# The classical Greek questions

October 18, 2011 4 Comments

Today we’ll answer these questions:

- Can we build a cube with twice the volume of a given cube?
- Can we trisect an arbitrary angle?
- Can we construct a square with the same area as the unit circle?

**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).

#### Proof:

Remember that . We’re considering as a vector space over the field , and want to know its dimension. Let be an irreducible polynomial with and . The vector space is definitely spanned by . I claim that form a basis. This would be a basis with elements, making .

We have a linear dependence , so any basis must have fewer than elements. Now suppose we have a linear dependence among . Say,

Then we have a polynomial of smaller degree with .

Why is this a problem? We can take combinations of and , and try to make the polynomial of smallest degree in this way. Let’s call it for some . This is the “greatest common divisor” of and . 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 is a divisor of . But we assumed was irreducible, so the only possible divisors are itself and . Since has smaller degree than , so does , so it must be that . But then . What? Contradiction!

This means we have is a basis, so .

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.

**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 . But is an irreducible polynomial (check that yourself) with as a root, so this would mean we have a degree 3 extension . But every extension of in has degree that is a power of two. Three is certainly not a power of two. Contradiction!**NO!**If we could construct arbitrary angles given , we could in general construct the length from . If , then , which is a root of the polynomial (this comes from the triple-angle formula for cosine).**NO!**If we could, we could construct the side length . This is the most absurd, because then we could construct which isn’t the root of any polynomial. Some might want to even write , which is accurate. We can’t get to infinity by only finite constructions.

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 should be . And in your answer to question 2, I think you mean “the length from .

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.

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.

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