Solvable polynomials ed

Introduction ed

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

(explicit example for \(a,b,c\)) Playing a game: you don't know any numbers, but know how to calculate. Start with knowing \(p\), so we know \(a,b,c\). Include \(b^2\) etc. and the root...

Fields ed

Basics ed

Same game, more extreme. You know \(0\) and \(1\). 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 \(\frac{1}{2}, \frac{1}{3}, \frac{-13}{5},\dots\), this will "fill" all spaces between integers and we get the rational number \(\mathbb Q\) of all fractions \(\frac{m}{n}\).

\(\mathbb Q\) is finally "self consistent". Any operation (\(+,-,\cdot,/\)) will only give you numbers, that you already know. \(\mathbb Q\) 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 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:\[ r \cdot \mathbb Q = \mathbb Q + s = \mathbb Q \]for real \(r \neq 0\) and \(s\). (proof left to the reader)

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

Roots ed

Square roots ed

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

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, we end up with \(\mathbb Q\). 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. This also forces us to add \(2\sqrt{2}, 3\sqrt{2},4\sqrt{2},\dots\) and \(-\sqrt{2},-2\sqrt{2},-3\sqrt{2},\dots\), but also \(\frac{15}{3}\sqrt{2},\dots\). So for every rational number \(r\) we have to add \(r \, \sqrt{2}\). 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 like \(5 \sqrt{2} \cdot \sqrt{2}\), because \(\sqrt{2} \cdot \sqrt{2} = 2 \in \mathbb Q\). 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 \(\mathbb Q(\sqrt{2})\).

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

This means: this gives us another field made from two copies of \(\mathbb Q\) and every number inside can be written uniquely in the form \(x + y \sqrt{2}\) with \(x,y \in \mathbb Q\). Because you need two rational number to describe the new number, you could say, \(\mathbb Q(\sqrt{2})\) 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 just defined \(\sqrt{-1} = i\). This implies \(i \cdot i = -1\).

(drawing of the complex plane)

If we play 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 \(x + i \, y\) with \(x,y \in \mathbb Q\). So, again \(\mathbb Q(i) = \mathbb Q + i \, \mathbb Q\) is two-dimensional.

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 \(x + \sqrt[3]{2} \, y\). 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:\[ x + \sqrt[3]{2} \, y + (\sqrt[3]{2})^2 \, z \]

So,\[\mathbb Q(\sqrt[3]{2}) = \mathbb Q + \sqrt[3]{2} \, \mathbb Q + (\sqrt[3]{2})^2 \, \mathbb Q\]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 \(x^3 - 1 = 0\) (replace the 2 by a 1). Obviously one solution is \(x=1\). The other two solutions are hiding in the complex numbers.

(drawing of complex plane)

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 number \(1, \epsilon, \epsilon^2\) are the three solutions of \(x^3 - 1 = 0\). Does this make \(\mathbb Q(\epsilon)\) three-dimensional? ...no...

It turns out \(\mathbb Q(\epsilon)\) is only two-dimensional, because we can build \(\epsilon^2\) already by \(\epsilon^2 = -\epsilon - 1\). Also\[\mathbb Q(\epsilon) = \mathbb Q + \epsilon \, \mathbb Q \, . \]

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 \(\epsilon, \epsilon^2, \epsilon^3,\dots\). This \(\epsilon\) is called a "primitive root".

(image)

Sadly, dimensions of \(\mathbb Q(\sqrt[n])\) 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 \(\mathbb Q(\sqrt[3]{2})\).

Polynomials ed

Solutions and dimensions ed

What about higher order polynomials like\[ p(x) = a \, x^n + b \, x^{n-1} + \dots \]

In school we learn, there should be \(n\) solutions. 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\[ p(x) = (x - Z_1) \cdot (x - Z_2) \cdot (x - Z_3) \dots \, . \]

Let's assume all coefficients of \(p(x)\) are rational and 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 ...(explain)...

Repeating this procedure, we can look at \(\mathbb Q(Z_1, Z_2) = \big( \mathbb Q(Z_1) \big)(Z_2)\). 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!

Modulo (*) ed

The next two sections are just for fun!

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 nothing (*) 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\[ p(\epsilon) = \epsilon^2 - 2 = x^2 + 2x \, p(x) + p(x)^2 + 1 + p(x) \, \mathbb Q[x] = 0 + p(x) \, \mathbb Q[x] = [0] \]