# Are existentially defined subsets of affine algebraic sets unions of a finite number of affine algebraic sets?

Consider a set of polynomials in $\mathbb{C}[x_1,\dots,x_n]$. The zero locus of these polynomials $Z$ is a subset of $\mathbf{A}^n$ and is an affine algebraic set.

Now, consider the following subset of $\mathbf{A}^{n-1}$: $$S = \left \{ (x_1,\dots,x_{n-1}) \, \middle| \, (x_1,\dots,x_{n-1}) \in \mathbf{A}^{n-1} \text{ s.t. } \exists \, x_n \text{ where } (x_1,\dots,x_n) \in Z \right \}$$

Does this operation have a name?

Is $S$ equal to the union of a finite number of affine algebraic sets (as sets of points)? Clearly if $S$ is finite this is true.

If this does not hold, are there any other useful ways to decompose $S$, or indeed can anything useful be said about such sets?

-

Here's an example showing that $S$ is not always a finite union of algebraic sets. Let $Z$ be the zero locus of the single polynomial $x_1x_2 - 1$. Then $S = \mathbb{A}^1\setminus \{0\}$.

What is true is that $S$ is always a finite union of sets defined by finitely many polynomial equations (basic Zariski closed sets) and negated equations (basic Zariski open sets). Such a set is called a constructible set, and the fact that the projection of a Zariski-closed set (or more generally a constructible set) is a constructible set is known as Chevalley's theorem in algebraic geometry.

The $S$ in the example above is defined by the single negated equation $x_1\neq 0$.

As a logician, I prefer to think of this in terms of quantifier-elimination in the theory of algebraically closed fields. This result, due to Tarski, says that a subset of $K^n$ ($K$ algebraically closed) defined by a first-order formula (built up from polynomial equations by finite Boolean combinations and quantifiers) can actually be defined without quantifiers. The set $S$ in your question is defined by the first-order formula $$\exists x_n\, \bigwedge_{i = 1}^k f_i(x_1,\dots,x_n) = 0.$$

Putting the quantifier-free formula we get from quantifier-elimination in disjunctive normal form, it looks like $$\bigvee_{i = 1}^n \bigwedge_{j = 1}^m \varphi_{ij}(\overline{x}),$$ where each $\varphi_{ij}(\overline{x})$ is $p_{ij}(\overline{x}) = 0$ or $p_{ij}(\overline{x})\neq 0$ for some polynomial $p_{ij}$. This is explicitly a finite union of sets defined by finitely many polynomial equations and negated equations.

-
You beat me to it! – Rob Arthan Feb 11 at 17:52
I guess I'm pretty quick on the draw :0) – Alex Kruckman Feb 11 at 17:57
1. I would just say "an equivalent quantifier-free formula." An because there will be many equivalent quantifier-free formulas. – Alex Kruckman Feb 12 at 16:13
2. Again there's an issue with uniqueness, since there may be many equivalent presentations with different "closed parts". But there is a canonical minimal choice for the "closed part": the Zariski closure of the set $S$. – Alex Kruckman Feb 12 at 16:14
The trouble is with things like "$xy=0$ and $y\neq 0$". The intersection is the y-axis, but discarding the negated equation gives the union of the x- and y- axes. – Alex Kruckman Feb 12 at 17:15

The operation is called projection. Tbe first-order theory of the complex field (which is the same as the first-order theory of algebraically closed fields of characteristic $0$) admits quantifier elimination. This means that $\exists x_n (x_1, \ldots, x_n) \in Z$ is equivalent to a propositional combination of primitive formulas of the form $p_j(x_1, \ldots, x_n) = 0$ for some finite set of polynomials $p_j$. Hence $A$ can be obtained from a finite set of algebraic sets using union, intersection and complement.

[Aside: analogous results hold over the real field, in which case the definable sets are called semi-algebraic sets and the primitive formulas also include formulas of the form $p_j(x_1, \ldots, x_n) > 0$].

-
Thank you for your answer - it was also useful. Unfortunately I'm unable to upvote as a new user! – Jason Feb 11 at 17:54
Thanks! You can always come back when you've earned some more! – Rob Arthan Feb 11 at 17:54
It's true that over $\mathbb{R}$, projections of algebraic sets are semi-algebraic. But it sounds like you're saying that semi-algebraic sets are obtained from algebraic sets by union, intersection, and complement. You also need to allow polynomial inqualities. – Alex Kruckman Feb 11 at 17:56
@AlexKruckman: thanks. Fixed. – Rob Arthan Feb 13 at 9:53