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Let $X$ be a closed irreducible subvariety of projective $n$ space. Let $k$ be the dimension of a maximal linear subspace not intersecting $X$. Then one can define $dim(X)=n-k-1$ (see Harris page 134). Why is $dim(Y) < dim(X)$ for two properly contained such varieties?

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This is also mentioned in Shafarevich (p.71), but the development may be different. The dimension of an irreducible quasi-projective variety $X$ is defined as the transcendence degree of the function field of $X$. Then the fact that one can equivalently define dimension as you mention is proved as a corollary of the following theorem: If a form $F$ is not $0$ on an irreducible projective variety $X,$ then $\operatorname{dim}(X_F)=\operatorname{dim}(X)-1.$ This is a bit of a work-around though, maybe you prefer a direct geometric proof!

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The way in which Harris phrased this conclusion made me think that there should be a very easy direct proof from the definition I gave. –  Wuggletruggle Jul 19 '12 at 19:01
    
I agree, I just happened to be holding Shafarevich when I saw the question. –  Andrew Jul 19 '12 at 19:17
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