# Proving that the algebraic reals form an infinite field extension over $\mathbb Q$

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!

• Are these proofs not essentially the same? You're just picking a particular irreducible polynomial, $x^n+2$, in the second proof. – Cheerful Parsnip Mar 23 '14 at 18:06
• They are similar, but the arguments differ a bit, don't they? I explicitly used a fact from linear algebra in the first, and not so much in the second. And in the second I used the result from the question I linked; not so much in the first. In either case, my goal was to ensure that both proofs are valid, not to come up with two nontrivially distinct proofs. – mb7744 Mar 23 '14 at 18:13
• Fair point! They both look valid to me. – Cheerful Parsnip Mar 23 '14 at 18:18
• In your second way, that "fact" is exactly what's being asked...! – DonAntonio Mar 23 '14 at 19:34
• I would disagree that it is "exactly" what is being asked. The degree of a field extension of Q and the number of elements it appends to Q are not the same. One has to use the fact that the degree of an extension is multiplicative, and the fact that no individual simple extension can have infinite degree (because any arbitrary polynomial must have finite degree.) But yes the two notions are very close and that is why that proof is so short. – mb7744 Mar 23 '14 at 21:00