Natural density of solvable quintics

Question

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 $f(N)$?

Answer 1

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.)

Answer 2

We can get a lower bound on the probability by condlsidering Cayley's resolvent for the case of depressed quintics (with the fourth degree term zeroed out). As described here, a depressed quintic

$y^5+py^3+qy^2+ry+s=0$

is guaranteed solvable if the expression

$-p^6+28p^4r-16p^3q^2-176p^2r^2-80p^2qs+224pq^2r-64q^4+400ps^2+320r^3-1600qrs$

equals zero, for then the Cayley resolvent is guaranteed to have zero as the required rational root.

If all coefficients in the depressed quintic are $O(N)$, then the above expression is $O(N^6)$ dominated by the $-p^6$ term. Thus the probability of hotting zero is $O(N^{-6})$ and this must be considered a lower boubd for the probability of the depressed quintic being soluble.

We can improve this bound by reckoning with dimensional consistency. The coefficients $p,q,r,s$ effectively have different units -- $p$ has units of $y^2$, $q$ has units of $y^3$, $r$ and $s$ respectively have unuts of $y^4$ and $y^5$. Thus means the terms in the expression we are targeting for zero remain $O(N^6)$ if

$p=O(N),q=O(N^{1.5}), r=O(N^2),s=O(N^{2.5}).$

For instance, picking the term $+400ps^2$, we get $O(N)[O(N^{2.5})]^2=O(N^6)$. So we have the lower bound:

$\color{blue}{p=O(N),q=O(N^{1.5}), r=O(N^2),s=O(N^{2.5})\implies \text{Probability of solvability}\ge O(N^{-6})}.$