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It is well-known that, as polynomials of degree exceeding 4, there exist quintics whose roots cannot be solved for by radicals (Abel-Ruffini theorem). So we can divide the set of rational quintics into those which are solvable by radials and those which not. Obviously, both subsets are infinite in size.

What is not obvious to me is whether both have the cardinality density in rational quintics (see update below). In other words, we can ask the question: "Are there more unsolvable rational quintics than solvable?" My naive expectation would be 'yes', but I've not the background to properly formulate or prove this claim. So I'd like to see a proof either way; useful references/citations are welcome.

UPDATE: My use of 'cardinality' in my original question didn't reflect my intention. Rather, it was as MikeMiller has indicated in comments: If the subset of solvable rational quintics is denoted as $S\subset \mathbb{Q}^5$, is either $S$ or $\mathbb{Q}^5-S$ dense in $\mathbb{Q}^5$? More quantitatively, can the following limit be computed:

$$\lim_{N \to \infty} \frac{\text{# of solvable quintics with } |a_i| < N}{N^5}=?$$

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    $\begingroup$ Since rational equations are numerable, both solvable and unsolvable equations are numerably infinite. $\endgroup$
    – Crostul
    Commented Nov 11, 2014 at 21:11
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    $\begingroup$ Denote quintics as $x^5+a_0x^4+\dots + a_4$ with integer coefficieints. Perhaps the right question is: what is the value of $$\lim_{N \to \infty} \frac{\text{# of solvable quintics with } |a_i| < N}{N^5}?$$ $\endgroup$
    – user98602
    Commented Nov 11, 2014 at 21:16
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    $\begingroup$ Alternatively, perhaps consider the subsets $S$ and $U$ of solvable and unsolvable quintics with rational coefficients (as subsets of $\Bbb Q^5$. Is either dense in $\Bbb Q^5$? $\endgroup$
    – user98602
    Commented Nov 11, 2014 at 21:17
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    $\begingroup$ Quintics with a rational root are solvable, and these are easily seen to be dense in $\mathbb Q^5$. $\endgroup$ Commented Nov 11, 2014 at 21:21
  • $\begingroup$ @RobertIsrael: Can you add that to your answer below? Mike's two formulations are both more in the spirit of what I had intended, and I'm editing my question to reflect that $\endgroup$ Commented Nov 11, 2014 at 21:26

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Just to get a data point, using Maple I took $2000$ random quintics with coefficients pseudo-random numbers from -100 to 100 (but the coefficient of $x^5$ nonzero). $1981$ of these were irreducible (of course the reducible ones are solvable). All $1981$ irreducible quintics were not solvable.

EDIT: Quintics with a rational root are solvable, and these are easily seen to be dense in $\mathbb Q^5$. Namely, take a rational approximation $r$ of a real root of the polynomial. Then $p(X) - p(r)$ has rational root $r$, and is arbitrarily close to $p(X)$.

EDIT: If I'm not mistaken, quintics with Galois group $S_5$ are dense in $\mathbb Q^5$. Consider the proof that $x^5 - x - 1$ has Galois group $S_5$. The same proof should apply to $p(X) = X^5 +\sum_{i=0}^4 \alpha_i X^i$ as long as

  1. All of the denominators of the $\alpha_i$ are congruent to $1 \mod 6$.
  2. The numerators of $\alpha_0$ and $\alpha_1$ are congruent to $5 \mod 6$, those of $\alpha_2, \alpha_3$ and $\alpha_4$ are congruent to $0 \mod 6$.

$5$-tuples satisfying these conditions are dense in $\mathbb Q^5$.

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  • $\begingroup$ Do you know if said density generalizes to polynomials of higher degree? In the sextic case, for example, having one rational root isn't sufficient. $\endgroup$ Commented Nov 11, 2014 at 21:47
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    $\begingroup$ In general, you can get all roots in $\mathbb Q + i \mathbb Q$, i.e. approximate $\prod_j (X - r_j)$ by $\prod_j (X - q_j)$ where $q_j$ is an approximation of $r_j$. So polynomials that factor over $\mathbb Q$ into linear and quadratic factors are dense. $\endgroup$ Commented Nov 12, 2014 at 1:10
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Yes, in fact, we can generalize Mike's reformulation (with integer coefficients, and allowing nonmonic polynomials, which ought to be inessential) and 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. Reformulating his statement with some trivial algebra gives (at least asymptotically) that $$\frac{\#\{p \in P_N : \text{Gal}(p) \not\cong S_n\}}{N^n} \ll \frac{\log N}{\sqrt{N}},$$ and 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.

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  • $\begingroup$ Hi Travis, I posted that part of the question as its own question here when it didn't seem like it would receive a response. There's a bounty on it that's set to expire sometime tomorrow; can I encourage you to post this answer there as well, so I can accept it and you can receive the bounty? $\endgroup$
    – user98602
    Commented Dec 17, 2014 at 3:44
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    $\begingroup$ Mike, sure, I've just done so, and thank you for alerting me to the bounty. I've added links to that version to some related questions on math.se, too, including one of my own. $\endgroup$ Commented Dec 17, 2014 at 3:53

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