But can I really solve this equation?

Important

This post is a work in progress. The second half is still unwritten, but I believe the existing first half is already large enough for an interesting reading and could be a post by itself.

Overture

Revolutionary mathematics. Pointless death.

You fancy writer, enough with the overblown intros, give me my math!

A young mathematician. A sleepless night. A mind well aware of the mathematical relevance of his fresh ideas. Years of frustration under self-imposed labels. Misunderstood. Ahead of his time. The eve of a fatal duel. A mathematical testament. A coquette de bas étage. Infatuation. A gunshot. An everlasting romanticized entry in the history books.

That’s the story you’ve heard, isn’t it?

History is harder. The figure of Évariste Galois as a precocious mind is undeniable. These stories are great engagements into mathematics, but are often exaggerated or fantasized. We’ll shortly try to get these down to earth in this introduction.

Figure 1: Sketch of Galois that you’re probably tired of seeing everywhere because there aren’t any others he looks less creepier than in the one his brother did from memory years after his death

Revolutionary mathematics. Évariste Galois essentially gave birth to two big branches of mathematics, group theory and Galois theory. What an unexpected name for the second one, right? We will soon start getting our hands dirty with the same mathematics that were driving the passion of a young Galois, and then you will have to judge whether they were really revolutionary mathematics. You will try to actively approach his life through his mathematics.

Galois had a short life. He died at the age of 20. Despite this short life, he wrote his arguably most important paper (the one which showcases his revolutionary ideas and often referred to as the first memoir) four years before his death, roughly at the age of 17. This is of course far from the common belief of him writing down all his theory just the night prior to his death. My intention in this post is going through this first memoir, to introduce you to Galois’ ideas as close as possible to how he himself viewed them back then. We will start doing this after a few shamefully long but I believe necessary motivation sections.

Pointless death. But why did he die so young? Ah, yes, these mathematicians needing some love and doing anything in their hands when they get a chance. And let’s just throw in some Reddit jokes because why not. In reality, this is an obscure part of Galois’ life and we can’t really tell for sure what happened. If you think dying in a duel for the woman you love is weird, wait until I tell you about this one.

In some accounts they say Galois staged his death. He was so involved into politics that he was willing to give his own life to the republicans, so that his death would help starting a riot and eventually, another revolution. I don’t know if you find this more believable than the love duel… But there’s some logic here. In his last letters, Galois was talking about his own death like it was something unavoidable. You tell me, if you were really going out there for a duel, you would try as hard as possible to win and thus live, right? You’re only sure about your death if you’re not willing to try in the first place.

This theory seems more reasonable to me, but who cares about my opinion? You can also have your own. Anyway, if this one was really the plan, history was evil. The riots on Galois’ death were supposed to start at his funeral, but just one day after his death, the one of the well-known general Lamarque followed. These news completely eclipsed those of the not so broadly known mathematician. Lamarque’s death did start a rebellion, but was soon stopped. If we look back at Galois, this was, indeed, a pointless death.

The ideas that originally led to the creation of new branches of math thanks to Galois can be summarized in a short phrase. The study of solutions of polynomial equations. If I give you an equation like \(x^{3} + 5x - 2 = 0\), can you find out which values of \(x\) satisfy this relation? This question quickly evolves into a more intriguing one… Can I always write down a formula for the solutions of any equation you give me? These are the sort of questions Galois and his math predecessors were interested in. Despite finding remarkable answers at such an early age, we shouldn’t feel discouraged. We should rather reframe our view. If the most basic form of this theory was originally developed by a high schooler, what prevents me from understanding it? As for almost anything, I think this barrier is just hard work. I like to think that Galois’ sleepless nights immersed into his mathematics speak for themselves.

We’re going to follow Galois on this journey.

You’ll have to do mathematics.

First degree

Here’s one for my kindergarten fans out there. Can you solve this equation?

\[ x + a = 0 \]

What’s \(x\)?

What’s \(a\)?

…

Just kidding.

Hint: Subtract \(a\) from both sides… And try clicking me!

You might see boxes like this one throughout the article. If they have an arrow at the right end, it means they have hidden content and you can expand to see it by clicking on them.

Another thing worth mentioning. Writing a blog with aesthetically pleasant math formulas is totally not easy. They have to look good in all kinds of devices. For this reason, the safest workaround so that I don’t risk someone seeing absolute rubbish is to use horizontal scroll bars when the formula might otherwise overflow the screen. This will rarely appear on a large monitor but will be quite frequent in mobile devices, and I think there’s no way to force a scroll bar actually showing on mobile browsers, so I had to warn you. If you feel like a formula is going out of the screen, please try to scroll it even if you don’t see a scroll bar! Here, an example:

\[ \text{Unfortunately what is little recognized is that the most worthwhile scientific books are those in which the author clearly indicates what he does not know; for an author most hurts his readers by concealing difficulties.} \]

Second degree

I know that one

Our good ol’ friends the Babylonians were already solving second degree equations… Ok, yeah, they didn’t formulate them in the same way, but they really could solve them. For us, this is a second degree equation in the modern sense:

\[ ax^{2} + bx + c = 0 \]

Now it’s your time to shine… You do remember that formula you were forced to memorize in high school, don’t you? Yes!

\[ x = \frac{-b \pm \sqrt{b^{2} - 4ac} }{2a} \]

Huh…

This is supposed to be an entry to Summer of Math Exposition #4. If you check this link, you’ll find one part that goes like this:

Most Frequently Requested Topics

Foundation-Level Concepts (Ages 11-16):

1. Variables and algebra - Students don’t understand that “x” represents an unknown value to find

…

Well, well, we just started and I’m already digging my own pit. I think I’ll just wait for someone else to address that in a whole post. If you happen to be one of those confused students and you end up here, well, good luck with that. You have all your time. Don’t think the post will be an easy read. But as for the prerequisites, I don’t expect from you anything more than the algebra and complex numbers you were taught in high school.

While you probably knew the above formula with the coefficients \(a, b, c\) just like that, I’m going to be using the same letter with subscripts, where \(a_i\) is the coefficient of the term \(x^{i}\). Moreover, note that the coefficient \(a_n\) of the highest degree term (here \(a_2\)) is useless: the equation obtained from dividing everything by that coefficient will have the same roots.

Why are the roots the same?

If \(x_1\) is a solution of \(a_2 x^{2} + a_1 x + a_0 = 0\), then we have:

\[ a_2 x_1^{2} + a_1 x_1 + a_0 = 0 \]

But zero divided by any number will still be zero, so we have:

\[ \frac{1}{a_2} \cdot (a_2 x_1^{2} + a_1 x_1 + a_0) = \frac{1}{a_2} \cdot 0 = 0 \]

This is the same as:

\[ x_1^{2} + \frac{a_1}{a_2} x + \frac{a_0}{a_2} = 0 \]

This last relation exactly means that \(x_1\) is also a solution of the new equation, which has no coefficient for the \(x^{2}\) term.

So the equation we’ll work with looks like this:

\[ x^{2} + a_1 x + a_0 = 0 \]

We haven’t talked about which values can the coefficients of the equation take. We could decide that they can be any natural number (\(0, 1, 2\dots\)), any integer (\(0, 1, -1, 2, -2\dots\)), any rational… From now on, we’ll assume they’re rational numbers.

What’s a rational number?

This is just a fancy term for fractions, the result of dividing one integer by another. They are things like \(0, \frac{1}{2}, -\frac{57}{3}, -2\)… Of course, integers are also rationals because any integer \(a\) can be written as \(\frac{a}{1}\).

And why not choosing, say, just integers? Because we need some freedom with the operations we do with coefficients. When we talked about dividing everything by \(a_2\), we were already considering that the new coefficients had to be at least rational. Otherwise, depending on which division we perform, if we had integers initially, we could now end up with something that isn’t an integer anymore. Thus, if we choose rational coefficients we’re allowed to do basic operations (sum, subtract, multiply, divide) with them and always be sure that the result will still be a rational number.

It’s dividing fractions… What’s dividing a fraction by a fraction anyway?

— Only yesterday (1991)

No, I’m not going to explain how to divide fractions today. I just wanted an excuse to include a part of one of my favourite movies, where Taeko, the protagonist, shows her frustration of not being able to understand the logic behind dividing fractions. Ok, I get what I have to do, but why is it like that? This is a recurring theme in many fields, and math is no exception. I also feel like this, even while doing my research to write this post. Many suggest that math is something you start understanding while you do it, meaning that you shouldn’t force yourself too far to get the logic behind some concepts until you have already spent quite some time working with them. While I do agree with this, and is partly how I accept my own knowledge gaps, I will try my best in this article to build some intuition.

