I believe I have proved the following, from the free textbook here, page 356. I have 2 different proofs and would greatly appreciate thoughts on them.
Show that the set of all elements in $\mathbb{R}$ that are algebraic over $\mathbb{Q}$ form a field extension of $\mathbb{Q}$ that is not finite.
Now, my understanding is that an infinite field extension $E$ is one that cannot be written as follows: $$E=F(\alpha_1, \ldots, \alpha_n),\space n\in \mathbb{N}$$
Proof 1
Suppose in order to derive a contradiction that the given field extension, denoted $E$, can be written as $E=F(\alpha_1, \ldots, \alpha_n),\space n\in \mathbb{N}$. Then let $[E:\mathbb{Q}] = N$. $N$ is a natural number because the $\mathbb{Q}$ is the product of the degree of $n$ finite field extensions.
Now, we know that there is no irreducible polynomial in $\mathbb{Q}$ of highest order, for we can construct one of arbitrary order using Eisenstein's criterion. Let $p(x)$ be an irreducible polynomial in $\mathbb{Q}$ of degree $N+1$, and let $\alpha$ be a root of $p(x)$. Then $[\mathbb{Q}(\alpha):\mathbb{Q}] = N+1$.
Because the degree of $\mathbb{Q}(\alpha)$ is greater than the degree of $E$, the dimension of $\mathbb{Q}(\alpha)$ when viewed as a vector space over $\mathbb{Q}$ is greater than that of $E$, so $\mathbb{Q}(\alpha) \not\subset E$. Therefore there exists an element of $\mathbb{Q}(\alpha)$ that is not in $E$, and because both of those sets contain $\mathbb{Q}$, this element is irrational. Therefore there is an algebraic real that is not in $E$, and the contradiction is reached.
Now I came up with that proof first, and this next one is definitely simpler. However I would still like to know if the above is valid, because I am new to field theory and I think finding an error in it would expose any difficulties I have with the subject.
Proof 2
Use the following fact:
The field of all algebraic reals over Q has infinite degree.
Now, that is discussed in this question. My understanding of the argument there is that because $\forall\,n\in\Bbb N\;,\;\;[\Bbb Q(\sqrt[n]2):\Bbb Q]=n$, the degree of an extension including all roots of 2 is unbounded.
Now using this fact, there cannot possibly exist an extension $E$ as above, for its degree would therefore have to be infinite, but because the degree of each simple extension that builds $E$ is finite, and there are only finitely many such simple extensions, the degree of $E$ must be finite!
Am I on the mark here? Thank you very much!