Walkthrough of Galois theory

Symmetries of polynomials; the Galois group

You have often seen that there is a certain fundamental symmetry between the roots of a polynomial. For example, there is an obvious symmetry between $i$ and $-i$, the roots of $x^2+1=0$. But even without imaginary numbers, there is a certain symmetry between $2+\sqrt{3}$ and $2-\sqrt{3}$, the roots of $x^2-4x+1=0$. 

What is this symmetry precisely? The idea is this: $i$ and $-i$ can only "relate" to the reals through squaring -- thus any equation in them with real coefficients remains true up to permutation of roots. In equations like $\alpha^2+1=0$, $\alpha+\beta=0$, etc. one may permute the roots $\alpha,\beta$ and maintain the equations. The only way you can break this symmetry is with an equation whose coefficients are themselves complex. 

It's not necessary that every permutation is a symmetry of the polynomial. For example, consider the equation $x^4-10x^2+1=0$, whose roots are $\sqrt2+\sqrt3$, $\sqrt2-\sqrt3$, $-\sqrt2+\sqrt3$, $-\sqrt2-\sqrt3$. In fact, the only permutations that are symmetries of these roots are: $e$ (which conjugates nothing), $(12)(34)$ (which conjugates $\sqrt3$), $(13)(24)$ (which conjugates $\sqrt2$), $(14)(23)$ (which conjugates both $\sqrt2$ and $\sqrt3$) -- these permutations are the Klein 4-group $\mathbb{Z}_2\times \mathbb{Z}_2$. 

As a more extreme example, consider $x^2-3x+2=0$, whose roots are $1$ and $2$. There is no non-trivial permutation of these roots that preserves e.g. the equation $\alpha-1=0$.

This group of symmetries is known as the Galois group.

How might we formalize this? Well, you may notice that none of what we described really depends on the polynomial -- just its roots. Fundamentally, we are "extending" the rational numbers with these roots -- e.g. to $\mathbb{Q}[i]$ or $\mathbb{Q}[\sqrt2]$ or whatever -- then, instead of saying "a permutation of these roots that preserves all $\mathbb{Q}$-algebraic equations in them", we can say "a field automorphism that fixes $\mathbb{Q}$" (because under such an automorphism, all arithmetical operations and $\mathbb{Q}$ coefficients remain the same); this is nice, because it is more natural to think of complex conjugation as a symmetry of the complex plane as a whole rather than just .

Thus our formal definition: the Galois group of some field extension $L/K$ is its automorphism group (the group of automorphisms of $L$ that fix $K$). 

Why is this actually useful or relevant? It's a natural construction so of course it's useful, go perforate your head IDC.

Why field theory? Constructible numbers

Geometrical constructability is defined as follows: given some initial geometric figure (collection of points with known distances), what geometric figure can be constructed with only a straightedge and compass? 

More formally, we have the following axioms -- to define constructible points, shapes and numbers. Where $G$ is is a collection of points in the plane:

  1. The points in $G$ are constructible points.
  2. A line through two constructible points is a constructible shape.
  3. A circle with its centre a constructible point and its radius a constructible line is a constructible shape.
  4. The intersections of two constructible shapes are constructible points. 
  5. The co-ordinate of a constructible point is a constructible number.


In fact, these are the axioms for constructability with a collapsible compass -- a compass whose legs collapse once it is taken off the paper -- so you cannot use it to mark an identical distance elsewhere. For constructability with a non-collapsible compass, Axiom 3 would instead read: A circle with its centre a constructible point and its radius equal to the length of a constructible line is a constructible shape. 

Exercise: show that these definitions are equivalent -- that under the axioms for a collapsible compass, you can move a line to an arbitrary point. Solution.

In particular, this means that constructability only depends on the initial set of distances, not the precise points themselves, because any arrangement of these lengths can be shifted around -- we can formulate the question as "what numbers are constructible from some initial set of numbers?".

