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In I. Martin Isaacs Algebra: A Graduate Course, Isaacs uses the field of algebraic numbers $$\mathbb{A}=\{\alpha \in \mathbb{C} \; | \; \alpha \; \text{algebraic over} \; \mathbb{Q}\}$$ as an example of an infinite degree algebraic field extension. I have done a cursory google search and thought about it for a little while, but I cannot come up with a less contrived example.

My question is

What are some other examples of infinite degree algebraic field extensions?

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One example which is quite natural is the field given by adjoining all roots of unity or all radicals to $\mathbb Q$. But I must say, I don't like this question: It is not that hard to give you any number of infinite algebraic field extensions (take a suitable collection of polynomials and consider the splitting field), but it seems like there is very little to be learned from such an exercise in thinking up any examples. – Sam May 30 '12 at 13:50
@MTurgeon $\pi$ is transcendental over $\mathbb{Q}$, hence $\mathbb{Q}(\pi)/\mathbb{Q}$ is not algebraic. – Holdsworth88 May 30 '12 at 14:07
How are the algebraic numbers a contrived example? They are the largest algebraic extension of $\mathbb{Q}$! Any infinite algebraic extension lies in them! – Qiaochu Yuan May 30 '12 at 14:08
@Holdsworth88 I misread the requirements for this infinite degree extension. – M Turgeon May 30 '12 at 14:28
See also…. – lhf Nov 14 '13 at 11:44
up vote 8 down vote accepted

Another simple example is the extension obtained by adjoining all roots of unity.

Since adjoining a primitive $n$-th root of unity gives you an extension of degree $\varphi(n)$ and $\varphi(n)=n-1$ when $n$ is prime, you get algebraic numbers of arbitrarily large degree when you adjoin all roots of unity.

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And, as is well-known, this is the largest abelian algebraic extension – M Turgeon May 30 '12 at 14:30

$\mathbb Q[\sqrt 2, \sqrt 3, \sqrt 5, \cdots]$, obtained by adjoining the square root of the primes, is an example because if you use just $n$ primes, you get an extension of degree $2^n$.

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This is far from obvious, though (see… for a proof and a link to other proofs). If you want to be lazy, take $\mathbb{Q}[2^{1/2}, 2^{1/4}, 2^{1/8}, ...]$. – Qiaochu Yuan May 30 '12 at 14:10
For a proof, see… – lhf May 30 '12 at 14:12
So I suppose $\mathbb{Q}(\sqrt{m_1},\sqrt{m_2},...)$, where the $m_i$ are square-free coprime integers would also be an example then. – Holdsworth88 May 30 '12 at 14:13
@QiaochuYuan: Dear Qiaochu, While you are right that this result is not obvious, it is not that hard; it follows from the most basic algebraic number theory (more precisely, computations of discriminants for quadratic extensions). Regards, – Matt E Jun 1 '12 at 18:35

The field of algebraic numbers is important, as is the field of real algebraic numbers. There are plenty of other examples of the same nature. The field of Euclidean constructible numbers is an extension field of the rationals, of infinite degree over the rationals, that comes up "naturally."

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How about the following example: for any field $k$, consider the field extension $\cup_{n\geq 1} k(t^{2^{-n}})$ of the field $k(t)$ of rational functions. This extension is algebraic and of infinite dimension. The idea behind is quite simple. But I admit it require some work to define the extension rigorously.

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Let $\{n_1,n_2,...\}$ be pairwise coprime, nonsquare positive integers. Then $\mathbb{Q}(\sqrt{n_1},\sqrt{n_2},...)$ is an algebraic extension of infinite degree.

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