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

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, 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 \(1\).

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 add \(\sqrt{2}\) to the numbers we know. This also forces us to know \(2\sqrt{2}, 3\sqrt{2},4\sqrt{2},\dots\) and \(-\sqrt{2},-2\sqrt{2},-3\sqrt{2},\dots\), but also \(\frac{15}{3}\sqrt{2},\dots\). So for every rational number\fraction \(\frac{n}{m}\) we have to add \(\frac{n}{m}\sqrt{2}\). We end up with adding exactly a whole copy of \(\mathbb Q\) here: for every rational number, we get a new number, but there is no "overlap".

The nice thing: we're finished now. There is no need to add things like \(5 \sqrt{2} \cdot \sqrt{2}\), because \(\sqrt{2} \cdot \sqrt{2} = 2 \in \mathbb Q\). The same is true for divisions. This means:

Adding \(\sqrt{2}\) to \(\mathbb Q\) gives us another field made from two copies of \(\mathbb Q\)

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 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 \(\mathbb Q(i)\) 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, \(\mathbb Q(\sqrt[3]{2})\) is three-dimensional? ...no quite...

The equation \(x^3 - 2 = 0\) has not just one solution, but three...

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)

The 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 \(x^3 - 1 = 0\). And in this case, \(\mathbb Q(\epsilon)\) is really just three-dimensional.

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.