Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 7 down vote accepted

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.

share|cite|improve this answer
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 $p_j(x_1, \ldots, x_n) > 0$].

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.