So just for the last time, let’s rewrite the equation and the solution with the new letters…

\[ x^{2} + a_1 x + a_0 = 0 \text{ where $a_0$, $a_1$ are rationals} \]

\[ x = \frac{-a_1 \pm \sqrt{a_1^{2} -4 a_0} }{2} \]

But how did you get there?

Shut up and memorize!

Just joking. Let’s see what we can do…

In the previous section we talked about equations and solutions. We might sometimes still use that terminology, but we’re also going to talk about polynomials and roots. For our purposes, an equation can be expressed as \(p(x) = 0\), where \(p(x)\) is a polynomial. In the quadratic example, the polynomial would be \(p(x) = x^{2} + a_1 x + a_0\). A root of the polynomial is any value that gives an output of \(0\) when plugged as \(x\) into that polynomial. Therefore, a solution of the equation is a root of the polynomial.

Suppose we give names \(x_1\), \(x_2\) to our hypothetical roots. If something is a root, as we just said, it should evaluate the polynomial to \(0\). This must work for both roots, so when factored out, the polynomial should look like \((x - x_1)(x - x_2)\). Here, if we change \(x\) by either \(x_1\) or \(x_2\), the result will be \(0\) as expected. Can you express the coefficients \(a_1\) and \(a_0\) in terms of the roots \(x_1\) and \(x_2\)?

Hint: try expanding the product \((x-x_1)(x-x_2)\)

Let’s expand the product:

\[ (x - x_1) (x - x_2) = x^{2} + (- x_1 - x_2) x + x_1 x_2 \]

But considering the original polynomial looked like \(x^{2} + a_1 x + a_0\), that means we have the following equalities for our roots:

\[ \left\{ \begin{aligned} x_1 + x_2 &= -a_1 \\ x_1 x_2 &= a_0 \end{aligned} \right. \]

The cool thing about these relations is that the result of adding or multiplying the roots is again a rational number (because our coefficients were rationals), even if the roots themselves might not be rational. In fact, knowing that, any weird expression made up of sums and products of roots, say \(\frac{(x_1 + x_2)^{3}}{3x_1 x_2} -4(x_1 + x_2)(x_1 x_2)^{6}\), will be rationally known. I’m using these words on purpose, because that’s how Galois himself referred to quantities that ultimately evaluate to a rational number, even if the operands themselves aren’t necessarily rationals (like our roots). Keep this cool observation in mind, you’ll see why in just a while.

So we have to figure out a formula for both roots. They’re just two now, yes, but you know this is going further, right? We will work with higher degree equations later. Wouldn’t it be cool if we could just find the value of one thing, and then get the value of all roots for free just from that? Of course, besides this weird value, we’re also allowed to use the original coefficients of the polynomial or any other known numbers.

I’ll help you with that. There are probably a lot of values we could choose to find, but I’ll give you this one:

\[ t = x_1 - x_2 \]

Of course you’re now supposed to check that this does encode everything about the roots \(x_1\) and \(x_2\). Can you write them in terms of \(t\)?

Hint: use also \(-a_1 = x_1 + x_2\)

Remember learning how to solve systems of equations in high school? Reduction method, is that you?

\[ \left. \begin{aligned} t &= x_1 - x_2 \\ -a_1 &= x_1 + x_2 \end{aligned} \right\} \Rightarrow \left\{ \begin{aligned} x_1 &= \frac{-a_1 + t}{2} \\ x_2 &= \frac{-a_1 - t}{2} \end{aligned} \right. \]

Do you see where this is going? Of course, we’ll now try to find the value of \(t\)! But why would this be easier?

Hint: try to find the value of \(t^{2}\) first

Well, instead of computing \(t\), let’s try with \(t^{2}\):

\[ t^{2} = (x_1 - x_2)^{2} = x_1^{2} + x_2^{2} - 2 x_1 x_2 = (x_1 + x_2)^{2} - 4 x_1 x_2 \overset{\text{cool obs.}}{=} a_1^{2} - 4 a_0 \]

You do know what happened in the last step, right? The cool observation we made earlier. We put everything in terms of \(x_1 + x_2\) and \(x_1 x_2\), so we can then use the known rational coefficients, and we’re done. We know what \(t^{2}\) is, so we can now pick a square root and have \(t = \sqrt{a_1^{2} - 4a_0}\).

Now that we know the value of \(t\), getting back the values of the original roots \(x_1\) and \(x_2\) is routine substitution in the equalities we stated earlier.

But before claiming our trophy in the form of a final result, let’s expand our perspective a bit. Suppose we craft another equation, of which we’re sure that \(t\) is a root, and we know how to solve this new equation. That way we would get the value of \(t\), just like above, hoping that it would be easier than trying to solve the original equation.

So… How does such an equation look like? The one I’ll show you won’t be too surprising, and you’ll even think I’m just making fun of you, but keep in mind I’m trying to prepare a recipe for future sections. Alright, the equation:

\[ (x - t)(x + t) = 0 \]

Of course this equation makes sense. We already made a useful observation before. If \(t\) has to be a root, there must be a factor like \((x - t)\) out there. The reason we also included \((x + t)\) is that now this product will give a neat result, as you might expect from before. Yes, we just happen to know that because we already calculated \(t^{2}\) before, and this seems to be some circular reasoning, but bear with me, this will be a useful reinterpretation later…

\[ \begin{array}{c} (x - t)(x + t) = 0 \\ \Downarrow \\ x^{2} - t^{2} = 0 \\ \Downarrow \\ x^{2} = t^{2} \\ \Downarrow \\ x = \sqrt{t^{2}} = \sqrt{a_1^{2} - 4 a_0} = t \end{array} \]

At this point you might be wondering why I never bother to write the typical sign \(\pm\) to represent that we could take either the positive or the negative value, in this case, when taking the square root. At least for our quadratic example, this is irrelevant! Why? Well, you just look back at the relations:

\[ \left\{ \begin{aligned} x_1 &= \frac{-a_1 + t}{2} \\ x_2 &= \frac{-a_1 - t}{2} \end{aligned} \right. \]

If we take the positive value, that is, \(t = \sqrt{a_1^{2} - 4 a_0}\), then we get the solutions:

\[ \left\{ \begin{aligned} x_1 &= \frac{-a_1 + \sqrt{a_1^{2} - 4 a_0}}{2} \\ x_2 &= \frac{-a_1 - \sqrt{a_1^{2} - 4 a_0}}{2} \end{aligned} \right. \]

If we instead take the negative value, i.e., \(t = -\sqrt{a_1^{2} - 4 a_0}\), then our solutions will be:

\[ \left\{ \begin{aligned} x_1 &= \frac{-a_1 - \sqrt{a_1^{2} - 4 a_0}}{2} \\ x_2 &= \frac{-a_1 + \sqrt{a_1^{2} - 4 a_0}}{2} \end{aligned} \right. \]

They’re the same but swapped! So who cares? We just wanted to find the roots in the first place, we didn’t even define what should be called \(x_1\) and what \(x_2\), there’s just no difference. This also somehow addresses another question you might have had. We want to find the value of the root \(t\) from the new equation. But how do we even know which one of the roots is \(t\)? Well, as we just said, it turns out any of them works… The more general answer to this question isn’t that easy, but will play a crucial role in our understanding of these equations, and we’ll get back to it later. No spoilers.

First symmetry

I bet you know what’s coming next… But before we try to tame that naughty cubic equation, we need some polynomial machinery. We’ll find some symmetries along the way, I hope you’re excited!

