The classical Greek questions

Today we’ll answer these questions:

1. Can we build a cube with twice the volume of a given cube?
2. Can we trisect an arbitrary angle?
3. Can we construct a square with the same area as the unit circle?
Here’s the first idea we will need. Let $\alpha$ be the root of a cubic polynomial with coefficients in $\mathbb Q$. When I say a cubic polynomial, I want one that can’t be factored. So, while $\sqrt{17}$ is a root of $P(x)=x^3-x^2-17x+17$, it’s silly, because $P(x)=(x^2-17)(x-1)$. We call polynomials like $P$ reducible because we can reduce it to smaller polynomials. We want an irreducible polynomial. In some sense, you should think of irreducible polynomials as “prime polynomials,” but don’t ever say this to an algebraist, because strictly speaking this is a different concept (but in our case they are the same).
I want to show that $[\mathbb Q(\alpha):\mathbb Q]=3$. In fact, if $\alpha$ is the root of an irreducible polynomial with coefficients in $\mathbb Q$ of degree $n$, then $[\mathbb Q(\alpha):\mathbb Q]=n$.

Proof:

Remember that $[\mathbb Q(\alpha):\mathbb Q]=\dim_{\mathbb Q}\mathbb Q(\alpha)$. We’re considering $\mathbb Q(\alpha)$ as a vector space over the field $\mathbb Q$, and want to know its dimension. Let $P(x)=c_0+c_1x+\dots+c_nx^n$ be an irreducible polynomial with $c_i\in\mathbb Q$ and $P(\alpha)=0$. The vector space $\mathbb Q(\alpha)$ is definitely spanned by $1, \alpha, \alpha^2,\alpha^3,\dots$. I claim that $1,\alpha,\dots, \alpha^{n-1}$ form a basis. This would be a basis with $n$ elements, making $[\mathbb Q(\alpha):\mathbb Q]=n$.

We have a linear dependence $c_0+c_1\alpha+\dots+c_n\alpha^n=0$, so any basis must have fewer than $n+1$ elements. Now suppose we have a linear dependence among $1,\alpha,\alpha^2,\dots,\alpha^{n-1}$. Say,

$\lambda_0+\lambda_1\alpha+\lambda_2\alpha^2+\dots+\lambda_{n-1}\alpha^{n-1}$

Then we have a polynomial $Q(x)=\lambda_0+\lambda_1x+\lambda_2x^2+\dots+\lambda_{n-1}x^{n-1}$ of smaller degree with $Q(\alpha)=0$.

Why is this a problem? We can take combinations of $P(x)$ and $Q(x)$, and try to make the polynomial of smallest degree in this way. Let’s call it $R(x)=A(x)P(x)+B(x)Q(x)$ for some $A,B$. This is the “greatest common divisor” of $P$ and $Q$. You thought GCDs only existed for integers? Nope. There’s a general structure (of which the integers are an example) which always has a concept of a GCD.

The important thing is that $R(x)$ is a divisor of $P(x)$. But we assumed $P(x)$ was irreducible, so the only possible divisors are $P$ itself and $1$. Since $Q(x)$ has smaller degree than $P(x)$, so does $R(x)$, so it must be that $R(x)=1$. But then $R(\alpha)\neq 0$. What? Contradiction!

This means we have $\{1,\alpha,\alpha^2,\dots,\alpha^{n-1}\}$  is a basis, so $[\mathbb Q(\alpha):\mathbb Q]=n$.

$\square$

This “back-and-forth” between bases for vector spaces and polynomials is a really cool trick, and is a fun part of Galois theory.

Now we can answer the questions.

1. NO! If we could construct a cube with twice the volume of a given cube, we could in particular construct a cube of volume 2, so we could construct the side length $\sqrt[3]2$. But $x^3-2$ is an irreducible polynomial (check that yourself) with $\sqrt[3]2$ as a root, so this would mean we have a degree 3 extension $\mathbb Q(\sqrt[3]2)\subseteq \mathbb K$. But every extension of $\mathbb Q$ in $\mathbb K$ has degree that is a power of two. Three is certainly not a power of two. Contradiction!
2. NO! If we could construct arbitrary angles $\theta$ given $3\theta$, we could in general construct the length $\cos\theta$ from $\cos3\theta$. If $3\theta=60^\circ$, then $\theta=20^\circ$, which is a root of the polynomial $4x^3-3x-\frac12$ (this comes from the triple-angle formula for cosine).
3. NO! If we could, we could construct the side length $\sqrt\pi$. This is the most absurd, because then we could construct $\pi$ which isn’t the root of any polynomial. Some might want to even write $[\mathbb Q(\pi):\mathbb Q]=\infty$, which is accurate. We can’t get to infinity by only finite constructions.
So there you have it. Field theory rocks.
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4 Responses to The classical Greek questions

1. JCummings says:

Nice post, Andy! I hadn’t seen how field extensions answer those questions before. That’s rally cool stuff!
Very minor typos: In the first centered equation of your proof, the $\alpha^n$ should be $\alpha^{n-1}$. And in your answer to question 2, I think you mean “the length $\cos(\theta)$ from $\cos(3 \theta)$.

• soffer801 says:

Yes, it is really cool stuff!

And thanks for finding the typos. They’re fixed now. I’m awful at catching these.

Also, do you have any tricks for drawing diagrams? Or will I need to do them by hand? Galois theory is starting soon, and you can’t talk about separability without diagrams.

• JCummings says:

I don’t. I have been using another program to draw graphs and saving them as JPEGs and then importing them. Which works fine. I am using a very simple program called Ipe that I already had on my computer. It’s works well though because the graphs I want to draw are very simple.