# rational normal curve of degree 3 not written by intersection of two quadrics

I'm learning about rational normal curves of degree n. And the book says that rational normal curves of degree 3 cannot be written by intersection of two quadrics. I can visualize the situation in my head, but cannot formulate a rigorous argument...Could anyone help me?

• Do you know about degrees of (embedded) projective varieties? Two quadrics will intersect in a curve of degree 4. Oct 20 '14 at 4:03

If one of the quadrics has maximal rank $4$ (or better one of the quadrics in the pencil, which is in fact true), then it is projectively equivalent to the Segre variety. In this case, the other quadric cuts out a curve of bidegree $(2, 2)$ on $\mathbb{P}^1 \times \mathbb{P}^1$. As the twisted cubic has bidegree $(1, 2)$, it follows that we get not only the twisted cubic, but a line, something of bidegree $(1, 0)$, so that the union has bidegree $(2, 2)$. Now, the line is a fiber of one of the rulings, and a general fiber meets the cubic in two points, since a quadratic polynomial has two roots in general.