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.