Let’s recall the cool observation from previous section:

\[ \left\{ \begin{aligned} x_1 + x_2 &= -a_1 \\ x_1 x_2 &= a_0 \end{aligned} \right. \]

The question is obvious… Can we find something similar for the cubic equation?

\[ x^{3} + a_2 x^{2} + a_1 x + a_0 = 0 \text{ where $a_0$, $a_1$, $a_2$ are rationals} \]

Yes we can! Hint: expand \((x-x_1)(x-x_2)(x-x_3)\)

We proceed the same way as for the quadratic. If the roots are \(x_1\), \(x_2\) and \(x_3\), we should be able to factor the equation like \((x - x_1)(x - x_2)(x - x_3)\). What do we get from here? You can check it yourself:

\[ (x - x_1)(x - x_2)(x - x_3) = x^{3} - (x_1 + x_2 + x_3) x^{2} + (x_1 x_2 + x_1 x_3 + x_2 x_3) x - x_1 x_2 x_3 \]

We got our relations!

\[ \left\{ \begin{aligned} x_1 + x_2 + x_3 &= -a_2 \\ x_1 x_2 + x_1 x_3 + x_2 x_3 &= a_1 \\ x_1 x_2 x_3 &= -a_0 \end{aligned} \right. \]

I know you’re seeing the pattern for the coefficients:

• Sum of all roots, negative sign
• Sum of all products of two different roots, positive sign
• Sum of all products of three different roots, negative sign
• …

Now pay attention to this. If we have an equation of degree \(n\), in coefficient \(a_i\), we have all possible different ways of choosing \(n-i\) roots. What happens if we swap any two of the roots? Of course, nothing! We rearranged some sums and products, but since the order doesn’t matter, and we were including all possible products, we still have the same result! We say that these expressions are symmetric polynomials. A polynomial on the roots \(x_1, x_2, \dots, x_n\) is symmetric if it doesn’t change no matter how we rearrange the roots.

The symmetric polynomials we’ve been working with actually have a special name: they’re called elementary symmetric polynomials. Just to make it clear, the elementary symmetric polynomials we’ve seen so far are the following:

For two roots

\[ \left\{ \begin{aligned} x_1 + x_2 \\ x_1 x_2 \end{aligned} \right. \]

For three roots

\[ \left\{ \begin{aligned} x_1 + x_2 + x_3 \\ x_1 x_2 + x_1 x_3 + x_2 x_3 \\ x_1 x_2 x_3 \end{aligned} \right. \]

Guess the ones for four roots

\[ \left\{ \begin{aligned} x_1 + x_2 + x_3 + x_4 \\ x_1 x_2 + x_1 x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4 \\ x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4 \\ x_1 x_2 x_3 x_4 \end{aligned} \right. \]

Any guess for why they’re called elementary? I’ll give you a few examples to think.

Let’s take for example this polynomial:

\[ x_1^{2} + x_2^{2} + x_3^{2} \]

It is obviously symmetric, since all three terms are squared and the order of the sum doesn’t matter. However, it’s not an elementary symmetric polynomial itself. The closest elementary one from above might be \(x_1 + x_2 + x_3\), but ours has each term squared, so it doesn’t fit that pattern exactly.

Can we actually know its value? Let’s rewrite it…

Hint: try to relate this value with \((x_1 + x_2 + x_3)^{2}\)

\[ x_1^{2} + x_2^{2} + x_3^{2} = (x_1 + x_2 + x_3)^{2} - 2(x_1 x_2 + x_1 x_3 + x_2 x_3) = a_2^{2} - 2 a_1 \]

Yes! That’s a known rational number.

Let’s try another symmetric polynomial:

\[ x_1^{3} x_2 + x_2^{3} x_3 + x_3^{3} x_1 \]

Oops! This is not a symmetric polynomial! Can you see why? It needs a small fix:

\[ x_1^{3} x_2 x_3 + x_1 x_2^{3} x_3 + x_1 x_2 x_3^{3} \]

Yes, much better now. Can you prove this one is indeed symmetric? Hint: it’s enough to check swaps of two roots. That’s because any rearrangement of roots can be explained as a series of swaps of two roots. Now, is this also a known value?

Hint: is this a product of two familiar expressions…?

\[ x_1^{3} x_2 x_3 + x_1 x_2^{3} x_3 + x_1 x_2 x_3^{3} = (x_1^{2} + x_2^{2} + x_3^{2} )(x_1 x_2 x_3) = (a_2^{2} -2 a_1) \cdot (-a_0) \]

Well, this one also worked. Do you see where I’m going?

We were already observing in a previous section that any expression written in terms of these elementary symmetric polynomials has a known rational value, no matter how complicated the expression may be. An expression written in terms of elementary symmetric polynomials is of course symmetric, because each of these terms is. But can we go further?

Any symmetric polynomial can be written in terms of elementary symmetric polynomials.

I hope you agree this is not obvious at all. And the statement just like that doesn’t seem too interesting. Because we know that elementary symmetric polynomials are known rational numbers, then we get this more insightful result for free:

Any symmetric polynomial on the roots is a known number.

We’ll use this result a few times later. Of course, we should prove the first statement about elementary symmetric polynomials. In Galois’ time, this was such a well-known fact that he didn’t even bother including it in the prerequisites for his memoir.

Unless you’re really really interested at this point, you could simply believe the statement is true and keep reading, and only come back later. Or you can skim through the proof and don’t stress yourself if something isn’t entirely clear. This is a really important result, but it’s just a building block for more things that are coming.

Why is the statement true?

This won’t be a super general rigorous proof, but you’ll get the idea and by the end you should be able to write down something yourself. The idea of the proof is constructive. What’s a constructive proof? We prove something exists by actually building a recipe that, if followed, will arrive at that thing. Sounds familiar? Yeah, I told you previously that we’re building another recipe, but more on that later. In this case, this thing is a way of writing the symmetric polynomial in terms of elementary symmetric polynomials.

To get an idea, let’s come up with a different, more naive way of finding that representation for one of the previous examples, the polynomial

\[ f = x_1^{3} x_2 x_3 + x_1 x_2^{3} x_3 + x_1 x_2 x_3^{3} \]

Suppose we would like to arrive at an expression that doesn’t include the first term \(x_1^{3} x_2 x_3\). Why that term you may ask? I promise, I’ll tell you later! We’ll try to find a clever symmetric polynomial that also has this term, so then we can get rid of it by subtracting this polynomial from the original \(f\) one. How would such a polynomial look like? Look at the coefficients of our term \(x_1^{3} x_2 x_3\). They’re \(3, 1, 1\). Now look at this cool expression:

\[ (x_1 + x_2 + x_3)^{3 - 1} (x_1 x_2 + x_1 x_3 + x_2 x_3)^{1 - 1} (x_1 x_2 x_3)^{1} = (x_1 + x_2 + x_3)^{2}(x_1 x_2 x_3) \]

This expression has a lot going on, but if you focus on each first term summing, that is, \(x_1\), \(x_1 x_2\) and \(x_1 x_2 x_3\), the expression certainly contains at least the following term:

\[ (x_1)^{3 - 1}(x_1 x_2)^{1 - 1}(x_1 x_2 x_3)^{1} = x_1^{3-1+1} x_2^{1-1+1} x_3^{1} = x_1^{3} x_2^{1} x_3^{1} \]

This is the term we wanted! We got it. We found a polynomial that doesn’t contain our term. How does it look like? Hopefully it’s simpler…?

\[ f - (x_1 + x_2 + x_3)^{2}(x_1 x_2 x_3) = -2 x_1^{2} x_2^{2} x_3 - 2 x_1^{2} x_2 x_3^{2} - 2 x_1 x_2^{2} x_3^{2} \]

Geez… Is this a joke? This seems as hard as the original polynomial! How did it get any better? Trust me… Let’s keep going! We now try to find another one that doesn’t contain the first term \(-2 x_1^{2} x_2^{2} x_3\). You already know how the trick works, don’t you? Except we now also have a \(-2\) in there, but that’s not a big deal, just multiply by \(-2\)… Recall our coefficients this time, they’re \(2, 2, 1\):

\[ -2 (x_1 + x_2 + x_3)^{2 - 2} (x_1 x_2 + x_1 x_3 + x_2 x_3)^{2 - 1} (x_1 x_2 x_3)^{1} = -2 (x_1 x_2 + x_1 x_3 + x_2 x_3)(x_1 x_2 x_3) \]

Again, this will at least contain the following term:

\[ -2 (x_1)^{2 - 2}(x_1 x_2)^{2 - 1}(x_1 x_2 x_3)^{1} = -2 x_1^{2-2+2} x_2^{2-1+1} x_3^{1} = -2 x_1^{2} x_2^{2} x_3^{1} \]

Got it! Let’s see what the new polynomial is…

\[ f - (x_1 + x_2 + x_3)^{2}(x_1 x_2 x_3) - (-2) (x_1 x_2 + x_1 x_3 + x_2 x_3)(x_1 x_2 x_3) = 0 \]

Oh, we must’ve been so lucky there, the result is \(0\)! Then we can just express \(f\) like this:

\[ f = (x_1 + x_2 + x_3)^{2}(x_1 x_2 x_3) -2 (x_1 x_2 + x_1 x_3 + x_2 x_3)(x_1 x_2 x_3) \]

These are all elementary symmetric polynomials! And we can even factor \(x_1 x_2 x_3\) so we get exactly the same expression we found earlier by just pure observation:

\[ f = ((x_1 + x_2 + x_3)^{2} - 2 (x_1 x_2 + x_1 x_3 + x_2 x_3))(x_1 x_2 x_3) = (a_2^{2} - 2 a_1)(-a_0) \]

What did we gain here? Remember, we’re trying to build a recipe that can work in general for any symmetric polynomial. To make sure this is the case, we need to revisit some questions:

1. Why did we choose to remove those specific terms?
2. Can we always construct the polynomial with fancy exponent subtractions?
3. Will this whole procedure always somehow end in a finite number of steps?

