Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am looking for the standard term for a system that consists of things of the form $p_i(x_1,\ldots ,x_n)=0$ and of the form $q_j(x_1,\ldots,x_n)\neq 0$ with the $p_i$ and $q_j$ polynomials. I have seen used the term "system of equations and inequalities" but to me inequality means "$\leq$" or "$\geq$" and not "$\neq$" so I would prefer not to use it. Is there some commonly accepted term?

I made this CW since I am not sure there is a correct answer.

share|improve this question
3  
See "semialgebraic set" ... en.wikipedia.org/wiki/Semialgebraic_set ... in that setting, of course, $q_j \ne 0$ is the same as $q_j^2 > 0$. –  Gerald Edgar Apr 7 '12 at 17:55
    
Also search for inequations. Gerhard "Ask Me About System Design" Paseman, 2012.04.07 –  Gerhard Paseman Apr 7 '12 at 18:19
    
I've also seen the term "disequality" or "disequation". –  Andreas Blass Apr 7 '12 at 18:23
3  
These are called "equations and inequations" in the literature related to first order theories (say, Tarski problem and such). –  Mark Sapir Apr 7 '12 at 21:21
1  
There's also the standard trick of introducing auxiliary variables $y_j$ and replacing each nequation $q_j(x_1,\ldots,x_n) \neq 0$ by the equation $q_j(x_1,\ldots,x_n) \cdot y_j = 1$. –  Noam D. Elkies Apr 8 '12 at 0:53
show 3 more comments

1 Answer

In algebraic geometry, it's called a quasi-projective algebraic set, which by definition is a Zariski open subset of a Zariski closed subset of projective space. (I'm assuming you only have finitely many $p_i$'s and finitely many $q_j$'s.) Since you're using non-homogeneous polynomials, you're starting in affine space, but that's simply projective space in the homogeneous variables $X_0,\ldots,X_n$ with the condition $X_0\ne0$, so it fits into your framework.

share|improve this answer
    
Thanks Joe. I'm really looking though for the name of the set of equations and "anti"-equations rather than the set of points in affine space which they define. –  Benjamin Steinberg Apr 7 '12 at 20:48
add comment

Your Answer

 
discard

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.