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?

  • $\begingroup$ 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. $\endgroup$ – random123 Dec 14 '18 at 10:54
  • $\begingroup$ Ah, good catch, thanks! I will try to work through the exercise. $\endgroup$ – red_trumpet Dec 14 '18 at 17:55

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