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I want to construct the real algebraic numbers from $\mathbb{Q}$ in a manner that sort of "looks like" the construction of the complex numbers from the reals, in a superficial manner. I can't define a single "irrational unit", but I instead tried to start with the set $\{ \sqrt{z} \mid z \in \mathbb{Q} \}$ and build the algebraic numbers from there.

This is what I'm thinking:

Let $A_{0} = \{ q + p \sqrt{z} \mid q, p, z \in \mathbb{Q} \}$.

In general, let $A_{n} = \{ q + p \sqrt{z} \mid q, p, z \in \mathbb{Q} \cup A_{0} \cup \dots \cup A_{n-1} \}$.

Let $A = \cup_{i=0}^{\infty} A_{i}$.

Is $A = \mathbb{A}$? If not, what's missing, and if so, can this description be simplified in some way?

Thanks all.

Jonathan Hebert
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    Is $1+\sqrt[3]2\in A$? – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Mar 17 '18 at 16:12
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    It is well known that it is not possible to construct all roots of polynomials by iterating square roots the way you propose, a first example being found in degree $5$. So no, you won't get all algebraic numbers that way. – Daniel Robert-Nicoud Mar 17 '18 at 16:13
  • Ah, right, so I in fact need all ${ z^{x} \mid z,x \in \mathbb{Q} }$, and for $A_{n}$, keep $x$ in the rationals only (and not the union of $A_{i}s)$? – Jonathan Hebert Mar 17 '18 at 16:17
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    Even that is not enough. There are some polynomials whose roots cannot be expressed using radicals. See https://math.stackexchange.com/questions/657168/understanding-non-solvable-algebraic-numbers – badjohn Mar 17 '18 at 16:21
  • I see, thank you. I was aware of no general solution for polynomials of degree 5 or higher, but I didn't really grasp that that meant that we literally can't "write them down" in terms of these operations even after they're known. So I take it that determining if a non-constructible (thank you Hurkyl) number is algebraic or not is actually a quite difficult problem. – Jonathan Hebert Mar 17 '18 at 16:25

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These are called the constructible numbers, so named because they are the numbers you can construct via Euclidean geometry with a compass and straightedge, starting from a unit interval.

Adjoining all square roots isn't enough to produce the algebraic numbers; you can never produce $\sqrt[3]{2}$ in this fashion.

Adjoining $n$-th roots isn't enough either; for example, there are explicitly known polynomials of degree five whose roots cannot be expressed in that fashion. (i.e. the "insolvability" of the quintic)