2
$\begingroup$

I read somewhere the following (and I quote it identically)

"Let $f_1,f_2,...,f_n$ be polynomials in $\mathbb{C}[x_1,...,x_n]$. Let $New(f_j)$ denote the Newton polytope of $f_j$. If $f_1,...,f_n$ are generic, then the number of solutions of the polynomial system of equations $f_1=...=f_n=0$ with no $x_i=0$ ..."

Now, my question is: What does the author mean when he writes the word generic in there? I know from my experience that the word generic means different things according the text/author/discipline that is being applied into (and sometimes is rather non-rigorous), but what does it mean in that context?

$\endgroup$
  • 1
    $\begingroup$ Probably the coefficients $c_{ij}$ of the $f_i$s are indeterminates. $\endgroup$ – Bernard Feb 1 '17 at 16:08
  • $\begingroup$ That's what I've been thinking as well, though it confuses me that says $f_i's$ are elements of $\mathbb{C}[x_1,...,x_n]$. $\endgroup$ – user321268 Feb 1 '17 at 16:11
  • 1
    $\begingroup$ Or it means informally the coefficients do not belong to some algebraic subvariety – which may depend on the problem at hand. $\endgroup$ – Bernard Feb 1 '17 at 16:22
1
$\begingroup$

Usually in algebraic geometry, one says that something is generic if it holds for every choice outside a well defined (but maybe not easy to determine) proper Zariski closed set depending on the problem.

For instance, the quadratic equation $a x^2 + bx + c = 0$ has two distinct solutions for generic choice of $a,b,c$ because the cases where it does not have two distinct solutions are the ones satisfying $b^2 - 4ac = 0$ that is a Zariski closed subset of the space (say $\mathbb{C}^3)$ where $(a,b,c)$ lives.

In your case, saying that generically the system has no solution with some $x_i = 0$ means that the condition "there is a solution with some $x_i = 0$" can be expressed as a Zarisi closed condition on the coefficients of the $f_j$'s.

Most of the times, over $\mathbb{C}$, "generic choice" can be thought as "random choice" where random is according to some honest probability distribution (technically, according to a measure that is absolutely continuous with respect to the Lebesgue measure in $\mathbb{C}$, or even better, a measure where proper Zariski closed subsets have measure $0$)

$\endgroup$
  • $\begingroup$ I think it's pretty clear to me what @fulges, wants to say. Nice answer! I did misread the whole text I think! $\endgroup$ – user321268 Feb 1 '17 at 18:08
  • $\begingroup$ Yes...proper Zariski closed. I fixed it! Thanks $\endgroup$ – fulges Feb 1 '17 at 18:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy