Solvable polynomials ed

Introduction ed

In school you learn: quadratic equations\[p(x) = ax^2 + bx + c = 0\]can be solved via\[ x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Similarly, formulas for equations/polynomials of degree 3 and 4 have been found, using more and higher roots. But the pattern breaks for degree 5 and beyond...

To undstand why, we need to know 3 incredients:

• fields - sets that keep track of the possible numbers that can show up when calculating the solutions
• the hidden symmetry between different solutions (like \(x_1\) and \(x_2\))
• how these two are related

Fields ed

How to keep track of numbers that might show up during a calculation?

Basics ed

Let's start with a simple game: The only numbers we know are \(0\) and \(1\), but we know how to perform simple operations like \((+,-,\cdot,/)\). What numbers can we encounter?

Obviously, we can write\[(1+1), (1+1+1), (1+1+1+1), \dots \, ,\]so we need to include \(2,3,4,\dots\) (all positive integers). Also the negative ones, because\[(0-1), (0-1-1), \dots \, .\]So, at least, we need all integers \(\mathbb Z\).

We can also look at multiplication, but whatever integers we multiply, the result is again an integer, so multiplication does not force us to include more numbers. What about division? We get all fractions \(\frac{m}{n}\), like\[\frac{1}{2}, \frac{1}{3}, \frac{-13}{5},\dots \, .\]This will "fill" all spaces between integers and gives us the rational numbers \(\mathbb Q\).

\(\mathbb Q\) is finally "self consistent". Any operation (\(+,-,\cdot,/\)) will only give us numbers, that we already know. That's why \(\mathbb Q\) is called a field. And it is the smallest field starting with \(0\) and \(1\).

Yes, the real numbers \(\mathbb R\) are also a field.

Lazy notation ed

To make things easier later on, let's write\[ 2 \mathbb Z \]meaning "take any number from \(\mathbb Z\) and double it". This gives the set of all even numbers. You can also write \(2 \mathbb Z + 1\) for the odd numbers.

For the real numbers, rescaling and moving only produces the same set again:\[ r \mathbb Q = \mathbb Q, \qquad \mathbb Q + s = \mathbb Q \]for real \(r \neq 0\) and \(s\). (try yourself)

On the other hand, \(\mathbb Q + s\) for \(s\notin \mathbb Q\) will be more interesting...

Roots ed

To see, how fields show up when solving equations, let's look at some simple examples.

Square roots ed

What is a square root, like \(\sqrt{2}\)? We're looking for a number \(x\) that gives \(2\), when squared. I.e. it solves the equation \(x^2 = 2\) or\[ \quad x^2 -2 = 0 \, .\]

So, what if we play the not-knowing-numbers game? We need to start by knowing \(0,1,2\), because they show up in the equation \(x^2 - 2 = 0\). Again, this brings us to \(\mathbb Q\). Also in school, we learn that \(\sqrt{2} \notin \mathbb Q\). This means, we need more than just the basic math operations and the given numbers to solve the problem.

Ok, so let's cheat and simply add \(\sqrt{2}\) to the numbers we know. Since we want a field again, this also forces us to add \[(2\sqrt{2}), (3\sqrt{2}), (4\sqrt{2}), \dots \quad \mathrm{and} \quad (-\sqrt{2}), (-2\sqrt{2}), (-3\sqrt{2}), \dots \, ,\]but also\[(\frac{15}{3}\sqrt{2}),(\frac{-7}{13} + \frac{15}{3}\sqrt{2}),\dots \, .\]

In the end, for every pair of rational numbers \(r, s\), we will have to add\[ r + s \, \sqrt{2} \]and there won't be any overlap, because no other pair \(r', s'\) will give the same number. Using our lazy notation, we end up with the whole set\[ \mathbb Q + \sqrt{2}\,\mathbb Q \, . \]

The nice thing: we're finished now. There is no need to add products, because\[(a + b \, \sqrt{2}) \cdot (c + d \, \sqrt{2}) = (ac + 2 bd) + (bc + ad) \sqrt{2} \quad \in \mathbb Q + \sqrt{2}\,\mathbb Q\]The same is true for divisions.