I didn’t tell you, but I wrote the polynomials above in a special order. Have you heard of lexicographical ordering? Such a fancy term. All it means is that you compare things one by one, from left to right. In our case, the terms are compared lexicographically based on their coefficients. We first compare the coefficients of \(x_1\). If they’re different we’re done. Otherwise we check \(x_2\), and so on… The constant numbers that multiply terms are ignored.

Now we can answer the first question. The chosen term is the largest one lexicographically. In the original polynomial, \(x_1^{3} x_2 x_3\) is larger than both \(x_1 x_2^{3} x_3\) and \(x_1 x_2 x_3^{3}\) because when comparing \(x_1\), we have from the coefficients that \(3 > 1\). In the polynomial from the second step, we got \(-2x_1^{2} x_2^{2} x_3\). This is larger than \(-2 x_1 x_2^{2} x_3^{2}\) for the same reason as before. It is also larger than \(-2 x_1^{2} x_2 x_3^{2}\), since in this case we have \(2 = 2\) for the coefficients of \(x_1\), but we have \(2 > 1\) for those of \(x_2\).

What about the second question? This is a matter of making sure the exponents are non-negative. Did you see the pattern? If we have a term with some coefficients \(x_1^{a_1} x_2^{a_2} x_3^{a_3}\), the exponents will be \(a_1 - a_2\), \(a_2 - a_3\) and \(a_3\). For the subtractions, this all means that the coefficients must form a non-decreasing sequence, that is, \(a_1 \ge a_2 \ge a_3\).

But why would this be true? Because the polynomials are symmetric! I mean, the original one is symmetric, and the ones at each step are too, because they’re the original one minus some other symmetric polynomials, so the whole result is also a symmetric polynomial. Now, suppose we chose the lexicographically largest term but, for example, we have \(a_2 < a_3\). Because the polynomial is symmetric, it is the same when swapping \(x_2\) and \(x_3\), so there should also be another term \(x_1^{a_1} x_2^{a_3} x_3^{a_2}\). But this term is lexicographically larger than the one we chose before! That would mean we failed at picking the largest one, so we should choose this one instead, which does satisfy that the coefficients are non-decreasing. Therefore, for the lexicographically largest term, we always have the non-decreasing coefficients and the subtractions can’t be negative.

That leaves us with the third question… Why would this procedure even finish? It turns out this is also a consequence of the lexicographical ordering! But be careful. Suppose we have a sequence of three numbers \((1, 1, 1)\). There can be infinitely many sequences that are lexicographically smaller. How come? Just choose \((0, 1, 1)\), \((0, 2, 1)\), \((0, 3, 1)\), \((0, 4, 1)\), … You get the idea.

But in our case, we’re getting a smaller one at each step of the procedure. This is not exactly the same. Imagine we start with \((0, 0, 3)\). In the worst case, it would take three steps to finish if we have the sequence:

\[ (0, 0, 3) \to (0, 0, 2) \to (0, 0, 1) \to (0, 0, 0) \]

What about starting with \((0, 1, 1)\)? The next one could be \((0, 1, 0)\), but then we can only keep going if we subtract from the second element. Even if we’re unlucky and the next one is something like \((0, 0, 100)\), we would still finish eventually. What about starting with \((1, 2, 3)\)? I think you get the idea…

But intuition isn’t always spot on. The rigorous answer as for why this ends relies on a property called well ordering principle for our lexicographical ordering. Again, just a fancy term. It means that, if I give you any bunch of different sequences, you can always find the smallest one. How come? Well, start by finding the minimum value among all the first elements. There might be more of those, just pick all of them. Now, among these, do the same for the second element, and so on… In the end we get a single one.

Now suppose our procedure never finishes. This is the same as saying we get an infinite list of sequences, each one smaller than the previous one. What’s the smallest one here? No matter which one we choose, there are always other smaller ones! This is a contradiction, because we said we can always find the smallest element. Then we failed at assuming the procedure never finishes.

In addition, if we keep going freely, the only time we can’t keep going is when we have \((0, 0, 0)\). And we must finish at some point because of the previous argument. So we must finish here. When talking about our polynomials, this means the term looks like \(k \cdot x_1^{0} x_2^{0} x_3^{0}\), where \(k\) is some constant known number. So we don’t necessarily end at \(0\), but at any known number.

We’re done.

The proof we saw gives a recipe that can be followed for any symmetric polynomial, and will give us a way of expressing it in terms of elementary symmetric polynomials, and thus, in terms of known numbers. Most of the times we’re only going to use this result to assert that some symmetric polynomial is a known number, without caring about what number it might be. In the cases we do want to know the specific number, we now have a way of finding it. It might be a stupidly long procedure, but it can be done. And because of that, we don’t have to do it ourselves! We could feed this set of instructions to a computer, and in theory it will do it for us. There’s actually one place in the next section where I tell the computer to do it because I’m also too lazy to do these calculations…

Third degree

Déjà vu

Time for the cubic! Just to recap, this is what we have to start with:

\[ x^{3} + a_2 x^{2} + a_1 x + a_0 = 0 \text{ where $a_0$, $a_1$, $a_2$ are rationals} \]

\[ \left\{ \begin{aligned} x_1 + x_2 + x_3 &= -a_2 \\ x_1 x_2 + x_1 x_3 + x_2 x_3 &= a_1 \\ x_1 x_2 x_3 &= -a_0 \end{aligned} \right. \]

Remember what we did in the Second degree section? We were discussing the possibility of finding a special value \(t\) which could encode all the information about our roots, in the sense that we could find their values just from \(t\) itself and other known values. For the quadratic we used \(t = x_1 - x_2\). Can we find such a value for the cubic?

Hint: Can you use the solutions of the equation \(x^{3} = 1\)?

Yes we can! Did you think about complex roots of unity? If you like following the pattern, for the quadratic we have \(t = \alpha_1 x_1 + \alpha_2 x_2\) where \(\alpha_1, \alpha_2\) are the solutions of the equation \(x^{2} = 1\). These are of course \(1\) and \(-1\).

What are the solutions of \(x^{3} = 1\)? They’re \(1\), \(\omega\) and \(\omega^{2}\), where \(\omega\) is a complex root of \(x^{2} + x + 1 = 0\). Then, again just following the pattern, we could think of the following special value:

\[ t = x_1 + \omega x_2 + \omega^{2} x_3 \]

This is just a hunch! In order to use this, we should actually show that all \(x_1\), \(x_2\), \(x_3\) can be expressed in terms of this \(t\) and other known values.

Now that we got the value, it’s time to show that the roots can be expressed in terms of this \(t\)… Or that’s what I thought while writing this. Too optimistic… I’m not saying it’s not possible, but it can be quite difficult to get those formulas. Instead of trying to narrow down to a single value \(t\), let’s try for a moment to use two values, the other being very similar to \(t\). I’ll help you with this one.

Check these relations

\[ \left\{ \begin{aligned} t &= x_1 + \omega x_2 + \omega^{2} x_3 \\ t' &= x_1 + \omega^{2} x_2 + \omega x_3 \\ -a_2 &= x_1 + x_2 + x_3 \end{aligned} \right. \]

Can you now find the roots by using \(t\), \(t'\), \(\omega\) and the coefficients?

Hint: Can you use the fact that \(1 + \omega + \omega^{2} = 0\)?

We can try to sum all three equations multiplied by specific coefficients so that our root remains there and the other two are cancelled out à la reduction method, because of the given relation of \(\omega\).

