Let $k$ be an infinite field and $R$ a homogeneous $k$-algebra, i.e. a $k$-algebra that is generated by linear forms. Let $s = \sup\left\{\dim_k h R_{n-1} : h \in R_1\right\}$, where $R_i$ denotes the homogeneous component of $R$ of degree $i$. Let $U$ be the subset of $R_1$ such that for every element $h \in U$ we have that $s = \dim_k h R_{n-1}$. The goal is to show that $U$ is open in the Zariski topology of $R$. To show this, let $a_1,\dots,a_m$ be a basis of $R_1$. Then any element $h$ of $R_1$ can be written as $h = \sum_{i=1}^m x_i a_i, \, x_i \in k$. Then multiplication by $h$ on $R_{n-1}$ is a map $R_{n-1} \rightarrow R_n$ that can be represented by a matrix $A$ of linear forms in $x_i$. Replacing $x_i$ by indeterminates $Y_i$, makes $A$ into a matrix whose elements are linear forms with coefficients in $k$. Then why is it true (note that $R \cong k[Y_1,\dots,Y_m]$) that $U$ is the complement of $V(I_s(A))$, where $V(J)$ is the set of all prime ideals of $R$ that contain ideal $J$ and $I_s(A)$ is the ideal generated by all $s \times s$ minors of $A$?
Reference: This appears in the first paragraph of the proof of theorem 4.2.12 in Bruns and Herzog, Cohen-Macaulay Rings.
PS: I find the writing of the argument somewhat abstract and thus confusing.