New notation: Adding a number like \(\sqrt{2}\) to \(\mathbb Q\) and including all necessary numbers to make it a field again, let's write that \(\mathbb Q(\sqrt{2})\).

Conclusion: we just found, that\[\mathbb Q(\sqrt{2}) = \mathbb Q + \sqrt{2}\,\mathbb Q \, . \]

Because every number inside can be written uniquely in the form \(r + s \, \sqrt{2}\) with \(r,s \in \mathbb Q\), we can think of the new numbers as points on a two-dimensional grid, with \((x,y)=(r,s)\) as coordinates. In that sense, \(\mathbb Q(\sqrt{2})\) is "two-dimensional".

Oops, did we forget the other solution of \(x^2 = 2\)? Nope, since that would be \(-\sqrt{2}\) and we already included that.

Complex numbers ed

Complex numbers come from the old question "what is \(\sqrt{-1}\)?". As an equation that is\[x^2 + 1 = 0\]and there are no solutions in \(\mathbb Q\), not even in the real numbers \(\mathbb R\). So, some crazy people simply defined the solution to be an "imaginary" number \(\sqrt{-1} = i\). This implies \(i \cdot i = -1\).

If we played the game again and add \(\sqrt{-1}\) into \(\mathbb Q\), we would get the same result as we did with \(\sqrt{2}\). I.e. every number in \(\mathbb Q(i)\) is just of the form \(r + s \, i\) with \(r,s \in \mathbb Q\). So, again \(\mathbb Q(i) = \mathbb Q + i \, \mathbb Q\) is two-dimensional.

Of course, the "official" complex numbers start from the real numbers:\[ \mathbb C = \mathbb R(i) = \mathbb R + i \, \mathbb R \]

Cube roots and beyond ed

What about \(\sqrt[3]{2}\)? That would be\[x^3 - 2 = 0 \, .\]

Well, since \(\sqrt[3]{2} \notin \mathbb Q\), we can play the game again and try to build \(\mathbb Q(\sqrt[3]{2})\) by all numbers \(r + s \, \sqrt[3]{2}\). But this time, we're not finished. The reason is, that \(\sqrt[3]{2} \cdot \sqrt[3]{2}\) is still not rational. Only after going up to \(\sqrt[3]{2} \cdot \sqrt[3]{2} \cdot \sqrt[3]{2}\) we're back to a rational number. So, this time we need three copies:\[ r + s \, \sqrt[3]{2} + t \, (\sqrt[3]{2})^2 \]

Does that mean\[\mathbb Q(\sqrt[3]{2}) = \mathbb Q + \sqrt[3]{2} \, \mathbb Q + (\sqrt[3]{2})^2 \, \mathbb Q\]is three-dimensional? ...yes, but we're missing the other two solutions of the equation \(x^3 - 2 = 0\)...

Unit circles ed

Let's go simple for now and first look at\[x^3 - 1 = 0\](replacing the 2 by a 1). Obviously one solution is \(x=1\). The other two solutions are hiding in the complex numbers.

There is a complex number, let's call it \(\epsilon\). It has distance \(1\) from the origin but is at a \(120^\circ\) angle upwards. Since multiplication of complex numbers can be done by adding angles... we get \(\epsilon^2\) below the real line and with \(\epsilon^3 = 1\) we're back on the real line. Together these three numbers \(1, \epsilon, \epsilon^2\) are the three solutions of \(x^3 - 1 = 0\). Does this make \(\mathbb Q(\epsilon)\) three-dimensional? ...no...

It turns out, there is a hidden relation \(\epsilon^2 = -\epsilon - 1\) and we don't even need to include \(\epsilon^2\), we only need \(\epsilon\). This makes \(\mathbb Q(\epsilon)\) only two-dimensional:\[\mathbb Q(\epsilon) = \mathbb Q + \epsilon \, \mathbb Q \]

What about higher \(\sqrt[4]{1}, \sqrt[5]{1}, \sqrt[6]{1}, \dots\)?