Doing that, we get these formulas for the roots:

\[ \left\{ \begin{aligned} x_1 &= \frac{t + t' - a_2}{3} \\ x_2 &= \frac{\omega^{2} t + \omega t' - a_2}{3} \\ x_3 &= \frac{\omega t + \omega^{2} t' - a_2}{3} \end{aligned} \right. \]

OK, so we can get the roots by using both \(t\) and \(t'\). Yes, I know what you’re thinking. We now have to find two values instead of one, but trust me, this is justified, since these values seem so similar that I hope we’ll be able to find both of them at the same time.

You already know how this goes, now we have to build a polynomial for which \(t\) is a root, and hopefully, just hopefully, \(t'\) will also be one of its roots. Let’s recall, or rather expand, the way we built the polynomial for the quadratic case. I told you it would be helpful later, and the time has come.

Remember that we had \(t = x_1 - x_2\). Our special polynomial for the quadratic was this:

\[ (x - t)(x + t) = (x - (x_1 - x_2))(x + (x_1 - x_2)) = (x - (x_1 - x_2))(x - (x_2 - x_1)) = x^{2} - ((x_1 - x_2) + (x_2 - x_1)) x + (x_1 - x_2)(x_2 - x_1) \]

You already know what the actual polynomial is, but I left it in this last form to notice something: the coefficients are symmetric polynomials on the roots! If you try swapping \(x_1\) and \(x_2\) in both expressions \(((x_1 - x_2) + (x_2 - x_1))\) and \((x_1 - x_2)(x_2 - x_1)\), you get the same result. In the previous section, we just saw that all symmetric polynomials on the roots are actually known values. They’re made from the coefficients of the original equation, which were rational, so we got a new equation with rational coefficients.

Of course, there is one concern. Perhaps this new equation is just as hard to solve as the original one, and then we did something completely useless. While that’s a possibility, we were hoping that our choice of \(t\) would make the new equation sufficiently easy to solve. In the case of the quadratic, you know that this was true because we were able to lose the term of \(x\), and get a simpler quadratic \(x^{2} - t^{2} = 0\), which only requires to take a square root to solve.

We saw the coefficients we got in the end are symmetric polynomials, but we could have deduced that from the start, from the way we defined the polynomial as a product. Yes, in the quadratic example it was easier to see why we would benefit from choosing both \(t\) and \(-t\) (sum times difference is the difference of squares). Another way to look at this is that \(-t\) is the result of swapping the roots in \(t\):

\[ (x - (\text{t with no swaps}))(x - (\text{t with roots swapped})) \]

This form is telling us that this is a symmetric polynomial on the roots, and therefore, that its final coefficients will be known rational values. Can you see why? I’ll let you figure that out, because now it’s your turn again… Can you use this idea to find a suitable new polynomial for the cubic, for which \(t\) and \(t'\) are two of its roots? Just to remember, these are their expressions:

\[ \left\{ \begin{aligned} t &= x_1 + \omega x_2 + \omega^{2} x_3 \\ t' &= x_1 + \omega^{2} x_2 + \omega x_3 \end{aligned} \right. \quad \text{The value $\omega$ is such that $1+\omega+\omega^{2} = 0$} \]

Come on, try to find the polynomial!

Hint: Why is the quadratic example a symmetric polynomial on its roots?

We defined it as the product of two terms. Note that the order of the product doesn’t matter. What happens to the expression if we swap the roots \(x_1\) and \(x_2\)? Then we get:

\[ (x - (\text{t with roots swapped}))(x - (\text{t with no swaps})) \]

And this is certainly the same polynomial. For the quadratic, all the possible rearrangements are only swapping both roots or leaving them the same way. Since all these rearrangements give the same polynomial, we can say the polynomial is symmetric in the roots \(x_1\) and \(x_2\). Do you see how this helps for the cubic?

Hint: You’ll get a wild expression for the cubic, but try to simplify as much as possible.

The previous hint talked about all rearrangements of the roots. This should already be a really good hint. Yes, we’re doing exactly that. Let’s start from \(t\) and find all possible expressions that come from rearrangements of \(x_1\), \(x_2\) and \(x_3\) in \(t\):

\[ \left\{ \begin{aligned} x_1 + \omega x_2 + \omega^{2} x_3 &= t \\ x_1 + \omega x_3 + \omega^{2} x_2 &= t' \\ x_3 + \omega x_1 + \omega^{2} x_2 &= \omega t \\ x_2 + \omega x_1 + \omega^{2} x_3 &= \omega t' \\ x_2 + \omega x_3 + \omega^{2} x_1 &= \omega^{2} t \\ x_3 + \omega x_2 + \omega^{2} x_1 &= \omega^{2} t' \end{aligned} \right. \]

There are quite some things going on there, so you’d better pause and check those relations hold. Note that we also included \(t'\) in there! We now see that \(t'\) was just one of the expressions obtained from the rearrangements of the roots in \(t\). Of course, now we’ll build a polynomial that has all of these as roots, so in particular it will have both \(t\) and \(t'\) as roots, just like we wanted! The reason I wrote the equalities above is so that we now don’t have to write the absolute beast that this polynomial would be as a product. We will instead just write the right side of the equalities:

\[ (x - t)(x - t')(x - \omega t)(x - \omega t')(x - \omega^{2} t)(x - \omega^{2} t') \]

Oh, yeah, we’re going to have fun trying to simplify this… But by now we should know that, no matter how complicated it looks, in the end, this should be a polynomial with rational coefficients. It’s a symmetric polynomial on the roots, we made it like that by multiplying all rearrangements of roots.

What happens if we try to multiply only those terms having \(t\)? I’ll let you check that, but the product simplifies nicely because of the relations \(1 + \omega + \omega^{2} = 0\) and \(\omega^{3} = 1\):

\[ (x - t)(x - \omega t)(x - \omega^{2} t) = x^{3} - t^{3} \]

We can of course do the same with the terms involving \(t'\):

\[ (x - t')(x - \omega t')(x - \omega^{2} t') = x^{3} - t'^{3} \]

So because of the useful relations that we got from the roots of unity, our product simplified into this much cleaner expression:

\[ (x^{3} - t^{3})(x^{3} - t'^{3}) \]

Yes, we still have some work to do to simplify that…

I ain’t doing that

I really like these boxes that help hiding spoilers and try to encourage you, my dear lazy reader, to work through some of the steps on your own. Of course, we can’t hide everything inside these boxes and the text must keep flowing. I’ll now have to show how the polynomial we were looking for in the previous section looks like, but I’m showing a simplified expression. The moral of the story is that the spoiler isn’t the polynomial itself, because it’s simplified. The real spoiler is the series of steps you needed to get there! Remember we’re trying to build intuition.

And in fact, we already got a lot of intuition from the previous section! This section is only meant to show you that we can actually end and get the solutions of the cubic, although it’s more of a rutinary calculation from here on. You and me are both lazy human beings that don’t bother doing those mundane calculations, do we? Let’s see what we can do…

We left at this point:

\[ (x^{3} - t^{3})(x^{3} - t'^{3}) \]

We can expand that product and get this:

\[ x^{6} - (t^{3} + t'^{3}) x^{3} + t^{3} t'^{3} \]

See, the values \(t^{3}\) and \(t'^{3}\) don’t have to be rationals themselves, but since this came from a symmetric polynomial on the roots, the coefficients must be rationals! And we want to try solving a polynomial with known coefficients… OK, OK, you don’t have to calculate those coefficients if you don’t want to. I’ll tell you what they really are, but we’ll give them short names \(u\) and \(v\) for a while after that:

\[ \left\{ \begin{aligned} u &= t^{3} + t'^{3} = - 27 a_{0} + 9 a_{1} a_{2} - 2 a_{2}^{3} \\ v &= t^{3} t'^{3} = (- 3 a_{1} + a_{2}^{2})^{3} \end{aligned} \right. \]

So you still want to know how I got them…

Let the computer do the boring stuff for you! We can use the sympy package available in Python. If you’re interested but never used Python, it’s fine, most of it reads more or less like math, so you can probably still read through the steps below.

