Let $I=(f_1,\dots,f_r)\subset k[x_1,\dots,x_n]$ be an ideal with generators homogeneous linear polynomials and furthermore we assume that $f_1,\dots,f_r$ form a minimal generating set of $I$. Is the information above enough to tell that the height of this ideal is $r$?
If the direct proof is hard and a nontrivial theorem is needed, please let me know the source. Thanks!
Let $H_i$ be the vanishing set of $f_i$, $H_i$ is a linear hyperplane of $k^n$. The assumption on the $f_i$ means that no $H_i$ contains the intersection of the others, ie that the $f_i$ are linearly independent linear forms.
Now, $R=k[x_1,\ldots,x_n]/I$ is clearly the space of polynomial functions on $H:= \cap_{i=1}^r{H_i}$, (if $k$ is infinite; I think the argument can be refined with bases for finite fields but let’s ignore it for the sake of simplicity) which is a vector space of dimension $n-r$. So it is an integral domain of dimension $n-r$, therefore $I$ is prime and its height is at most $r$.
But better, actually, for the same reason $(0)$, $(f_1), \ldots , (f_1,\ldots,f_{r-1})$ is an increasing chain of prime ideals contained in $I$. So $I$ has height at least $r$, QED.