In general, the solutions of \(x^n - 1 = 0\) will always be \(n\) points on the unit circle in \(\mathbb C\), equally spaced. And there is always a "smallest angle" element \(\epsilon\) giving all other solutions via \(\epsilon, \epsilon^2, \epsilon^3,\dots\).

Sadly, dimensions of \(\mathbb Q(\sqrt[n]{1})\) are a bit messy. At least, the dimension will always be \(\le n\).

Back to cube roots ed

Now we know, how to deal with \(\sqrt[3]{2}\) and \(x^3 - 2 = 0\). Yes, we need the real numbers \(\sqrt[3]{2}, (\sqrt[3]{2})^2\), but we also need the complex \(\epsilon = \sqrt[3]{1}\), because the other two solutions of \(x^3 - 2 = 0\) are \(\epsilon \sqrt[3]{2}\) and \(\epsilon^2 \sqrt[3]{2}\).

If we want to fully solve the equation \(x^3 - 2 = 0\), we first need to go through the three-dimensional \(\mathbb Q(\sqrt[3]{2})\). But when also including \(\epsilon\), we end up with\[\mathbb Q(\sqrt[3]{2}) + \epsilon \, \mathbb Q(\sqrt[3]{2})\]or\[ \mathbb Q + \sqrt[3]{2} \, \mathbb Q + (\sqrt[3]{2})^2 \, \mathbb Q + \epsilon \, \mathbb Q + \epsilon \sqrt[3]{2} \, \mathbb Q + \epsilon (\sqrt[3]{2})^2 \, \mathbb Q \, . \]That is a \(6\)-dimensional space!

The product \(6=3 \cdot 2\) here reflects the two steps: First we include \(\sqrt[3]{2}\), giving 3 copies of \(\mathbb Q\). Then we include \(\epsilon\), so we get two copies of \(\mathbb Q(\sqrt[3]{2})\).\[ \mathbb Q \stackrel{\times 3}{\longrightarrow} \mathbb Q(\sqrt[3]{2}) \stackrel{\times 2}{\longrightarrow} \mathbb Q(\sqrt[3]{2}, \epsilon) \]

Polynomials ed

Solutions and dimensions ed

What about higher degree polynomials, like\[ p(x) = a \, x^n + b \, x^{n-1} + c\, x^{n-2} + \dots \]with rational coefficients?

In school we learn, there should be \(n\) solutions, but those might be complex numbers. We are now searching for the simplest set of numbers (field) that includes all the solutions.

If we somehow know one solution, calling it \(Z_1\), we can always rewrite \(p(x)\) as\[ p(x) = (x - Z_1) \cdot q(x) \]with a new polynomial \(q(x)\) of degree \(n-1\). If we could repeat this same idea for a solution \(Z_2\) of \(q(x)=0\) and so on, we would end up with\[ p(x) = a \cdot (x - Z_1) \cdot (x - Z_2) \cdot (x - Z_3) \dots \, . \]

Let's look at the first solution \(Z_1\) and construct \(\mathbb Q(Z_1)\). If \(Z_1 \in \mathbb Q\), then \(\mathbb Q(Z_1) = \mathbb Q\). If not, then \(\mathbb Q(Z_1)\) will be at most \(n\)-dimensional. The reason being, that \(Z_1\) solves the equation \(p(Z_1) = 0\), or\[ (Z_1)^n = -\frac{1}{a} \big(b (Z_1)^{n-1} + c (Z_1)^{n-2} + \dots \big) \]And so, an \(n\)-th power of \(Z_1\) can be replaced by lower powers.

Repeating this procedure, we can also include \(Z_2\) and look at \(\mathbb Q(Z_1, Z_2)\). This is at most \(n \cdot (n-1)\)-dimensional. After going through all the solutions, we end up with \(\mathbb Q(Z_1,\dots,Z_n)\) being at most \(n(n-1)(n-2)\dots = n!\)-dimensional.