We first define all the symbols we need in our expressions:

import sympy

x1, x2, x3, a0, a1, a2, w = sympy.symbols("x1 x2 x3 a0 a1 a2 w")
t = x1 + w * x2 + w**2 * x3
t_prime = x1 + w**2 * x2 + w * x3

Here I tried to guess that \(t \cdot t'\) is already a symmetric polynomial. If that’s true, then we only need to raise that to the third power.

from sympy.polys.polyfuncs import symmetrize

expr, _ = symmetrize(t * t_prime, [x1, x2, x3])

expr = expr.subs(
    {
        x1 + x2 + x3: -a2,
        x1 * x2 + x2 * x3 + x3 * x1: a1,
        x1 * x2 * x3: -a0,
    }
)

expr

\(\displaystyle a_{1} \left(w^{2} + w - 2\right) + a_{2}^{2}\)

Looks like poor sympy needs some help with \(\omega\)… Of course! It doesn’t even know what \(\omega\) is yet.

expr.subs(1 + w + w**2, 0)

\(\displaystyle - 3 a_{1} + a_{2}^{2}\)

And \(t \cdot t'\) was indeed a symmetric polynomial! Then we also got the desired value:

\[ t^3 t'^3 = (-3 a_1 + a_2^2)^3 \]

Let’s do the same for \(t^{3} + t'^{3}\):

expr, _ = symmetrize(t**3 + t_prime**3, [x1, x2, x3])

expr = expr.subs(
    {
        x1 + x2 + x3: -a2,
        x1 * x2 + x2 * x3 + x3 * x1: a1,
        x1 * x2 * x3: -a0,
    }
)

expr

\(\displaystyle - a_{0} \left(12 w^{3} - 9 w^{2} - 9 w + 6\right) - a_{1} a_{2} \left(3 w^{2} + 3 w - 6\right) - 2 a_{2}^{3}\)

We can first get rid of the \(\omega^{3}\) part:

expr = expr.subs(w**3, 1)
expr

\(\displaystyle - a_{0} \left(- 9 w^{2} - 9 w + 18\right) - a_{1} a_{2} \left(3 w^{2} + 3 w - 6\right) - 2 a_{2}^{3}\)

expr.subs(1 + w + w**2, 0)

\(\displaystyle - a_{0} \left(- 9 w^{2} - 9 w + 18\right) - a_{1} a_{2} \left(3 w^{2} + 3 w - 6\right) - 2 a_{2}^{3}\)

You naughty sympy, why won’t you listen to me? Well, let’s try simplifying first…

expr.simplify().subs(1 + w + w**2, 0)

\(\displaystyle - 27 a_{0} + 9 a_{1} a_{2} - 2 a_{2}^{3}\)

Yes! We got a really neat expression:

\[ t^{3} + t'^{3} = - 27 a_{0} + 9 a_{1} a_{2} - 2 a_{2}^{3} \]

We’ll use the actual values of \(u\) and \(v\) later, just to not write too long equations. It turns out we end up with a surprisingly cool equation:

\[ x^{6} - u x^{3} + v = 0 \]

Another way to look at it is this:

\[ (x^{3})^{2} - u (x^{3}) + v = 0 \]

Can you see it? This is really just a quadratic equation!

\[ x^{3} = \frac{u \pm \sqrt{u^{2} - 4v}}{2} \]

We can obviously just take a cube root now and obtain a solution:

\[ x = \sqrt[3]{\frac{u \pm \sqrt{u^{2} - 4v}}{2}} \]

No, this isn’t one solution, these are a lot of solutions! Yes, that’s right. We already had this kind of discussion before, right? In this case, we have two possible values for the inner square root, and three possible values for the outer cube root. The six possible solutions! Which ones are \(t\) and \(t'\)? Unfortunately, in this case, we can’t just choose any combination.

By now I think it’s time we get rid of \(t'\). How come? Well, we already derived a useful relation:

\[ \begin{array}{c} t \cdot t' = - 3 a_{1} + a_{2}^{2} \\ \Downarrow \\ t' = \frac{- 3 a_{1} + a_{2}^{2} }{t} \end{array} \]

Let’s go back to our original equations for the roots. We originally wanted to get back the roots by just using one quantity \(t\) and other known values. We finally made it!

\[ \left\{ \begin{aligned} x_1 &= \frac{t + \frac{k}{t} - a_2}{3} \\ x_2 &= \frac{\omega^{2} t + \omega \frac{k}{t} - a_2}{3} \\ x_3 &= \frac{\omega t + \omega^{2} \frac{k}{t} - a_2}{3} \end{aligned} \right. \quad \text{where } k = -3 a_1 + a_2^{2} \]

Did the question change in any way? How do we know which root of the new polynomial is the \(t\) we want? Of course, we could always try all six possible values in these equations for \(x_1\), \(x_2\) and \(x_3\), and double check if they really are solutions of the original cubic equation.

Will you do that for me? Well…

Before we die in the attempt, let’s try to give some short expressions for all six roots. We will fix one of them and get the others in terms of that one. We’ll call this one \(t\), but this is just a name for a fixed root, it could be any of them. Suppose then that \(t\) is this:

\[ t = \sqrt[3]{\frac{u + \sqrt{u^{2} - 4v}}{2}} \]

When taking cube roots, one way of writing all three of them is to fix one and multiply by roots of unity to get the others (because of course you get the same if you again raise to the third power). That means another two roots are \(\omega t\) and \(\omega^{2} t\).

What about the other three? We should take the negative sign for the square root inside. If we call that one \(t'\), we then have the relation we defined earlier:

\[ t \cdot t' = k \Rightarrow t' = \frac{k}{t} \quad \text{where } k = -3 a_1 + a_2^{2} \]

So another root is \(\frac{k}{t}\). Since there’s also a cube root there, we can pick another two roots to be \(\omega \frac{k}{t}\) and \(\omega^{2} \frac{k}{t}\). So after fixing one root \(t\), all six of them can be expressed like this:

\[ \left\{ t, \omega t, \omega ^{2} t, \frac{k}{t}, \omega \frac{k}{t}, \omega^{2} \frac{k}{t} \right\} \]

Remember this \(t\) was just some fixed root. If we use this one, in the equations above for the roots, we would get \(x_1\), \(x_2\) and \(x_3\) in that order. That’s basically just the definition. How about trying to input \(\omega t\) where \(t\) appears in the equations?

Hint: Try to relate with the original expressions in \(t\)

\[ \left\{ \begin{aligned} \frac{(\omega t) + \frac{k}{(\omega t)} - a_2}{3} = \frac{\omega t + \omega^{2} \frac{k}{t} - a_2}{3} &= x_3 \\ \frac{\omega^{2} (\omega t) + \omega \frac{k}{(\omega t)} - a_2}{3} = \frac{t + \frac{k}{t} - a_2}{3} &= x_1 \\ \frac{\omega (\omega t) + \omega^{2} \frac{k}{(\omega t)} - a_2}{3} = \frac{\omega^{2} t + \omega \frac{k}{t} - a_2}{3} &= x_2 \end{aligned} \right. \]

In the first one we used the equality \(\frac{1}{\omega} = \frac{\omega^{2}}{\omega^{2}} \cdot \frac{1}{\omega} = \omega^{2}\).

Perhaps unexpectedly, we got the same set of solutions, but in a different arrangement! Let’s try with another one, for example, \(\frac{k}{t}\):

Hint: same idea as before

\[ \left\{ \begin{aligned} \frac{(\frac{k}{t}) + \frac{k}{(\frac{k}{t})} - a_2}{3} = \frac{\frac{k}{t} + t - a_2}{3} &= x_1 \\ \frac{\omega^{2} (\frac{k}{t}) + \omega \frac{k}{\frac{k}{t}} - a_2}{3} = \frac{\omega^{2} \frac{k}{t} + \omega t - a_2}{3} &= x_3 \\ \frac{\omega (\frac{k}{t}) + \omega^{2} \frac{k}{(\frac{k}{t})} - a_2}{3} = \frac{\omega \frac{k}{t} + \omega^{2} t - a_2}{3} &= x_2 \end{aligned} \right. \]

We got another rearrangement of the roots! What kind of witchery is this? Well, I don’t know about you, but I’m starting to become tired of writing these equations… I claim that each one of the roots of the new polynomial gives a different rearrangement of the roots \(x_1\), \(x_2\) and \(x_3\) when introduced in the formulae of the roots. Thus, any root of the new polynomial will give us valid solutions for the original equation, no matter which one we take. Of course, you don’t need to trust me, you can check it yourself!

All the rearrangements

\[ \left\{ \begin{aligned} t \mapsto (x_1, x_2, x_3) \\ \omega t \mapsto (x_3, x_1, x_2) \\ \omega^{2} t \mapsto (x_2, x_3, x_1) \\ \frac{k}{t} \mapsto (x_1, x_3, x_2) \\ \omega \frac{k}{t} \mapsto (x_2, x_1, x_3) \\ \omega^{2} \frac{k}{t} \mapsto (x_3, x_2, x_1) \\ \end{aligned} \right. \]

Yes, we have the cool result that any root of the new polynomial is valid for getting the original solutions of the cubic. But what do you think about these neat rearrangements? Are they only useful for the above conclusion, or can we do more with them? Was this a coincidence, or a more general result? Save some popcorn for later. I warned you.

