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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?

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2 Answers

up vote 6 down vote accepted

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.

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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 '10 at 1:18
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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.

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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. –  Bruno Joyal Jun 17 '10 at 3:52
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@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". –  Boyarsky Jun 17 '10 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 '10 at 4:41
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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. –  Ewan Delanoy Jun 17 '10 at 5:03
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@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. –  David Speyer Jun 17 '10 at 12:20
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