For \(x^3 - 2 = 0\), we found exactly the \(3! = 6\) dimensions!

Strategy for solving ed

For \(n=2\), the famous formula\[ x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]tells us to calculate a strange, but rational number \(b^2 - 4ac\). Then we have to take the square root. This is the step, where we (potentially) leave \(\mathbb Q\) and have to expand into \(\mathbb Q(\sqrt{b^2 - 4ac})\). All further steps will stay in that set of numbers and we find the solutions in the two-dimensional\[ x_{1,2} \in \mathbb Q(\sqrt{b^2 - 4ac}) \, . \]

For \(n=3\), the approach is similar (but uglier). Again, we calculate strange, rational numbers. Then a square root brings us into the two-dimensional \(\mathbb Q(\sqrt{\dots})\). But then, there are more steps, involving cube roots, pushing us into the \(2 \cdot 3 = 6\) dimensional \(\mathbb Q(\sqrt{\dots}, \sqrt[3]{\dots})\). And for \(n=4\), there is another step into \(2 \cdot 3 \cdot 4 = 24\) dimensions via \(\sqrt[4]{\dots}\).

The step to \(n=5\) now seems obvious and we already know, there will be \(5!\) dimensions. So, what's the problem?

In short, the pattern, in which we grow from \(\mathbb Q\) through the different dimensions has some hidden structure that is encoded into "symmetries". And this structure will create problems for \(n \ge 5\).

Symmetry ed

Degree 2

For the equation \(x^2+1=0\) we picked one of the two solutions \(\pm \sqrt{-1}\) to write all elements of \(\mathbb Q(\sqrt{-1})\) as \(\alpha = x+y\,\sqrt{-1}\). We could randomly exchange\[\sqrt{-1} \; \leftrightarrow \; -\sqrt{-1} \, .\]This would also replace any \(\alpha\) with\[ \alpha \; \leftrightarrow \; \overline \alpha = x-y\,\sqrt{-1} \, . \]More interestingly, this operation keeps all calculations we do with these numbers correct, i.e. you flip the numbers, do your calculations and then flip back:\[ \alpha + \beta = \overline{\overline \alpha + \overline \beta}, \qquad \alpha \cdot \beta = \overline{\overline \alpha \cdot \overline \beta} \]

The same is true for \(x^2-2=0\) with \(\sqrt{2} \leftrightarrow -\sqrt{2}\).

Degree 3

For \(x^3-2=0\) it is still true, but since there are 3 solutions, you have more options. You can do 3 basic flips:\[ 1 \leftrightarrow \epsilon \quad \mathrm{or} \quad 1 \leftrightarrow \epsilon^2 \quad \mathrm{or} \quad \epsilon \leftrightarrow \epsilon^2 \]Actually you can also do 2 multi-flips\[ (1 \rightarrow \epsilon), (\epsilon \rightarrow \epsilon^2), (\epsilon^2 \rightarrow 1) \quad \mathrm{or} \quad (1 \leftarrow \epsilon), (\epsilon \leftarrow \epsilon^2), (\epsilon^2 \leftarrow 1) \]as well as the most boring flip: keep everything the way it is. Together, there are 6 possible flips.

In general

Galois ed

A suspicion: "Are the \(n!\) dimensions and the \(n!\) permutations related?"

Mr Galois found out: Yes, but the relation is even deeper!

The 6-dimensional space from \(x^3-2=0\) had a sub-structure, because the \(6=2 \cdot 3\) dimensions were built up step-by-step. On the other hand the set of 6 permutations also has a sub-structure:

More about permutations

You can perform several permutations in a row, and the overall action can itself again be seen as a single permutation. Also all permutations can be reversed and the reverse is a permutation. Mathematicians call a set like this a "group".

