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There is a small result I don't understand.

To preface, 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 $V(f_1,\dots,f_r)$ the set of zeroes in $\mathbb{A}^n$ of some homogeneous forms $f_i$ in $F[X]$.

As an algebraic set, we know $V(f_1,\dots,f_r)$ can be written as a finite union of irreducible components without inclusion relations. Now let $M(f_1,\dots,f_r)$ be the maximum of the dimensions of the irreducible components. Apparently for any nonnegative $d$, the set of points $(f_1,\dots,f_r):=(w)_f$ where $M(f_1,\dots,f_r)>d$ is also an algebraic set.

I identify the points $(f_1,\dots,f_r)$ as a subset of $\mathbb{A}^n$ in the following way. For a finite set of forms $(f)=(f_1,\dots,f_r)$, let $d_1,\dots,d_r$ be the degrees, with $d_i\geq 1$ for all $i$. Each $f_i$ can be written as $$ f_i=\sum w_{i,(v)}M_{(v)}(X) $$ where $M_{(v)}(X)$ is a monomial in some set of indeterminates $(X)$ of degree $d_i$, and $w_{i,(v)}$ is a coefficient. Let $(w)=(w)_f$ be the point obtained by arranging the coefficients $w_{i,(v)}$ in some definite order, and consider this point in some affine space $\mathbb{A}^n$, where $n$ is the number of coefficients, determined by the degrees $d_1,\dots,d_r$. So given such degrees, the set of all forms $(f)=(f_1,\dots,f_r)$ with these degrees is in bijection with the points of $\mathbb{A}^n$.

So if given a fixed $d\geq 0$, then why is the set of $(f_1,\dots,f_r):=(w)_f$ such that $M(f_1,\dots,f_r)>d$ an algebraic set? So given $d\geq 0$, I want to find all sets of forms $f_1,\dots,f_r$ such that the maximum dimension of the irreducible components of their set of zeroes in $\mathbb{A}^n$ has degree greater than that fixed $d$. Then for each possible set of forms meeting that condition, I identify with a point in $\mathbb{A}^n$ as described above. Why are those points in $\mathbb{A}^n$ an algebraic set?

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I haven't thought about algebraic geometry for a while so apologies if this question is trivial: when you write $(f_1,...,f_m)$, in where do the coordinates live? In the polynomial ring $F[X]$ or in $F(x)$? –  Jerry Gagelman Feb 11 '12 at 22:58
    
@JerryGagelman Sorry if I was unclear, they are in the polynomial ring $F[X]$. –  Waldott Feb 12 '12 at 4:26
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Your notation is either weird or wrong. The «set of$(f_1,\dots,f_m)$ such that [...]» cannot be an algebraic set if the $f$s are polynomials... simply because such things do not belong to $\mathbb A^n$. –  Mariano Suárez-Alvarez Feb 12 '12 at 4:29
    
It may help to tell us where you got this result from... –  Mariano Suárez-Alvarez Feb 12 '12 at 4:31
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I don't know what you mean by «the space of coefficients o the forms». Again: it may be most efficient to tell us what you are reading that contains this result! –  Mariano Suárez-Alvarez Feb 12 '12 at 21:14
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1 Answer

What you denote by $M(f_1, \dots, f_r)$ is just the dimension of $V(f_1,\dots, f_r)$.

Let $\mathbb A^N$ be the space of the coefficients $w_{i,(v)}$ of the $f_i$'s and consider the subvariety $X:=V(f_1, \dots, f_r)$ in $\mathbb A^N\times \mathbb A^n$. Here $f_i$ is viewed as a polynomial with coefficients in the polynomial ring $F[w_{i, (v)}]_{\{ v\}}$. Consider the projection $p : X\to Y:=\mathbb A^N$. You are looking at the set of $y\in Y$ such that $\dim p^{-1}(y)\ge d+1$. This is the image by $p$ of a closed subset of $X$ according to the theorem of semi-continuity of Chevalley (see EGA, IV.13.1.3). So your set is a constructible subset of $\mathbb A^N$ by another theorem of Chevalley (EGA, IV.1.8.4).

To really get a closed subset of $Y$, as you supposed the $f_i$ homogeneous, one can work with $\mathbb A^N \times \mathbb P^{n-1}$ with a projection $q$ to $Y:=\mathbb A^N$. Then $q$ is projective, hence a closed map. So the set of points $y$ such that $q^{-1}(y)$ has dimension $\ge d+1$ is closed in $Y$ (EGA, IV.13.1.5). To go back to the affine variety $V(f_1,\dots, f_r)$, notice that its dimension is the dimension of the projective variety defined by the $f_i$ minus $1$.

EDIT Let me add an example on the construction of $X\to Y$.

Consider the case of two polynomials $f_1=W_1T^2+W_2TS+W_3S^2$ and $f_2=W_4T^2+W_5TS+W_6S^2$. Then the variety $X$ is the common zero locus of $f_1, f_2$ in $\mathbb A^6\times\mathbb A^2$, and the projection $p$ maps a point $(w_1,\dots, w_6, t,s)$ of $X$ to $(w_1, \dots, w_6)\in Y$. When we specify a set of coefficients $w_1, \dots, w_6\in F$, we get a point $y\in \mathbb A^6$ and $p^{-1}(y)$ is the common zero locus of $w_1T^2+w_2TS+w_3S^2, w_4T^2+w_5TS+w_6S^2\in F[S, T]$ in $\mathbb A^2$.

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