# Degree and codimension of nondegenerate varieties

A projective variety $$X \subset \mathbb{P}^r$$ (i.e. reduced, irreducible closed subscheme) is called nonegenerate, if it is not contained in any hypersurface $$H \subset \mathbb{P}^r$$. Equivalently, the smallest linear subspace containing $$X$$ is $$\mathbb{P}^n$$ itself.

I'm having trouble regarding the proof of the following standard lemma.

Lemma If $$X\subset \mathbb{P}^r$$ is a nondegenerate variety, then $$\text{deg } X \geq \text{codim }X + 1$$.

This can be found in the paper On Varieties of Minimal Degree by Eisenbud and Harris. The proof is as follows:

1. If $$\text{codim }X = 1$$, then $$\text{deg }X = 1$$ if and only if $$X$$ is a hyperplane. $$X$$ is nondegenerate, so this is not the case. Hence $$\text{deg }X \geq 2 = \text{codim }X + 1$$.
2. If $$\text{codim }X \geq 2$$, choose a "general point of $$X$$", and project $$X$$ from that point to $$\mathbb{P}^{r-1}$$.
3. Observe that the degree of the projection of $$X$$ is $$< \text{deg }X$$, and similarly the codimension of the projection is $$< \text{codim }X$$.
4. Proceed by induction on $$\text{codim } X$$.

Question How can I see that codimension and degree of $$X$$ both drop by at least one?

In a class on surfaces I'm taking my professor proved the Lemma even in the case that $$X$$ is not irreducible, by intersecting $$X$$ with a general hyperplane, reducing the codimension of $$X$$ by one, while leaving the degree as is is. Here the induction goes over the dimension of $$X$$, and the basis is $$n+1$$ (or more) points, which need to have degree $$\geq n+1$$.

Question How can I see that there exists a hyperplane, such that the codimension drops, and the degree stays the same?

I thought maybe this has something to do with the generalized Bézout's theorem, but I only know this for irreducible varieties. Is there a version of this for only reduced ones?

• How about following the exercise 7.7 from hartshorne Chapter 1 section 7? It seems that this exercise does actually help you complete the induction step. – random123 Dec 14 '18 at 10:54
• Ah, good catch, thanks! I will try to work through the exercise. – red_trumpet Dec 14 '18 at 17:55