I’m trying to understand the proof that there will never be a quintic formula. Some questions:

  1. What is the name for a number which is formed from progressive radicals and algebraic operations, i.e. a number which could be described with a formula? Example $\sqrt{\sqrt{2}+\sqrt{3}}$.

  2. How do you prove that when you adjoin an element like this to the field, that that field extension contains the component radicals? For are example $\sqrt{2}$ and $\sqrt{3}$?

I have been trying to work this out for ages but not making much progress. Be gentle with me I dont have a maths degree.

  • 6
    $\begingroup$ 1. You say this number is expressible by radicals. 2. In general this is very hard. See theorem 2.11 here $\endgroup$ – Wojowu Mar 27 at 8:11

I don't know of a name for this extension. When you only append square roots, it's called the pythagorean closure or the constructible numbers. It contains all field extensions made from iteratively appending square roots and is the smallest subfield of $\mathbb{C}$ such that this holds. The minimal polynomial of any element will have degree a power of $2$.

It sounds like you're adjoining $\sqrt{ \sqrt{2} + \sqrt{3}}$ to the field, and want to know that $\sqrt{2}$ and $\sqrt{3}$ are contained in it. Before doing this, it's good to keep track of the size of each of the relevant fields (dimension over $\mathbb{Q}$). This dimension $\text{dim}_{\mathbb{Q}} (K)$ can be calculated by introducing intermediate fields $L$ whenever necessary, and using that $\text{dim}_{\mathbb{Q}}(K) = \text{dim}_L (K) \cdot \text{dim}_{\mathbb{Q}}(L)$. So the method is to introduce as many intermediate fields as needed to make it easy to calculate the minimal polynomial. When the minimal polynomial is calculated, its degree will be the size of the extension.

Here we will calculate the degree of $\mathbb{Q}(\sqrt{\sqrt{2}+\sqrt{3}})$ to be $8$ and the degree of $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{\sqrt{2} + \sqrt{3}})$ to be $8$. Then we know that they are the same.

Claim: $\mathbb{Q}(\sqrt{2} + \sqrt{3})$ has dimension $4$ as a $\mathbb{Q}$-vector space.

Proof: The mininal polynomial of $\sqrt{2} + \sqrt{3}$ with respect to $\mathbb{Q}$ has degree either $2$ or $4$, since this degree is the dimension of the subfield $\mathbb{Q}(\sqrt{2} + \sqrt{3})$ of the field $\mathbb{Q}(\sqrt{2}, \sqrt{3})$. The second has dimenison $4$, so any subfield will have to have degree dividing $4$. If it had degree $2$, then there would be a monic polynomial $ax^2 + bx + c$ of degree $2$ in $\mathbb{Q}[x]$ evaluating to $0$ at $\sqrt{2} + \sqrt{3}$. Then $a(2 + \sqrt{2} \sqrt{3} + 3) + b(\sqrt{2} + \sqrt{3}) + c = 0$. This gives us $a \sqrt{2} \sqrt{3} + b (\sqrt{2} + \sqrt{3}) + d = 0$ for some rational number $d$. This forces $a, b, d$ to be $0$- there are a number of ways of showing this- think of modifying ye-old proof that $\sqrt{2}$ is irrational. We get $a, b, c= 0$. All this amounts to showing that there is no such minimal polynomial of degree $0$ (we showed any candidate for a polynomial would have to be $0$). So it has got to have degree $4$ (another argument shows that $\sqrt{2} + \sqrt{3}$ is not rational). The goal of this paragraph was to show that $\mathbb{Q}(\sqrt{2} + \sqrt{3})$ has degree $4$, and we have shown exactly that, since $\sqrt{2} + \sqrt{3}$ is forced to have a degree $4$ minimal polynomial, even though we haven't shown what it is (that requires a little more cleverness).

Now the degree of $\mathbb{Q}(\sqrt{\sqrt{2} + \sqrt{3}})/ \mathbb{Q}$ is the product of the degrees of $\mathbb{Q}(\sqrt{\sqrt{2} + \sqrt{3}})/\mathbb{Q}(\sqrt{2}+ \sqrt{3}) $ and $\mathbb{Q}(\sqrt{2}+ \sqrt{3})/ \mathbb{Q}$. You can show the first is $2$ and we just showed the second is $4$. So the degree of $\mathbb{Q}(\sqrt{\sqrt{2} + \sqrt{3}})/ \mathbb{Q}$ is $8$.

Next, $\mathbb{Q}(\sqrt{2}, \sqrt{3})$, it has an intermediate field $\mathbb{Q}(\sqrt{2})$, so we can calculate the degree of $\mathbb{Q}(\sqrt{2}, \sqrt{3}) / \mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$. You can show that each of these is $2$. So the degree of $\mathbb{Q}(\sqrt{2}, \sqrt{3})/ \mathbb{Q}$ is $4$. It follows that the degree of $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{2} + \sqrt{3})/ \mathbb{Q}(\sqrt{2}, \sqrt{3})$ is $8$. This shows that the extensions are the same.

There is a generalization of this result, called the primitive element theorem. It says that, in any finite separable extension $L/K$, there is an element $a \in L$ such that $L = K(a)$. Finding this $a$ is not easy, however, and not every choice of $a$ works. In fact, there can be only finitely many choices for $a$.

  • 1
    $\begingroup$ NB: The OP only said "radicals", not "quadratic radicals", so the Pythagorean closure (which I know as the constructible numbers) may and may not be the right name. $\endgroup$ – darij grinberg Apr 20 at 2:52
  • $\begingroup$ Right, sorry for that. $\endgroup$ – Dean Young Apr 20 at 2:53

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