Fourth degree

Ha! Did you really think I was going to do this again? Nope.

The journey is challenging, but if you’re willing to venture further… You know what to do!

Which special value \(t\) can we use for an equation of fourth degree?

You see the pattern. We should use the roots of \(x^{4} = 1\). These are \(1, i, -1, -i\), where \(i\) is the imaginary number (\(i^{2} = -1\)). Then the special value you should try to use is the familiar:

\[ t = x_1 + i \cdot x_2 - x_3 -i \cdot x_4 \]

Trust me, it works!

For those who don’t find excitement in the too specific formula for a quartic equation, just keep reading…

Second symmetry

Cubic, u still there?

In the third degree section we found ourselves solving the general cubic equation thanks to this special value \(t = x_1 + \omega x_2 + \omega^{2} x_3\). I think we managed to justify why this special one is quite helpful for solving the cubic. I mean this in the sense that, the new equation we have to solve is simple enough, so that finding one of its roots is easier than finding a root of the original equation, even if the new one has sixth degree.

Anyway, a natural question would be: Are there other special values that allow us to solve the original equation in the same way? And we’re now only interested in the theoretical question. We don’t mind if the new equation is hard to solve or not. We really only want to know if there are other special values which encode all information about the roots.

Let’s try ourselves! I’m a bit tired of writing monstrous equations though. We’ll try with a specific example, of which we could even know the roots in advance. Remember this is just a sort of thought experiment. OK, so our cubic equation will be this:

\[ x^{3} - x^{2} - x - 2 = 0 \]

We can quickly check if there is an integer solution by only trying the values \(\pm 1, \pm 2\) for \(x\). Maybe you even recall doing this in high school!

Can you see why?

We can rewrite the equation like this:

\[ x \cdot (x^{2} - x - 1) = 2 \]

If \(x\) was an integer, we would have \(x \cdot \text{other integer} = 2\), so \(x\) must be a divisor of \(2\). The only possible values would then be \(\pm 1, \pm 2\).

It turns out that \(x = 2\) is a valid solution (you can verify it). That means the cubic polynomial will have a factor \((x-2)\). Were you told how to perform polynomial long division? If so, go ahead, factor this cubic! If not, don’t worry, we have some workarounds. We can try to write the factorization like this:

\[ x^{3} - x^{2} -x - 2 = (x-2)(x^{2} + a_1 x + a_0) \]

Can you find \(a_1\) and \(a_0\)? Hint: expand the product and compare coefficients

\[ x^{3} + (a_1 - 2) x^{2} + (a_0 - 2 a_1) x - 2 a_0 \]

Following the coefficients of the original polynomial, we must have \(-2 a_0 = -2\), so \(a_0 = 1\). Likewise, \(a_1 - 2 = -1\), so \(a_1 = 1\). These also give the expected result for \(a_0 - 2 a_1 = -1\), so we got the coefficients. The factorization is then:

\[ x^{3} - x^{2} - x - 2 = (x-2)(x^{2} + x + 1) = (x-2)(x-\omega)(x-\omega^{2}) \]

We also know the last equality! The polynomial \(x^{2} + x + 1\) is already familiar to us, and we defined its solutions as \(\omega\) and \(\omega^{2}\).

So we now know the three solutions of this equation. We’re cheating! Who cares? Let’s try a different special value! If you recall, we used \(t = x_1 - x_2\) for the quadratic equation. Well, let’s just try it with the cubic as well, and see what happens.

Suppose we like defining \(x_1 = 2\), \(x_2 = \omega\), \(x_3 = \omega^{2}\). Then we can have these relations:

\[ \left\{ \begin{aligned} t &= x_1 - x_2 \\ \frac{7}{t} &= x_1 - x_3 \\ 1 &= x_1 + x_2 + x_3 \end{aligned} \right. \]

Where did those come from?

The first one is just the current definition for \(t\). The last one comes from the coefficient relation \(x_1 + x_2 + x_3 = -a_2\) we already saw in a previous section. The second one comes from a product:

\[ (x_1 - x_2)(x_1 - x_3) = (2 - \omega)(2 - \omega^{2}) = 4 -2\omega -2\omega^{2} + \omega^{3} = 6 - 2 (1 + \omega + \omega^{2}) + 1 = 7 \]

Then we have

\[ (x_1 - x_3) = \frac{7}{x_1 - x_2} = \frac{7}{t} \]

Using the above relations, can you obtain all roots in terms of \(t\)?

Hint: Use reduction method!

As in other examples, this is a matter of summing all equations, each one multiplied by a certain number to cancel out the roots we don’t want. After you find the correct numbers you get:

\[ \left\{ \begin{aligned} x_1 &= \frac{t + \frac{7}{t} + 1}{3} \\ x_2 &= \frac{-2t + \frac{7}{t} + 1}{3} \\ x_3 &= \frac{t - 2 \cdot \frac{7}{t} + 1}{3} \end{aligned} \right. \]

Yes! It turns out we can use a different special value, and still be able to encode the information of all roots. Now that we have the roots in terms of \(t\), we only need the values of \(t\). You know what to do now, don’t you? The new equation from which we get the values of \(t\) is that one with products of all possible rearrangements of \(x_1\), \(x_2\) and \(x_3\) in \(t\). Since we’re cheating and we already know the roots of the original polynomial, we can save ourselves from building that new polynomial monstrosity and directly write down all its six roots. In case you’re a bit lost, I’ll tell you the first one, which comes from no rearrangement: \(t = x_1 - x_2 = 2 - \omega\).

Which are the rearrangements?

\[ \left\{ \begin{aligned} t &= x_1 - x_2 = 2 - \omega \\ \frac{7}{t} &= x_1 - x_3 = 2 - \omega^{2} \\ -t &= x_2 - x_1 = \omega - 2 \\ -\frac{7}{t} &= x_3 - x_1 = \omega^{2} - 2 \\ -t + \frac{7}{t} &= x_2 - x_3 = \omega - \omega^{2} \\ t - \frac{7}{t} &= x_3 - x_2 = \omega^{2} - \omega \end{aligned} \right. \]

The right side of each equation is the actual rearrangement of the roots. I only wrote the left side to show you they can be written in terms of \(t\), and in case these alternative expressions are useful to you to more easily plug them into the root relations to calculate \(x_1\), \(x_2\) and \(x_3\).

Now for the final step! Let’s try to get those roots back from the values of \(t\). I said values in plural on purpose. In the general cubic, we somehow got the striking result that all values of \(t\) are valid because they just give back a different arrangement of \(x_1\), \(x_2\) and \(x_3\). Here we’re interested in seeing if this still holds for a different special value \(t\). Alright, I’ll also include this one in a spoiler just in case you’re in the mood for some calculations.

Find the roots for each possible value of \(t\)

\[ \left\{ \begin{aligned} t = 2 - \omega &\mapsto (2, \omega, \omega^{2}) \\ \frac{7}{t} = 2 - \omega^{2} &\mapsto (2, \omega^{2}, \omega) \\ -t = \omega - 2 &\mapsto \left(-\frac{4}{3}, \frac{2}{3} - \omega, \frac{5}{3} + \omega \right) \\ -\frac{7}{t} = \omega^{2} - 2 &\mapsto \left(-\frac{4}{3}, \frac{2}{3} - \omega^{2}, \frac{5}{3} + \omega^{2} \right) \\ -t + \frac{7}{t} = \omega - \omega^{2} &\mapsto \left(\frac{1}{3} - \frac{4}{9} (\omega - \omega^{2}), \frac{1}{3} - \frac{13}{9} (\omega - \omega^{2}), \frac{1}{3} + \frac{17}{9}(\omega - \omega^{2})\right) \\ t - \frac{7}{t} = \omega^{2} - \omega &\mapsto \left(\frac{1}{3} + \frac{4}{9} (\omega - \omega^{2}), \frac{1}{3} + \frac{13}{9} (\omega - \omega^{2}), \frac{1}{3} - \frac{17}{9}(\omega - \omega^{2})\right) \end{aligned} \right. \]

Huh… What is this absolute rubbish? We expected some neat rearrangements of the roots \(2\), \(\omega\), \(\omega^{2}\), and, well, we got that for the first two possible values of \(t\), but we got some ugly results for the others… Unfortunately, we’ll have to conclude from this that not all solutions of the new equation can be used to get the solutions of the original one.

