9
$\begingroup$

I know that not all quintics are solvable. But how do I identify the class of solvable ones?

$\endgroup$
  • 1
    $\begingroup$ The ones with at least one rational root can be identified quickly using the rational root theorem and then the remaining quartic always has an exact solution in radicals. Beyond that, I'm sure someone else will answer with deeper theory. $\endgroup$ – Deepak Jun 12 '16 at 13:20
  • $\begingroup$ You have to find out the so-called galois-group over $\mathbb Q$ of the given quintic. This is not so easy by hand, but for example PARI/GP and GAP can identify it. The quintic can be solved by radicals if and only if the galois group is solvable. $\endgroup$ – Peter Jun 12 '16 at 13:23
  • 1
    $\begingroup$ First check, whether the polynomial is irreducible over $\mathbb Q$. If no, the quintic can be solved by radicals because polynomials with degree less than $5$ can always be solved by radicals. If yes, you have to find out the galois-group. $\endgroup$ – Peter Jun 12 '16 at 13:26
  • 1
    $\begingroup$ It also depends on what you need the solution for. If it's for practical purposes, the numerical methods are usually better (like the Newton's method) and they work for any polynomial equation $\endgroup$ – Yuriy S Jun 12 '16 at 13:39
7
$\begingroup$

If you haven't reached that part of Galois theory yet, there is still a "simple" elementary way to test if a quintic is solvable. First, reduce it to depressed form (without the $x^4$ term).

Method 1:

Theorem (by Watson, 1930s): "Given an irreducible quintic with rational coefficients,

$$x^5 + 10c x^3 + 10d x^2 + 5 e x + f = 0\tag1$$

If the sextic,

$$3125p^6 - 625(3c^2 + e)p^4 + 25(15c^4 + 8c d^2 - 2c^2e + 3e^2 - 2d f)p^2 + p\sqrt{D} + (-25c^6 - 40c^3d^2 - 16d^4 + 35c^4e + 28c d^2e - 11c^2e^2 + e^3 - 2c^2d f - 2d e f + c f^2)=0$$

and discriminant $D$,

$$D=-3200c^3d^2e^2 - 2160d^4e^2 + 6400c^4e^3 + 5760c d^2e^3 - 2560c^2e^4 + 256e^5 + 5120c^3d^3f + 3456d^5f - 11520c^4d e f - 10080c d^3e f + 4480c^2d e^2f - 640d e ^3f + 3456c^5f^2 + 2640c^2d^2f^2 - 1440c^3e f^2 + 360d^2e f^2 + 160c e^2f^2 - 120c d f^3 + f^4$$

has a root $p$ such that $p^2$ is rational, then $(1)$ is a solvable quintic."

See "Commentary on an unpublished lecture by G. N. Watson on solving the quintic" by Bruce Berndt. This is easily implemented in Mathematica and is useful when dealing with parametric quintics.

Method 2:

If you are in a rush and just want to determine if a particular equation is solvable, you can find the order of its Galois group using this online Magma calculator. For example, to test the solvable but irreducible $x^5-5x+12=0$, copy and paste the command,

Z := Integers(); P < x > := PolynomialRing(Z); f := x^5-5*x+12; G, R := GaloisGroup(f); G;

One then finds the order is $10$, hence that quintic is solvable. All groups with order $<60$ are, though there are solvable groups with order $>60$.

Note: Don't forget the asterisk (*) between the numerical coefficient and the variable, like this: 5*x.

Addendum on Method 1:

We can reduce Method 1 to a simple rational root test by "root squaring".

Render the sextic equation above as

$\alpha p^6 + \beta p^4 + \gamma p^2 + p\sqrt{D} + \delta = 0$

Separate the odd degree term from the even degree ones:

$\alpha p^6 + \beta p^4 + \gamma p^2 + \delta = - p\sqrt{D}$

And square. Only even powers of $p$ now appear so we have a polynomial equation in $p^2$::

$(\alpha p^6 + \beta p^4 + \gamma p^2 + \delta)^2=p^2D$

$\alpha^2(p^2)^6 + 2\alpha \beta(p^2)^5 + (2\alpha \gamma + \beta^2)(p^2)^4 + 2(\alpha \delta + \beta \gamma)(p^2)^3 + (2\beta \delta + \gamma^2)(p^2)^2 + (2 \gamma \delta - D)(p^2) + \delta^2 = 0$

You may then apply the standard rational root test to this equation. If it passes, the conditions for solvability of the quintic are satisfied.

$\endgroup$
  • $\begingroup$ Is the condition in method $(1)$ both necessary and sufficient? $\endgroup$ – MathematicsStudent1122 Dec 15 '16 at 2:32
  • $\begingroup$ @MathematicsStudent1122: Yes, there is actually a relatively simple transformation that transforms this 1930s sextic to the modern version derived by Dummit in his well-known paper. $\endgroup$ – Tito Piezas III Dec 15 '16 at 2:35
  • $\begingroup$ Minor point here: when you say all groups with order $<60$ are solvable, do you mean $\le 60$? $\endgroup$ – Oscar Lanzi Mar 15 '18 at 10:54
  • $\begingroup$ @OscarLanzi: Strictly $<60$. Note that $A(5)$ has order $=60$ and is unsolvable. $\endgroup$ – Tito Piezas III Mar 15 '18 at 12:47
  • $\begingroup$ OK. What prompted me to ask was, you did not include order=60 in your discussion. $\endgroup$ – Oscar Lanzi Mar 15 '18 at 13:06

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.