Difference in extension of finite fields and extension of infinite fields

Question

Let $F\subset K$ be finite field extension. It is clear $[K:F]$ is finite saying $K$ is a finite dimensional vector space. Consider the root of $x^2-a=0,a\in F$.

Here is a statement saying if $F,K$ finite fields, $[K:F]=2l,l\in N-\{0\}$, $x\not\in F$ and $x^2-a=0$ with $a\in F$, then $x\in K$. I cannot prove this directly using degree formula as this does not give me any quite useful information. I want to show $[K(x):K]=1$. So $[K(x):F]=[K(x):K][K:F]=[K(x):F(x)][F(x):F]\leq 4l$. So $[K(x):K]\leq 2$

However consider $Q\to Q[2^{\frac{1}{4}}]$ degree 4 extension by $x^4-2$ being irreducible through eisenstein criterion where $Q$ is rational number and the map is embedding $Q$ as the subfield of $Q[2^{\frac{1}{4}}]$. Say $x^2+1=0,1\in Q$ for sure. I do not have $i\in Q[2^{\frac{1}{4}}]$.

What is the reconcillation here? It seems finite field and infinite field behaves quite differently here.

@RobertLewis Ops. Typo. I am thinking $K$,$L$ as I was taught in different notation for larger field. — user45765, Jul 20 '17 at 23:17
It would be good if you corrected it; your question is unclear to me. — Robert Lewis, Jul 20 '17 at 23:38
It is not corrected. $L$ still appears on the first line. What is $Q$ in the third paragraph? Something that is mapped to $Q$ multiplied by... Is it the rationals? The arrow is strange.. — Jyrki Lahtonen, Jul 22 '17 at 07:24
@JyrkiLahtonen $L$ should be $K$ here. Q is the rationals. The arrow is embedding by identifying $Q$ as the subfield of $Q[2^{\frac{1}{4}}]$. — user45765, Jul 22 '17 at 15:57

score 2 · Answer 1 · answered Jul 20 '17 at 23:49

2

The difference between finite and infinite fields here is that for any given $k$ an infinite field may have many extensions of degree $k$, but a finite field has only one.

answered Jul 20 '17 at 23:49

Gerry Myerson

185,413

That is a quite nice answer. If an extension is simple, I would expect finite number of intermediate fields. I have 2 follow up questions. 1. The reason why I do not expect finite number of extensions is that $[K:F]$ may not be simple? 2. Finite extension does not corresponds to adjoining finite number of elements? If so, what is the example? – user45765 Jul 20 '17 at 23:55
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Finite extensions of finite fields are always simple. By "finite extension" I mean an extension of finite degree. A finite extension of a finite field is a finite field, hence is obtained by adjoining a finite number of elements (but in fact can be obtained by adjoining a single element). – Gerry Myerson Jul 20 '17 at 23:58
I see. Thanks a lot. – user45765 Jul 20 '17 at 23:59

score 1 · Accepted Answer · edited Jun 12 '20 at 10:38

The structure theorem of finite fields and finite extensions of finite fields is quite strong.

There is one and only one finite field of order $p^n$ for every prime power $p^n$.

Every extension of finite fields is a Galois extension and the Galois group is cyclic.

If $m \mid n$ then $\operatorname{GF}(p^m) \subseteq \operatorname{GF}(p^n)$.

The first one is key here. Let's say $F = \operatorname{GF}(p^m)$.

If $p = 2$ then the map $x \mapsto x^2$ is an automorphism. Thus every element of $F$ has a square root.
If $p > 2$ then half the elements of $F^\times$ are squares.

(See The number of elements which are squares in a finite field. for details.)

So for $p = 2$ the equation $x^2 - a$ has all of its roots already in $F$. Next, suppose $p \ne 2$ and $a, b \in F^\times$ are two non-squares. Then

$$ F[x]/(x^2 - a) \text{ and } F[x]/(x^2 - b) $$

are fields of order $p^{2m}$ and hence by point (1.) above they are isomorphic.

Difference in extension of finite fields and extension of infinite fields

2 Answers2