# Field extensions of finite degree and primitive elements

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?

• 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. – CWsl2 Feb 15 '14 at 4:21
• 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. – DC 541 Feb 15 '14 at 4:23
• @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? – KReiser Feb 15 '14 at 4:23
• @KReiser right! Good question I'll have to think about this a little bit more – CWsl2 Feb 15 '14 at 4:29

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 $[K:F]\leq n$.
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.
• 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. – KCd Feb 15 '14 at 5:41