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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.

Manos
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  • isn't this just det = 0 is closed ? – meh Apr 14 '14 at 18:13
  • @aginensky: Maybe. Could you elaborate? – Manos Apr 14 '14 at 18:36
  • @aginensky: what do you mean by "det=0 is closed"? – Manos Apr 14 '14 at 19:01
  • Sure, the space of linear forms acts as (a vector space) of linear transformations of Hom( $R_{n-1}$, $R_n$ ). As long as the field is infinite, it is possible (by taking appropriate scalars) to get a linear transformation of maximal rank ( take any $h$ which hits a desired element of $R_n$ adding some scalar multiple of $h$ to an existing transformation will add the new element to the image. The determinant conditions you mention (and I mention) are exactly the condition that an element of $R_1$ is not of maximal rank. hth – meh Apr 14 '14 at 19:22
  • @aginensky: This helps indeed. I understand that if multiplication by a linear form $h=\sum_i x_i a_i$ achieves maximal rank $s$, then some $s \times s$ minor of the representing matrix $A$ will be non-zero. What confuses me is the fact that we substitute the coefficients $x_i$ with indeterminates $Y_i$. Why are we doing that? I can't follow the argument. – Manos Apr 15 '14 at 15:44
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    Think of the $Y_i$ as 'universal' linear forms. If one has an equation that says $\Sigma a_iY_i$ is less than maximal rank, then it is true that, if we substitue $x_i$ for the $Y_i$ , we still have the same identity being satisfied. The fact that not maximal rank is determined with specific equations being zero for the universal case ($Y_i$ ) means it is also true that for any substitution (read ring homomorphism) that those equations. – meh Apr 15 '14 at 18:12

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