Exercises: Given numbers $a$ and $b$, show that the following numbers are constructible from them:

  1. $a+b$
  2. $|a-b|$
  3. $ab$
  4. $a/b$
  5. $\sqrt{a}$

Numbers expressible through only these operations: $+,-,\times,/,\sqrt{\cdot}$ are called algebraically constructible from some base set of numbers. 

The above exercise demonstrates that all algebraically constructible numbers are geometrically constructible.

Conversely, as all geometrically constructible lengths can be constructed using these operations (you know, Pythagoras theorem and stuff), all geometrically constructible numbers are algebraically constructible.

Thus algebraical and geometrical constructability are equivalent. 

When no base figure is specified, the tacit assumption is that the base figure is a line segment of length 1, i.e. the base set is "1". We simply call the numbers constructible from this to be the constructible numbers

Other forms of constructability besides straightedge-and-compass are known -- such as conic constructability, solid constructability, neusis constructability, origami constructability. We want a theory that handles not only straightedge-and-compass (i.e. square roots), but also these more general forms.

This immediately demonstrates why:

  • It is impossible to double the cube.
  • It is impossible to trisect an arbitrary angle (because that is equivalent to constructing a cube root).
  • It is impossible to square the circle (because $\pi$ cannot be constructed with radicals). 
These are not proofs per se ... it still remains to be shown that you literally can't construct a cube root with some finite nesting of square roots, and so on. 

A set closed under $+,-,\times,/$ is a field. So constructible numbers are some type of fields. What kind of fields, and what do we do with them? Yeah, yeah, something.

Computing the Galois group

So how do we actually figure out the Galois group of a thing?

First of all, we said that it's really the roots that matter, not the polynomial. Does that mean that we can adjoin any elements to a field and calculate the Galois group of that extension?

Not really. There's nothing interesting to be said about the automorphism group of $\mathbb{Q}[\sqrt[4]{2}]$, for example. $x^4-2=0$ has imaginary roots and stuff, and we want to be able to permute them. The automorphism group of this field -- with only one root adjoined -- says nothing of the properties of $x^4-2=0$. 

The fundamental theorem of Galois theory -- which we haven't yet stated -- will not apply to such a shitty field extension.

No -- the field extensions we are interested in are those generated by adjoining all the roots of a polynomial. This is a "normal extension", "Galois extension", "splitting field", whatever (yeah, yeah, a Galois extension must be normal and separable but do I look like the kind of guy who'd bother with things that aren't separable? -- what even is separable, your FACE is separable, I'll split it in two).

We can "intuit" out the Galois group of things.

The splitting field of $x_4-1$ is $\mathbb{Q}[i]$; then the image of $i$ determines the Galois group, which is thus $S_2$.

The Galois group of $x^4+1$ is $K_4$. Prove it.

The splitting field of $x^4-2$ is $\mathbb{Q}[\sqrt[4]{2},i]$; then the images of $\sqrt[4]{2}$ (for which there are 4 options) and $i$ (for which there are 2 options) are sufficient to determine the Galois group, which is thus $D_4$.

The Galois group of $\mathbb{Q}[\sqrt{5},\sqrt{7}]$ is $K_4$. Prove it.

The Galois group of $x^3-1$ is $S_2$. Prove it.

The Galois group of $x^3+1$ is $S_2$. Prove it.

The Galois group of $x^3-2$ is $S_3$. Prove it.

The Galois group of $x^5-1$ is $C_4$. Prove it.

The Galois group of $x^5+1$ is $C_4$. Prove it.

The Galois group of $x^5-2$ is ... okay, just see this Math.SE question. It's some semidirect product, and you should be able to see why by now.

If you worked through the examples above, you should have a good intuitive grasp of the Fundamental theorem of Galois theory by now. We've been constructing the Galois group of a field extension by determining some number of "basis elements" whose images suffice to determine the automorphism; extensions by these basis elements (intermediate field extensions) correspond to specific subgroups (intermediate subgroups of the subgroup lattice).

Solvable groups


1 comment:

  1. AMAZING! It makes everything so much easier. Can I write Solvable group section for your blog?