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Over a field $F$ of characteristic $0$, if every every element of an extension field $K$ has degree less than $n$ over $F$, does this tell us that $K$ is a finite degree extension of $F$?

So it would seem like if every element of $K$ has minimal polynomial of at most degree $n$ over $F$, I would be able to say that the extension $K/F$ is finite.

But, it seems like maybe I am missing something.

Can every element of a field have finite degree, yet the extension as a whole be infinite?

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    $\begingroup$ The question you are asking seems to be if every algebraic field extension over a field of characteristic is zero is finite. This is not in general true. For instance, one can look at the field $K$ of all algebraic numbers over $\mathbb{Q}$, and then every element has finite degree (which is the same thing as saying is a root of a polynomial of finite degree over the rationals, or equivalently, is algebraic), but the extension is infinite. $\endgroup$
    – CWsl2
    Commented Feb 15, 2014 at 4:21
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    $\begingroup$ But, there are algebraic numbers of arbitrarily large degree. I am placing a bound on the degree of the elements. No element of my field has degree larger than $n$ suppose. $\endgroup$
    – DC 541
    Commented Feb 15, 2014 at 4:23
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    $\begingroup$ @CurtisW There's actually two questions here. The one that you answered, and the following: If every element of uniformly bounded degree, is the extension necessarily finite? $\endgroup$
    – KReiser
    Commented Feb 15, 2014 at 4:23
  • $\begingroup$ @KReiser right! Good question I'll have to think about this a little bit more $\endgroup$
    – CWsl2
    Commented Feb 15, 2014 at 4:29

1 Answer 1

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Let $K\supseteq F$ be fields of characteristic $0$ such that any element of $K$ has degree $\leq n$ over $F$.

If $[K:F]=1$ we are done, and if not then there is some $r_1\in K\setminus F$. Let $E_1=F(r_1)$. Observe that $[E_1:F]\leq n$ by hypothesis. If $[K:E_1]=1$ then $[K:F]=[E_1:F]$ and we are done, and if not then there is some $r_2\in K\setminus E_1$, and let $E_2=E_1(r_2)=F(r_1,r_2)$. Note that because $E_2$ is a finite separable extension of $F$, it is simple, say with $E_2=F(s)$. Thus, again by hypothesis, we have $[E_2:F]\leq n$. It is clear that we can repeat this argument any finite number of times. If we have not found an $m$ for which $K=E_m$ by the time we reach $E_d$ where $d=\lceil\log_2(n)\rceil+1$, then $$[E_d:F]=\underbrace{[E_d:E_{d-1}]}_{\geq 2}\cdots\underbrace{[E_1:F]}_{\geq 2}>n$$ but this contradicts $[E_d: F]\leq n$. Thus $K=E_m$ for some $m$, so that $[K:F]\leq n$.

(I've edited out my previous, unnecessarily-complicated argument using Zorn's lemma. Thanks to Mariano for catching that there was a simpler way.)

As Curtis points out, it is possible for every element of $K$ to have finite degree over $F$, while the extension $K/F$ has infinite degree. For example, take $F=\mathbb{Q}$ and $K=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5},\ldots)$, or $K=\overline{\mathbb{Q}}$. Note that finite is different than bounded.

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    $\begingroup$ Pick an element x_1 in K not in F, an element x_2 in K and not in F(x_1), x_3 in K and not in F(x_1, x_2), and so on. At each step the degree grows strictly, and is equal to the degree of one element, so this has to stop. At that rth step, K has to be equal to F(x_1,..., x_r). Since this last field has a primitive element, its degree is also at most n. $\endgroup$ Commented Feb 15, 2014 at 5:33
  • $\begingroup$ Oh. My whole excursion with Zorn's lemma was totally unnecessary. Oops... $\endgroup$ Commented Feb 15, 2014 at 5:34
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    $\begingroup$ This result, including the conclusion that $[K:F] \leq n$, is proved in Lang's Algebra (and in his Undergraduate Algebra) as part of his treatment of Galois theory. $\endgroup$
    – KCd
    Commented Feb 15, 2014 at 5:41
  • $\begingroup$ @ZevChonoles Very nice argument, thanks for posting this! I figured that the result would probably hold here but couldn't think of a nice way of showing it. Appreciate it! $\endgroup$
    – CWsl2
    Commented Feb 16, 2014 at 7:13

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