But there are certain sub-sets of permutations that already form groups. For example from the 6 permutations of 3 items \((A,B,C)\), pick the single flip \(A \leftrightarrow B\). If you perform this action twice, you get the leave-everything-alone permutation. But these two together can be performed as often as you want without ever creating any other permutations.

Btw. this sub-group can also be found by asking, which permutations leave item \(C\) unchanged.

Another group hidden inside the permutations are the even flips consisting of all permutations you can perform by an even number of single flips.

Galois relation

The Problem ed

Solving \(p(x)=0\) by adding roots would involve multiple steps\[ \mathbb Q \stackrel{\times k}{\longrightarrow} \mathbb Q(\sqrt[k]{A}) \longrightarrow \mathbb Q(\sqrt[k]{A}, \sqrt[l]{B}) \longrightarrow \dots \]and we already know, that each step of adding a simple root \(\sqrt[k]{A}\) involves a "minimal-angle" complex number \(\epsilon\) and some powers \(\epsilon^2, \epsilon^3,\dots,\epsilon^{k-1}\).

This shouls be reflected in the subgroups of the permutations. In each step, there should be a subgroup of permutations. And like the powers of \(\epsilon\), there should be a single permutation, that when performed repeatedly, creates all permutations in that subgroup.

For \(n\le 4\), this is always the case. But permutations have the subgroup of even-flips, that for \(n \ge 5\) breaks this behaviour.... argh

Some extra fun ed

We're done! The next two sections are just for fun!

Modulo (*) ed

The modulo operation is very simple but can twist the brain... We had the even numbers \(2 \mathbb Z\) and the odd numbers \(1 + \mathbb Z\). What about the set \(2 + 2\mathbb Z\)? Clearly, we just get the even numbers, again. In general,\[ n + 2\mathbb Z \]give either the even or the odd numbers, depending \(n\) being even/odd.

Another situation: the (analog) clock has 12 hours. When the clock says "3", it could mean any hour of the form \(3+12m\). The "3" represents the whole set of \(3 + 12 \mathbb Z\). And if we take all the possible hours for all eternety (\(\mathbb Z\)), the clock will sort them into 12 distinct categories/sets, labeled "1"..."12" and these sets are\[ (1 + 12\mathbb Z), \dots, (12 + 12\mathbb Z) \]Higher labels are redundant, because "13" \(=13 + 12\mathbb Z = 1 + 12 \mathbb Z=\) "1".

More generally, instead of 2 or 12, we can take any integer \(N \neq 0\) and look at all the possible sets\[ n+N\mathbb Z \]Of course, most of these sets will turn out to be the same:\[ n+N\mathbb Z = n'+N\mathbb Z \quad \mathrm{whenever} \quad n-n' \in N\mathbb Z \]We will call the smaller sets "classes" and write\[ [n] = n + N \mathbb Z \]The set of all these classes is\[ \mathbb Z / (N \mathbb Z) \quad "\mathbb Z \, \mathrm{modulo} \, N\mathbb Z"\]Like the labels on the clock, each class \([n]\) can have a "representative" \(n\). But, different representatives can give the same class.

Constructing solutions out of thin air (*) ed

The set of all polynomials \(p(x)\) with real coefficients: \(\mathbb Q[x]\).

What, if we don't have a solution of \(p(x) = 0\)? Well, there is an evil trick. Let's pick \(p(x) = x^2 - 2\) and look at the (quite complicated) set\[ \mathbb Q[x] / \big(p(x) \, \mathbb Q[x]\big) \]Any element is actually a whole class of polynomials of the form\[ [q] = q(x) + p(x) \, \mathbb Q[x] \]

In our case, let us define a class of polynomials\[ \epsilon = [x] = x + p(x) \, \mathbb Q[x] \]Then...TODO\[ p(\epsilon) = \epsilon^2 - 2 = x^2 + 2x \, p(x) + p(x)^2 + 1 + p(x) \, \mathbb Q[x] = 0 + p(x) \, \mathbb Q[x] = [0] \]