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I'm hoping to see how the following bound is reached.

For an algebraic variety $V\subset\mathbb{A}^n$ over some field $F$, one defines $\dim V=\operatorname{trdeg}(F(x)/F)$ for a generic point $(x)$ of $V$. Also, I denote by $V(f_1,\dots,f_m)$ the set of zeroes in $\mathbb{A}^n$ of some homogeneous forms $f_i$.

All right, so $V:=V(f_1,\dots,f_m)$ is a homogeneous algebraic variety. How does it follow that the dimension of $V$ is bounded below by $n-m$? That is, why is $\dim V\geq n-m$? Thank you.

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    It really boils down to Krull's principal ideal theorem. See Prop 7.1 in Hartshorne, page 48.2012-02-11
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    Dear @Parsa, you mean generalization of Krull's theorem since there are no principal ideals in sight here. Moreover the part describing the dimension of $V$ in terms of the transcendence degree of the field of rational functions on $V$ has nothing to do with Krull's principal ideal theorem and yet has to be taken care of before answering Waldott's question (the reference you give doesn't answer it) .2012-02-11

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