6
$\begingroup$

there is a theorem on Galois theory that

if F is a finite dimensional extension field of K (say [F:K]=n) then F is algebraic over K

I'm thinking about the converse of this statement. is the converse false? can anyone give me an example that contradict the converse

$\endgroup$
1
  • 2
    $\begingroup$ the set of all algebraic numbers over $\Bbb Q$. $\endgroup$
    – Angina Seng
    Nov 17, 2017 at 19:51

3 Answers 3

Reset to default
8
$\begingroup$

An extension $K/F$ is said to be algebraic if all of the elements of $K$ are algebraic over $F$, i.e. the solutions of polynomials in $F[X]$. Let's take the field extension $\mathbf{Q}(\sqrt{2},\sqrt{3},\ldots, \sqrt{n},\ldots)$. Every element we adjoin is algebraic. Indeed, $\sqrt{n}$ satisfies $x^2-n=0$. So, this extension is algebraic over $\mathbf{Q}$, but it certainly isn't finite.

$\endgroup$
2
  • $\begingroup$ is {$\sqrt 1$, $\sqrt 2$, ..., $\sqrt n$,...| where n is not a perfect square } a basis for $\Bbb Q(\sqrt 2, \sqrt 3, ...)$ $\endgroup$
    – Bad English
    Nov 17, 2017 at 19:58
  • 1
    $\begingroup$ Yes, you can prove that. I probably should have said $\sqrt{n}$ for $n$ not a perfect square, but writing it changes nothing, since when $n$ is a perfect square $\sqrt{n}$ is in $\mathbf{Q}$. $\endgroup$
    – Alekos Robotis
    Nov 17, 2017 at 19:59
3
$\begingroup$

$\Bbb F$ is an algebraic extension of a field $\Bbb K$ if every element $\phi \in \Bbb F$ satisfies some polynomial $k(x) \in \Bbb K[x]$; $k(\phi) = 0$. As such, we need not have $[\Bbb F: \Bbb K] < \infty$, since different elements of $\Bbb F$ may satisfy polynomials over $\Bbb K$ of different degrees. The most ready example is probably the field of algebraic numbers over $\Bbb Q$, which contains elements satisfying polynomials $p(x) \in \Bbb Q[x]$ of every degree. See this wikipedia entry for more.

$\endgroup$
3
$\begingroup$

First example: Let $K:=\mathbb{Q}$ be the field of rationals;
and let $K:=\bar{\mathbb{Q}}$ be it's algebraic clousure.
Then $F/K$ is algebraic but it is not finite dimentional.

Second example: Let $K:=\mathbb{Q}$ be the field of rationals;
and let's consider an infinite set of algebraic elements $A$; such that any finite set of $A$ is algebraicly independent. Then let $K:=\mathbb{Q}(A)$.

If you let $A= \bigg{\{} \sqrt{p} \ : \ p \ \text{is a prime number} \ \bigg{\}}$;
and let $K:=\mathbb{Q}(A)$; then you will get exactly the answer of Antonios-Alexandros Robotis.
Note that $\mathbb{Q}(A)=\mathbb{Q}(A')$; where $A'= \bigg{\{} \sqrt{n} \ : \ n \in \mathbb{N} \ \bigg{\}}$.

Also if you let
$A=$ set of all algebraic elements;
then you will get the first example.

Now you can construct another examples;

  • $A= \bigg{\{} \sqrt[3]{p} \ : \ p \ \text{is a prime number} \ \bigg{\}}$;

  • $A= \bigg{\{} \sqrt[2p+1]{p} \ : \ p \ \text{is a prime number} \ \bigg{\}}$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .