How is this lower bound of the dimension of a homogeneous algebraic variety reached? 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.
 A: There are two different ingredients  to the solution of your problem.
Consider  a finitely generated domain $A$ over a field $k$, and its associated variety $V= Spec (A)$
I) The Krull dimension of $V$ is equal to the transcendence degree of its field of rational functions: 
 $$\text {dim }(V) =\text {trdeg} _k(\text {Frac }(A))$$  
II) If $V\subset \mathbb A_k^n$ is given by a prime ideal $\mathfrak p=(f_1,...,f_m)\subset  k[T_1,...,T_n]$ generated by $m$ polynomials , that is $V=Spec(A)=Spec(k[T_1,...,T_n]/(f_1,...,f_m))$, then
$$       \text {dim }(V)\geq n-m             $$
Notes
1) You will find the proof of I) in Eisenbud's Commutative Algebra, Ch.13, §1, Theorem A.
And II) is Theorem 10.2 of the same book.  
2) There is no need to suppose thar the $f_i$'s are homogeneous in II) 
3) If $\text {dim }(V)= n-m $, the variety $V$ is said to be a scheme-theoretic complete intersection.
Beware that this equality is not true in general: the variety    $V \subset \mathbb A^3_k$ given parametrically by $x=t^3, y=t^4, z=t^5$ is of dimension $1$ but  has  an ideal $I(V)=\mathfrak p\subset k[T_1,T_2,T_3 ]$  which  cannot be generated by $ 3-1=2$ generators.
