12
$\begingroup$

Let $f = x^n + a_{n-1}x^n + \cdots + a_0$ be a monic polynomial of degree $n \geq 2$ with integer coefficients. By $\text{Gal}(f)$ we mean the Galois group over $\mathbb{Q}$ of the Galois closure of $f$. Define $H(f) = \max\{|a_i|\}$ denote the naive or box height of $f$. Hilbert's irreducibility theorem asserts that for most integer $n$-tuples $(a_0, \cdots, a_{n-1})$ with $H(f) \leq B$ say, the corresponding polynomial $f$ is irreducible. In fact it is conjectured that for all but $O(B^{n-1})$ many such polynomials, the Galois group of $f$ is isomorphic to the symmetric group $S_n$.

The best bounds to date, due to Rankin, assert that all but $O(B^{n-1/2})$ many monic polynomials with box height bounded by $B$ has Galois group isomorphic to $S_n$. David Zywina proved that one can sharpen this bound if one excludes those polynomials whose Galois group is the alternating group. Indeed, he showed that for every $\epsilon > 0$ there exists a number $N$ such that for all $n \geq N$, there are $O(B^{n-1+\epsilon})$ many monic polynomials of height at most $B$ which have Galois group distinct from $S_n$ and $A_n$.

What about lower bounds? Is it obvious that the alternating group should be the second most common Galois group, at least for all sufficiently large degrees? It is apparently a conjecture of Narkiewicz (from Carl Pomerance's talk https://math.dartmouth.edu/~carlp/PDF/fieldstalk.pdf) that the number of fields of degree $n$ over $\mathbb{Q}$ whose Galois group contains a transposition will constitute a positive proportion of all number fields of degree $n$. The alternating group conspicuously does not have any transpositions and so under this conjecture, $0\%$ of fields have Galois group equal to the alternating group. This to me makes it not so obvious why one would expect more polynomials with alternating Galois group as opposed to some other isomorphism class.

Are there known lower bounds for the number of polynomials with Galois group isomorphic to a subgroup of $A_n$, either with respect to the box height or otherwise?

$\endgroup$
17
$\begingroup$

Firstly, the conjecture whereof you speak (with an $\epsilon$ in the exponent) has been proved by yours truly (there is an arXiv.org preprint as of about six months ago).

Secondly, the most common "exceptional" situation is when the polynomial is reducible. It is clear that at least $O(B^{n-1})$ polynomials are reducible, and this is the truth, asymptotically, for $n>2.$

Thirdly, the Galois group is a subgroup of $A_n$ if and only if the discriminant is a perfect square. The obvious heuristic is that the probability that the value of a polynomial of degree $d$ is a perfect square is something like $1/B^{d/2}$ The degree of the discriminant is $2(n-1),$ which would indicate that alternating group is pretty thin on the ground.

ADDED LATER Experimental data (for the probability that a monic irreducible polynomial of degree $n$ and coefficients bounded by $B$ in absolute value has discriminant a perfect square) is consistent with the heuristic above when $n>3$ - the results are not clear for $n=3,$ and the question is vacuous for $n=2$ (a polynomial whose discriminant is a square is reducible).

$\endgroup$
  • $\begingroup$ Did your experimentation reveal an infinite family of quartic polynomials whose Galois group is $A_4$? $\endgroup$ – Stanley Yao Xiao May 29 '16 at 21:37
9
$\begingroup$

The answer to this question really depends on how you count. The conjecture that "Galois groups containing transpositions make up a positive proportion of number fields and are the only groups that do" applies to counting fields by discriminant; when you count polynomials in a box instead, you get 100% S_n, as you say. So in the discriminant-counting setting, the alternating group is not in fact the second-most-common one.

If you want a lower bound for the number of alternating extensions, one way is just to construct a bunch of them out of S_n-extensions (Wikipedia) More geometrically; the reason we know the inverse Galois problem has a positive answer for A_n is that we can construct a parameterized family of polynomials whose Galois group is contained in A_n; then you get that there are B^a such polynomials with coefficients at most B, for some probably smallish rational number a.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.