# Nondegenerate subvariety of prime degree not contained in smooth hypersurface

If $Z\subset \mathbb{P}_{\mathbb{C}}^{n+1}$ is a variety of dimension greater than $\frac{n}{2}$ and $X\subset\mathbb{P}_{\mathbb{C}}^{n+1}$ is a smooth hypersurface of degree $d$ containing $Z$, then an application of Lefschetz hyperplane theorem shows $\deg(X)|\deg(Z)$. This is given in Corollary C.10 of 3264 & All That by Harris and Eisenbud.

In addition, the same corollary claims that in particular, if $Z$ is of prime degree $p$ and nondegenerate, then it is not strictly contained in a smooth hypersurface.

How do we rule out the case where $\deg(X)=\deg(Z)=p$?

Edit: cross posted in MO in a different form (https://mathoverflow.net/questions/251504/hypersurface-containing-nondegenerate-subvariety-of-same-degree-and-large-dimens)

• If $X$ and $Z$ have the same degree, then $Z$ must be a section of $X$ by a hyperplane $\mathbf P^n$. But then $Z$ is degenerate. – Nefertiti Sep 29 '16 at 6:41
• I can show that if Z is contained in X, then Z and the hyperplane section are homologous. Why does this imply Z is contained in a hyperplane? – DCT Sep 29 '16 at 14:41
• Sorry for not following up on this --- I was assuming $Z$ had codimension 1 in $X$, but that was a misreading of the question on my part. – Nefertiti Oct 6 '16 at 9:41