# Why do generic polynomials work in reality?

I understand that a generic $G$-polynomial $f(t_1,...,t_n)[X]$ over field $k$ has Galois group $G$ over $k(t_1,...,t_n)$. And basically any $G$ extension of $k$ should be generated by a realization of $f$.(even a bit stronger but that is not the point here).

Now as much as I understand, our motivation for hunting these polynomials is that in real (constructive) life, we would like to plug random elements of $k$ into $t_1,...,t_n$ and get a $G$-extension. However, it's obvious that the definition doesn't guarantee it. For example as a trivial failure, we know that $X^n + t_1X^{n-1} + \cdots + t_n$ is generic for $S_n$, but not all values for $t_1, ..., t_n$ (basically all polynomials) lead to an $S_n$-extension.

So, basically, my question is this: what is the constructive value of the definition of generic polynomial. Is there any (although I know I'm saying nonsense) high probabilistic/statistic success rate in getting a $G$-extension when a random realization is chosen. Is there some kind of definition of "odd" that says those times that we don't get a $G$-extension are somehow odd and not normal?

Adding unto Boyarsky's answer: Stephen Cohen has given quantative bounds for how often generic polynomials work. If I've skimmed his paper correctly, when the coefficients are integers chosen from the interval $[-N, N]$, the probability that the Galois group comes out wrong is $O(N^{-1/2} \log N)$, with an explicitly computable constant which depends on the group and the precise parameterization being used.

• I think this probability is fundamentally important in understanding the value of generic polynomial. I'm amazed how all sources that I looked in constructive inverse Galois, didn't bother to even give a hint about it, specially that they call themselves constructive.
– Syed
Jun 18, 2010 at 1:18

Serre introduced a notion of "thin set" in the $k$-rational points of a $k$-variety (such as $k^n$ viewed as the $k$-rational points of affine $n$-space, or likewise for a Zariski-dense open locus in affine $n$-space over $k$, depending on denominators in the coefficients of $f$ in your motivating example) as a mild generalization of "nowhere Zariski-dense" precisely to quantify issues related to Hilbert irreducibility, exactly as in your question. So the answer to your question is the concept of thin sets in the $k$-rational points of $k$-varieties (with $k$ an infinite field). See the Wikipedia entry on "thin set" for more specific information and references to the literature.

• There are many examples of this kind of thing. For instance, invertible matrices are dense in the vector space of $n \times n$ complex matrices. Separable polynomials are dense in the vector space of polynomials over $\mathbb{C}$. Serre's construction does appear quite encompassing. Jun 17, 2010 at 3:52
• @Bruno: The examples of invertible matrices and separable polynomials are of course much "better" in the sense that their complementary loci are contained in a nowhere-dense Zarsiski-closed subset. The notion of "thin set" is meant to go beyond Zariski-dense open sets, as is needed to capture the idea of squares in $\mathbb{Q}$ or loci of reducibility of a monic polynomial with "generic" lower-degree coefficients being "sparse" in an arithmetic sense. So these latter kind of examples are more characteristic of the need for the notion of "thin set". Jun 17, 2010 at 4:23
• @Boyarsky: 1. Do you say "nowhere Zariski-dense" isn't enough for this problem? 2. Part of the question is that how I can prove that "the bad extension generators" are thin in k.
– Syed
Jun 17, 2010 at 4:41
• I don't see why you find this counter-intuitive : if the degree n poly is reducible, there is some equation satisified by the coefficients, so that the uple of coefficients in a variety of dimension strictly lower than the original variety. It is certainly intuitive that lines are "thin" in the plane, and that planes are "thin" in space. Jun 17, 2010 at 5:03
• @Ewan: It is not true that having a polynomial be reducible implies that there is a (polynomial) relation between the coefficients. For example, $a x^2+bx+c$ is reducible if and only if $b^2-4ac$ is a square. The set of such polynomials is not contained in any hypersurface in $(a,b,c)$ space. Jun 17, 2010 at 12:20