17
$\begingroup$

A recent question asked about the topological density of solvable monic quintics with rational coefficients in the space of all monic quintics with rational coefficients. Robert Israel gave a nice proof that both solvable and unsolvable quintics are dense in $\Bbb Q^5$. A natural (non-topological) way to ask about the relative density of solvable quintics in all quintics is to ask about their natural density, as follows:

Write our quintics as $x^5+a_0x^4 + \dots + a_4$, with $a_i \in \Bbb Z$. Write $$f(N) := \frac{\text{# of solvable quintics with } |a_i| < N}{(2N+1)^5}.$$ Robert Israel's answer to the linked question gives some data that supports our intuition that yes, the solvable quintics are in fact extremely rare. Do we indeed have $\lim_{N \to \infty} f(N) = 0$? Are there known nice asymptotics for $f(N)$?

$\endgroup$
  • 1
    $\begingroup$ I feel like it has been proved that 100% (in the limit) of quintics have Galois group equal to $S_5$, hence in particular are unsolvable. $\endgroup$ – Greg Martin Nov 13 '14 at 2:12
  • 1
    $\begingroup$ @GregMartin I would be surprised if this wasn't the case, especially given the computational evidence for it. I'm especially interested in the asymptotics, though, as $f(N)$ seems to go to $0$ extremely quickly. $\endgroup$ – user98602 Nov 13 '14 at 2:14
  • $\begingroup$ What are you looking for that isn't already in math.stackexchange.com/a/1027334/448 and the mathoverflow questions it links to mathoverflow.net/questions/58397 mathoverflow.net/questions/28453 ? $\endgroup$ – David E Speyer Dec 16 '14 at 13:36
  • $\begingroup$ Nothing, @DavidSpeyer; I wasn't aware of that answer. $\endgroup$ – user98602 Dec 16 '14 at 14:53
10
+100
$\begingroup$

We can in fact give a stronger result:

Let $P_N$ denote the set of monic polynomials of degree $n > 0$ in $\mathbb{Z}[x]$ whose coefficients all have absolute value $< N$. S. D. Cohen gave in The distribution of Galois groups of integral polynomials (Illinois J. of Math., 23 (1979), pp. 135-152) asymptotic bounds for the ratio in the above limit, and reformulating his statement with some trivial algebra gives (at least asymptotically) that $$\frac{\#\{p \in P_N : \text{Gal}(p) \not\cong S_n\}}{N^n} \ll \frac{\log N}{\sqrt{N}};$$ note that the limit of the ratio on the right-hand side as $N \to \infty$ is $0$. This implies a fortiori for $n = 5$ that $$\lim_{N \to \infty} \frac{\#\{p \in P_N : \text{Gal}(p) \text{ is solvable}\}}{N^n} = 0,$$ since for quintic polynomials $p$, $\text{Gal}(p)$ is unsolvable iff $\text{Gal}(p) \cong A_5$ or $\text{Gal}(p) \cong S_5$.

Some similar results were produced a few decades earlier: B. L. van der Waerden showed in Die Seltenheit der Gleichungen mit Affekt, (Mathematische Annalen 109:1 (1934), pp. 13–16) that the above ratio has limit zero (at least when one allows nonmonic polynomials and adjusts the denominator accordingly, which is probably inessential).

For more see this mathoverflow.net question and this old sci.math question. Closely related questions on math.se include Is the Galois group associated to a random polynomial solvable with probability 0? and (my own) How often are Galois groups equal to $S_n$? .

(This answer is more-or-less a duplicate of my answer to the question linked in OP's question.)

$\endgroup$

Your Answer

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