Why is $\sqrt [n] 1$ not an expression "in radicals" of a root of unity? In Edwards' Galois Theory, in the chapter on Cyclotomic polynomials, the author devotes a lot of effort to proving that prime order primitive roots of unity can be expressed "by radicals", and gives the example:
$$\sqrt[3] 1 = \frac {-1 \pm \sqrt {-3}} {2}$$
While I agree that this expresses $\sqrt[3] 1$ using radicals, isn't $\sqrt[3] 1$ already such an expression?
 A: I think this is a good question. 
Roots of unity are special - they are units of finite multiplicative order. It is sometimes necessary to know whether your context has "enough" (or "the right") roots of unity, so knowing their properties in terms of potentially more primitive concepts does help. Knowing when extensions by radicals include roots of unity can be very valuable. The key question is more about which radical extensions contain which roots of unity than whether roots of unity are themselves radical.
As an example of how we can get in a mess, we have $i^4=1$, but if you write $\sqrt[3]1=\frac {-1\pm i\sqrt 3}2$ what allows you to use $i$ in the expression for $\sqrt[3]1$? $i$ is itself a radical expression over $\mathbb Q$ or $\mathbb Z$ (and the ground ring or field tends to matter a bit here).
A: "The number $\alpha$ is expressed in terms of radicals" means that there is a chain:
$$F_n\supset F_{n-1}\supset\cdots\supset F_0=\mathbb Q$$
where there exists an irreducible polynomial $q_i(x)\in\mathbb F_{i-1}[x]$ of the form $q_i(x)=x^{n_i}-\alpha_i$ such that $F_i\cong F_{i-1}[x]/\langle q_i(x)\rangle$, and $\alpha\in F_n$.
That's a fairly technical definition, but the key word here is "irreducible." $x^3-1$ is not an irreducible polynomial.
A: the meaning is not the number $1$, but all of the solutions to the equation $x^3-1=0$, or $x^n-1=0$ in the general case. 
A: The fundamental theorem of algebra tells us that an expression like $x^3 - 1$ has three roots. But most of the time, when we write something like "$\root 3 \of 7$", we only want one root, a root that is in some way more "principal" than the others, such as, for example, being a purely real positive number. So some of us agree explicitly that $\root b \of a$ represents the principal $b$th root of $a$, while others don't even realize they're agreeing to this.
And so, just as $\root 3 \of 8$ means 2 and not $-1 + \sqrt{-3}$ nor $-1 + \sqrt{-3}$, $\root 3 \of 1$ means 1, and not $-\frac{1}{2} + \frac{\sqrt{-3}}{2}$ nor $-\frac{1}{2} - \frac{\sqrt{-3}}{2}$. So 1 is its own principal cubic root, but the complex cubic roots are of course more interesting, which is why there is often the assignment $\omega = -\frac{1}{2} + \frac{\sqrt{-3}}{2}$ ($w$ is sometimes used, but it's a poor substitute, in my opinion).

It was pointed out in a comment that $\sqrt{-3}$ could be considered ambiguous. I suppose it is, unless we agree to the following definition: the principal square root of a negative real number is a positive imaginary number. So, the principal square root of $-3$ is approximately $\frac{433}{250}i$, and $-\frac{433}{250}i$ is the other root.
