This morning at the Godsil 65 conference, Cheryl Praeger was wearing a T-shirt she was a bit ashamed of. It had been produced by the Australian Mathematics Trust to commemorate Niels Abel’s anniversary, and featured a quintic equation (he was the mathematician who first showed conclusively that the quintic is not soluble by radicals). Unfortunately the quintic they had chosen has an integer root!

So here is my suggestion for a problem which should be easy if you have done any Galois theory. The other thing Abel is remembered for is abelian groups, so why not an irreducible quintic with abelian Galois group?

**Exercise:** Find a simple example of such a quintic.

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## About Peter Cameron

I count all the things that need to be counted.

x^5-x^4-4*x^3+3*x^2+3*x-1 has Galois group of order 5. (stolen from http://world.std.com/~jmccarro/math/GaloisGroups/degree5.html#5T1). Here is a check with Sage:

sage: var(‘x’);

sage: K=NumberField(x^5-x^4-4*x^3+3*x^2+3*x-1,’s’)

sage: K.galois_group().order()

5

No need for the internet or Sage! Take the polynomial satisfied by the five quantities u+u

^{−1}as u ranges over the 11th roots of unity. Takes half a dozen lines to work out by hand, and can be done even during a talk that you are listening to!But never mind, you get full marks.

$f(x)=x^5+2$ can also be worked out pretty quickly. Obviously its splitting field over $\Q$

is $K=\Q(\zeta_5,2^{1/5})$, where $\zeta_5$ is any primitive fifth root of unity. It is then easily seen that the group of $\Q$-automorphisms of $K$ is isomorphic to the group of affine transformations of the affine line over the field of $5$ elements. In particular $K/\Q$ is an abelian extension.

I don’t think this would really do for a T-shirt, where you can write down the polynomial but not so easily the base field.

In fact the problem itself is slightly misconceived, since there is no quintic polynomial which both has an abelian Galois group and is not soluble by radicals.