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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 $F(x)$?2012-02-11
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    @JerryGagelman Sorry if I was unclear, they are in the polynomial ring $F[X]$.2012-02-12
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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 $\mathbb A^n$.2012-02-12
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    It may help to tell us where you got this result from...2012-02-12
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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!2012-02-12
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    @MarianoSuárez-Alvarez I'm terribly sorry, I simply meant $\mathbb{A}^n$. This was an aside in some old course notes. I suppose I can TeX them up and try to upload somewhere.2012-02-12
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    It would be really helpful if you did not denote scalars and forms with the same letters! Now, my mind reading machine suggests that if $(x_1,\dots,x_n)\in\mathbb A^n$ what you *mean* is that $M(x_1,\dots,x_n)$ is the maximum of the dimensions of the irreducible components of the algebraic set $X=V(f_1,\dots,f_n)$ which contain $(x_1,\dots,x_n)$ (notice that this is not what you *wrote*...) If my machine is right, you are asking why the union of the irreducible components of $X$ of dimension greater than $d$ is an algebraic set. this should be obvious, since the irreducible components ...2012-02-12
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    involved in that union are closed and finitely many, so their union is also closed.2012-02-12
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    What is the question ?2012-02-15
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    @Lierre My question is about the result in the third paragraph. Thanks for pointing out it was unclear.2012-02-15
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    Are you asking *why the subset $\{x\in V\mid \dim_x X \geq d \}$ is a closed subset of the algebraic variety V* ? These subsets are just union of some of the irreducible components. So they are closed.2012-02-15
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    @Lierre I'm asking 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?2012-02-15

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