But we can certainly use some of them. You see, the ones corresponding to \(2 - \omega = t\) and \(2 - \omega^{2} = \frac{7}{t}\) did give the intended solutions. Now the next question should be: Is there some rule to know exactly which solutions of the new equation do retrieve the solutions for the original one?

In this case, let’s consider a smaller part of the new equation. Let’s compute the product of only those two factors including the roots that worked:

\[ (x - (2 - \omega))(x - (2 - \omega^{2})) = x^{2} -(2 - \omega + 2 - \omega^{2})x + (2-\omega)(2-\omega^{2}) \]

In fact, this is already a polynomial with known coefficients! The coefficient for \(x\) simplifies thanks to \(1+\omega+\omega^{2} = 0\), and we already computed the product \((2-\omega)(2-\omega^{2})\) before. The polynomial is then:

\[ x^{2} -3x + 7 \]

The whole point of computing a huge product of all rearrangements was that we were sure the resulting polynomial would have known coefficients. We didn’t expect that, at least in this case, a smaller polynomial containing \(t\) would already have known coefficients. We could already find \(t\) from this even easier polynomial, since it’s just quadratic. And thanks to our half-failure from before, we also know that the other four roots were completely useless.

Following on that idea, remember when we factored the original polynomial because one of the roots was an integer. The other part would then be \(x^{2} + x + 1\), and we know from quadratics that \(x_2 + x_3 = -1\) and \(x_2 \cdot x_3 = 1\). In the First symmetry section we learned that:

Any symmetric polynomial on the roots is a known number.

In this example, because of the root \(x_1\) already being rational, we can use the same reasoning for only the other two roots, so we can be even more specific:

Any symmetric polynomial on the roots \(x_2\) and \(x_3\) is a known number.

This shows that, at least sometimes, if we confirm a polynomial on the roots is unchanged by some of the rearrangements, not necessarily all of them, we can already be sure that its result is a known number.

What about the other way around? If we know a polynomial on the roots is a known number, will it always be unchanged by those specific rearrangements?

In our specific example, the answer is yes! Why?

Any polynomial on the roots \(x_2 = \omega\) and \(x_3 = \omega^{2}\) can actually be written as \(a + b \omega\), where \(a\) and \(b\) are rationals. How? We can just rewrite \(\omega^{2}\) as \(-1 -\omega\) because of the relation \(1 + \omega + \omega^{2} = 0\), and then group all terms based on whether they multiply an \(\omega\) or not.

If the polynomial is written like that, then being a known rational value means that \(b\) must be an expression that simplifies to zero. This again means that swapping \(\omega\) and \(\omega^{2}\) in this expression doesn’t change the result, since that would only change \(a + b \omega\) to \(a + b \omega^{2}\), but we still have \(b\) simplifying to zero anyway.

So what have we learned?

• The roots of our original polynomial \(x^{3} - x^{2} -x - 2\) are \(2\), \(\omega\) and \(\omega^{2}\).
• Any polynomial expression on these roots satisfies a special property related to rearranging the roots of the expression.
• This property is that the operations of “not rearranging any root” and “swapping \(\omega\) and \(\omega^{2}\)” together completely determine when a polynomial on the roots simplifies to a known rational number.
• We got these two operations from the rearrangements obtained by plugging the roots of the polynomial with known coefficients \(x^{2} - 3x + 7\) (which was smaller than expected) into the expressions for \(x_1\), \(x_2\) and \(x_3\) in terms of \(t\).
• From the monstrous polynomial with all rearrangements of roots in \(t\), the polynomial \(x^{2} - 3x + 7\) is actually the factor with lowest degree that has known rational coefficients and includes \(t = 1 - \omega\) as one of its roots. It can’t have degree 1 because that would mean \(1-\omega\) is a rational number. Maybe this “polynomial with lowest degree that has \(t\) as a root” is a good rule of thumb in general?

But we already learned how to solve cubics… Why should I care about all of this? There’s a high chance you’re asking yourself that question. I’ll clarify shortly. We’re on the verge of finding an invariant of the original polynomial. This is how mathematicians refer to some special property that is inherent to an object. Something that doesn’t change.

But what could change here? Well, in the above example we chose a specific labeling of the roots (we said that \(x_1 = 2\), \(x_2 = \omega\) and \(x_3 = \omega^{2}\)) and we also chose a certain special value \(t = x_1 - x_2\). And we now know there might be more than one such value that works. Just imagine being able to create the same object, no matter which choice we wake on the order of the roots or the value of \(t\).

// Start of TV-drama cliffhanger.

Will we succeed? Find out in the next episode!

// End of TV-drama cliffhanger.

Descending into abstractness

Well, it seems like this descent will take quite some time, let’s wait…

You can’t solve it

Of course I can’t, you haven’t written anything here!

Afterword

I’ll tell you a bit about this post. After all, this is a personal blog.

I studied math and computer science. Because of that, the curriculum for both was shortened as compared to studying each one individually. I didn’t have a Galois theory class, so I decided to write my End of Degree Project on this topic. It was basically like studying a more advanced math class, nothing else. Just some hand-waving of complex mathematical structures like groups, rings and fields. Surely it was a good way to develop my math skills, but looking back at my final writing, it has no particular interest. No attempt to make the topic more appealing. Too generic and boring writing style.

That was three years ago, when I finished my studies. I convinced myself that I had enough with the math side and I’d better get some software engineering jobs and call it a day. Three years later, math is waiting for me to catch up. I’m starting a master’s degree in math soon.

This post is basically me saying I could’ve done better. Writing is fun.

My initial intention was writing a kind of summary of the content I covered in my End of Degree Project. This was reinforced after finding a surprisingly short (just six pages!) introduction to the topic titled Galois Theory for Beginners, written by John Stillwell. I still think it’s a good read, but making a text short is usually the opposite of an easy read. I was also finding myself again trying to understand the concept of a Galois group, looking for something a bit more intuitive than its modern definition using automorphisms.

I also wanted to take a look at some existing introductions of Galois theory out there, be it YouTube videos or written posts. These are contents I watched/read myself, I won’t judge if they’re good introductions or not, you can do that yourself. Some of them are:

• (video) Why you can’t solve quintic equations (Galois theory approach)
• (video) But why is there no quintic formula? | Galois Theory
• (video) The Insolvability of the Quintic
• (video) Galois Theory Explained Simply
• (written) What is Galois Theory Anyway?
• (written) An introduction to Galois theory (NRICH)

Surprisingly, my favourite one is Why There’s ‘No’ Quintic Formula (proof without Galois theory), which, as the name says, is not really an introduction to Galois theory, but deals with the same topic as this post (the unsolvability of the quintic), from a different perspective. This is a video originally based on the article Arnolds’s elementary proof of the insolvability of the quintic, which is by itself a worth read. To my knowledge, there’s still no introduction based on John Stillwell’s text. I think that would be a great addition to the world of Galois theory exposition intros.

At the same time, I also started reading Laura Toti’s biography of Galois. While reading a lot of mentions to his first memoir in there, I was starting to consider the idea of reading that memoir myself, just out of curiosity. Before this, I never really tried reading such old original works. And I started reading Galois’ first memoir. And once I started understanding the basic ideas, I decided that was the way to go in this post. Just do what Galois did.

The original works of Galois are translated to English in the book The mathematical writings of Evariste Galois by Peter M. Neumann. Galois himself also has a rather verbose writing style. Maybe this was true in general for works from that time, compared with modern math works, I’m not sure. I can just tell I like his writing style. The problem is, there are some parts where he isn’t very clear or he just leaves empty gaps to fill, and I needed some help to follow them.

That’s how I arrived at what I believe is the main source for this post, Galois Theory by Harold Edwards. In his book, he basically does the same we tried to do here, that is, to try to present Galois theory in the same way Galois himself did it. This book covers a larger content than what I specifically wanted for this post, and he also doesn’t mind using some modern concepts to draw the connections with modern Galois theory. To keep the post as elementary as possible, I tried to avoid that.

And that’s how we got here. Unfortunately, most of the still unwritten second half of the post covers the real content from Galois’ first memoir. But there’s already a lot of intuition in the existing portion. Until I manage to finish writing everything, I encourage you to try more polynomial examples similar to how we did in Cubic, u still there? and try to develop your own intuition. Maybe you end up proving some results yourself! As we said, you really learn math by doing it.

I hope you enjoy the content.