Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there any example of a field extension $K/F$ where degree of extension $[K:F]$ is finite but the number of intermediate field is infinite ?

share|improve this question

1 Answer 1

up vote 11 down vote accepted

A field extension of finite degree has only finitely many intermediate extensions if and only if there is a primitive element. So if we can find a finite extension that has no primitive element then the number of intermediate fields must be infinite.

Consider $K = \mathbb F_p(X,Y)$, the field of rational functions in two variables over the finite field with $p$ elements, and the extension $L = K(X^{1/p}, Y^{1/p})$. Then $L$ is a field extension of degree $p^2$ over $K$. However, there cannot be a primitive element since $\alpha^p \in K$ for every $\alpha \in L$ but a primitive element must have degree $p^2$ over $K$.

The fundamental theorem of Galois theory doesn't apply here since the extension is not separable.


EDIT: By looking at the proof of the primitive element theorem, an infinite number of intermediate fields can explicitly given by $K(\alpha X^{1/p}+Y^{1/p})$ for $\alpha \in K$. To show that these fields are distinct assume $E = K(\alpha X^{1/p}+Y^{1/p}) = K(\beta X^{1/p}+Y^{1/p})$ with $\alpha,\beta \in K$ and $\alpha \neq \beta$. Then $E$ contains $(\alpha-\beta)X^{1/p}$, hence it contains $X^{1/p}$ and $Y^{1/p}$. But then $E = K(X^{1/p},Y^{1/p})$ which is a contradiction since $\alpha X^{1/p}+Y^{1/p}$ would be a primitive element of $K(X^{1/p},Y^{1/p})$ over $K$ which is impossible.

share|improve this answer
    
Thank you very much for the help. Will you kindly provide infinite class of intermediate fields? –  user12290 Mar 11 '12 at 11:13
    
Thank you again. –  user12290 Mar 11 '12 at 13:52
    
My first instinct was "No, Galois theory," but as you pointed out it doesn't work when the extension is not Galois :) –  you Mar 11 '12 at 23:29
    
Yes, if the extension is Galois then such an example cannot exist because of the fundamental theorem of Galois theory. In fact, if the extension is only separable (not necessarily normal) then you can still look at the Galois closure. That's why we had to choose an extension that is not even separable. –  marlu Mar 11 '12 at 23:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.