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This is a question from a course in Galois Theory and I am quite confused.

In general, the degree of a field extension $E/F$ is the dimension of the vector space $E$.

What would $E$ and $F$ be in these cases? Would $E$ be any field extension, $F$ be $\mathbb{Q}$.. where does $\mathbb{C}$ come into it - does this just mean that all elements in our field extensions over $\mathbb{Q}$ belong to the complex numbers?

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    $\begingroup$ For every $n \in \mathbb N$, you are looking for $E$ such that $\mathbb Q \subseteq E \subseteq \mathbb C$ where $[E:\mathbb Q] = n$. In other words, you want to prove that there is at least one algebraic number field of degree $n$ for every natural number $n$. This is equivalent to finding $\mathbb Q$-irreducible polynomials of degree $n$ for every $n$, which is probably the easiest way to show this. $\endgroup$ – flawr May 30 '16 at 16:07
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    $\begingroup$ What criteria do you know that can be used to prove that a rational polynomial is irreducible? Do you see the link between the irreducibility of a polynomial and the construction of extensions ? $\endgroup$ – PseudoNeo May 30 '16 at 16:11
  • $\begingroup$ What you need to do is to find, for any possible $n\in 1,2,\ldots$, a field $E_n$ satisfying $\mathbb{Q}\subseteq E_n\subseteq \mathbb{C}$ and $[E_n:\mathbb{Q}]=n$. For instance, for $n=2$ we may take $E=\mathbb{Q}(\sqrt{2})$. $\endgroup$ – Darío G May 30 '16 at 16:12
  • $\begingroup$ @PseudoNeo I know Eisenstein's criterion, also trial and error checking factors... I think that is all I know for irreducibility. I think that if say $\mathbb{Q}(a)$ is a splitting field then $x-a$ is an irreducible polynomial.. although my knowledge is quite sketchy $\endgroup$ – amiz9 May 30 '16 at 16:15
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Consider the family of polynomials $f_n:=x^n-2$. Can you show that for each $n\geq 2$, $f_n$ is irreducible? What degree extension will adjoining a root of $f_n$ produce?

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  • $\begingroup$ yes by Eisenstein's criterion with $p=2$. Each $f_n$ will correspond to an extension of degree $n$, and since $n$ can be arbitrarily large this leads to extensions of any finite degree. Is this correct? $\endgroup$ – amiz9 May 30 '16 at 16:21
  • $\begingroup$ Yep! The below answer is very good too, I just thought this might be the easiest to come up with. $\endgroup$ – J.G May 30 '16 at 17:04
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By Eisenstein's criterion, if $p$ is prime, then the polynomial $f(x):=x^n+px^{n-1}+\cdots+px+p$ is irreducible, and hence the field extension $\mathbb Q[x]/(f)$ over $\mathbb Q$ is of degree $n.$
Hope this helps.

P.S. This extension is the same as adjoining a root of $f$ to $\mathbb Q,$ and hence is a sub-extension of $\mathbb C.$

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  • $\begingroup$ Thanks. So is $(f)$ a field? Also in the question when it says in $\mathbb{C}$, does this just mean that $\mathbb{Q}$ is a sub-extension of $\mathbb{C}$, as you say? $\endgroup$ – amiz9 May 30 '16 at 16:23
  • $\begingroup$ $(f)$ is not a field, but its residue field $\mathbb Q[x]/(f)$ is, as $f$ is an irreducible polynomial. Also, I think the question wants the extension of $\mathbb Q$ that we are constructing here to be a sub-extension of $\mathbb C,$ and my last sentence meant to say this: the subject of the subordinate clause is the same as the main clause, thus omitted. $\endgroup$ – awllower May 30 '16 at 16:33

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