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I'm trying to prove that given an irr. hypersurface $H$ and an irr variety X of dim $\geq 1$ in $\mathbb{P}^n_k$, that $X\bigcap H$ is non-empty. I know one way of showing this using the veronese embedding, but I want to prove it by considering the cone over $X,H$ in affine space. I suppose the best way to do this would be a dimension argument, something along the lines of if $X,H$ do not intersect then the complement of $H$ is open and irreducible, and properly contains $X$, so $dim(X) < dim(\mathbb{A}_k^{n+1}-H)$. My question is what is the dimension of the complement of H, is it $n$? It is not the same as the codimension. Any tips on how to approach this would be appreciated.

Andrew
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bijection
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Im a bit confused, because the equations defining the cone over $X$ are the same as those defining $X$, so shouldn't the transcendence degree of the field of rational functions be the same? i.e. why is the dimension of the cone over $X = dim(X) +1$.

caspar
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  • Welcome to MSE! I do not believe you have enough reputation to leave comments, but this area is intended for answers. Regards – Amzoti Mar 22 '13 at 05:48
  • The affine cone of $X$ and $X$ don't live in the same space. –  Apr 09 '13 at 14:02