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I am trying to prove that a certain class of polynomials have symmetric galois group.

Using the Newton polygon, I have shown that the galois groups of these polynomials are transitive on $k$-sets for all $k$ less than the degree of the polynomial. Beaumont and Peterson proved that this condition holds only for symmetric and alternating groups, along with 4 sporadic exceptions. I have discounted the 4 exceptions, so I am left with the task of showing that the alternating group does not occur.

I am looking for a way to do this. I have tried without success to show that there must be a transposition in the group; likewise showing that the determinant is not a square does not seem to be possible.

Does anybody know of any tricks for showing that one of the above sufficient conditions holds? Alternatively, do you know of any other way to prove that a group is the symmetric group, or not the alternating group?

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    $\begingroup$ Is this over Q? At least in that case, the Chebotarev Density Theorem gives you a quick probabilistic test by factoring the polynomial mod p, and seeing whether the pattern of irreducible factors is even or odd. This may not help for the whole class at once, though. $\endgroup$
    – Douglas Zare
    Commented Feb 26, 2010 at 16:09
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    $\begingroup$ Unfortunately not! They are multivariate polynomials in $\mathbb{Q}(x,\textbf{y})$, where $\textbf{y}=(y_1,y_2,..,y_n)$. I am considering the galois group: Aut$(K/\mathbb{Q}(\textbf{y}))$, where $K$ is the splitting field. Sorry, I should have said this in the first place. I have tried to find some generalisation of the Chebotarev Density Thoerem to a function field, but there doesn't seem to be one. $\endgroup$
    – Adam
    Commented Feb 26, 2010 at 17:33

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There is a version of the Chebotarev density theorem for finitely generated fields, or more precisely, after spreading out, for an étale Galois cover of schemes of finite type over a ring of $S$-integers. This is a consequence of work surrounding the Weil conjectures. See Lemma 1.2 in Torsten Ekedahl, An effective version of Hilbert's irreducibility theorem, Séminaire de Théorie des Nombres, Paris 1988-89, Birkhäuser 1990.

But what I would recommend trying first is to use the monodromy groups. If you spread out your Galois extension to a finite branched cover of a dense open subscheme $U$ of $\mathbf{P}^n_{\mathbf{Q}}$, then the monodromy group (inertia group) associated to any irreducible divisor on $U$ is contained in your Galois group. In particular, if there is a divisor that ramifies in the simplest possible way (ramification indices $2,1,1,\ldots,1$), then you know that your group contains a transposition.

The other approach to try, already mentioned by others in the comments, is specialization, which amounts to spreading out as above and then restricting to the cover above an irreducible closed subscheme. More concretely, you could plug in rational numbers for some or all of $y_1,\ldots,y_n$ such that you get a separable polynomial (of the same degree as the original polynomial). Or you could spread out to a scheme of finite type over $\mathbf{Z}$ and then restrict to an irreducible closed subscheme; this includes reduction modulo primes. If specialization results in a separable polynomial of the same degree over the new function field, then the Galois group of the specialization is a subgroup of the original Galois group, so it suffices to show that the specialized Galois group is not contained in $A_n$.

For more techniques for computing Galois groups, I recommend the book Topics in Galois theory by Jean-Pierre Serre.

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The Galois group of $K[x]/f(x)$ is contained in $A_n$ if and only if the discriminant of $f$ is square in $K$. (Because $\sqrt{f}$ is preserved by $A_n$.)

As Mariano points out, you have already tried this. Why is it not reasonable to compute the dsciriminant? Is the size of the determinant too big for your computer algebra system, or do you get the polynomial and then not know how to check whether it is a square, or is there some other problem?

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    $\begingroup$ Aparently he cannot, as he remarked in the question :) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 26, 2010 at 19:41
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    $\begingroup$ If you're finding it difficult to check wether or not $f$ is a square because it is multivariate, try to specialize $y$ to a fixed vector and prove that this is not a square in the rationals. $\endgroup$
    – Dror Speiser
    Commented Feb 26, 2010 at 19:43
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    $\begingroup$ The problem seems to be that he has a whole family of polynomials. Checking whether the discriminant is a square for a given polynomial is easy, but for a whole family it might be a nontrivial diophantine problem. $\endgroup$
    – Franz Lemmermeyer
    Commented Feb 26, 2010 at 21:35
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    $\begingroup$ It's hard to comment without knowing the specific problem, but there are lots of ways to check that a polynomial is not a square. Show that its zero locus has a smooth point. Show that its value is not a square at some rational point. Show that its Newton polygon is not twice a smaller lattice polygon. $\endgroup$
    – David E Speyer
    Commented Feb 26, 2010 at 23:45
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    $\begingroup$ @David: K(x)/f(x) should be K[x]/f(x). $\endgroup$
    – Chandan Singh Dalawat
    Commented Feb 27, 2010 at 2:35

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