbb{P}^1}(1),
$$
and a collection of $d$ pencils to a morphism
$$
f \colon V \otimes \mathcal{O}\_{\mathbb{P}^1} \to \mathcal{O}\_{\mathbb{P}^1}(1)^{\oplus d}.
$$
The condition imposed on the pencils means that this is an epimorphism, therefore its kernel $\mathrm{Ker}(f)$ is a line subbundle in $V \otimes \mathcal{O}\_{\mathbb{P}^1}$, and computing its determinant it is easy to see that
$$
\mathrm{Ker}(f) \cong \mathcal{O}\_{\mathbb{P}^1}(-d).
$$
The embedding $\mathrm{Ker}(f) \to V \otimes \mathcal{O}\_{\mathbb{P}^1}$ induces a degree $d$ map $\mathbb{P}^1 \to \mathbb{P}(V)$, and its image is precisely the curve in question. By this construction, it is a rational normal curve of degree $d$.


EDIT. To show that the curve is nondegenerate, consider the exact sequence
$$
0 \to 
\mathcal{O}\_{\mathbb{P}^1}(-d) \to 
V \otimes \mathcal{O}\_{\mathbb{P}^1} \stackrel{f}\to
\mathcal{O}\_{\mathbb{P}^1}(1)^{\oplus d} \to 
0
$$
obtained above. Dualizing it, we obtain
$$
0 \to 
\mathcal{O}\_{\mathbb{P}^1}(-1)^{\oplus d} 
\stackrel{f^\vee}\to
V^\vee \otimes \mathcal{O}\_{\mathbb{P}^1} \to
\mathcal{O}\_{\mathbb{P}^1}(d) \to 
0.
$$
Since the sheaf $\mathcal{O}\_{\mathbb{P}^1}(-1)$ has no cohomology, the induced morphism
$$
V^\vee = 
H^0(\mathbb{P}^1, 
V^\vee \otimes \mathcal{O}\_{\mathbb{P}^1}) \to
H^0(\mathbb{P}^1, \mathcal{O}\_{\mathbb{P}^1}(d))
$$
is an isomorphism. This means that the curve is nondegenerate.


