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?