It is well-known that is $E$ is a simple finite extension over $F$, then it has finite many intermediate fields.

Actually Minle's notes says that if an finite extension has finite intermediate fields, then it is simple! I am curious about this fact, could someone give a proof? Thanks!


Let $E/F$ is a finite extension (say $[E:F]=n$). If $F$ is a finite field, the result is fairly trivial, so assume $F$ is infinite.

At this point, it is clear that we need only check the case $E=F(a,b)$ is simple i.e. $n=2$, and we can conclude the general result inductively.

Now consider all field extensions of $F$ of the form $F(ax+b)$, for $x \in F$. Since $F$ must have an infinite number of elements, there are infinitely many choices for $x$. Since $F$ has only finitely many intermediate fields, there exists $x\neq y \text{ in } F$ such that $F(ax+b)=F(ay+b)$. Set $c=ax+b$. Since $F(c) =F(ax+b)\subseteq F(a,b)$, it suffices to show $a,b \in F(c)$. Since $x\neq y$ and $ay+b \in F(c)$, we have $$a=\frac{a(x-y)}{x-y}=\frac{(ax+b)-(ay+b)}{x-y} \in F(c)$$

Similarly, $b=c-ax\in F(c)$, so therefore $$F(c)=F(a,b)$$

  • 2
    $\begingroup$ A field can have positive characteristic and not be finite. $\endgroup$ Aug 3 '16 at 20:36
  • 1
    $\begingroup$ This proof only uses that $F$ is infinite, not that it has characteristic $0$, so the fix is straightforward. $\endgroup$
    – Slade
    Aug 3 '16 at 21:23
  • $\begingroup$ Thank you for your comments. I'll correct it right away $\endgroup$
    – JasonM
    Aug 4 '16 at 4:37

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