Exercise A-30 in Milne's Fields and Galois Theory notes is:
Let $L/K$ be a separable algebraic extension of degree $d$. Show that the number of fields between $K$ and $L$ is at most $2^{d!}$.
Milne's answer (I'm adding the details):
By the primitive element theorem, we can write $L=K[\alpha]$. Let $f(X)\in K[X]$ be the minimal polynomial of $\alpha$ over $K$. Then, we have $\deg(f)=d$. Therefore, the splitting field $E$ over $L$ of $f$ is of degree at most $d!$ over $K$. Therefore, $G=\text{Gal}(E/K)=[E:K]\leq d!$. So, $G$ has at most $2^{d!}$ subsets, and therefore, at most $2^{d!}$ subgroups. By the fundamental theorem of Galois theory, this means that there are at most $2^{d!}$ intermediate fields.
My answer:
A better bound is possible! There are at most $2^d$ intermediate fields. Proof: Again, by the primitive element theorem, we can write $L=K[\alpha]$. Let $f(X)$ be the minimal polynomial of $\alpha$ over $K$. Again, it is of degree $d$. We now imitate the proof of Proposition 5.3 of the notes: Take an intermediate field $K\subset M\subset K[\alpha]$. Let $g(X)\in M[X]$ be the minimal polynomial of $\alpha$ over $M$ with coefficients $a_0,\dotsc,a_n$. Let $M'=K[a_0,\dotsc,a_n]$. Then $M'\subset M$ and $g(X)$ is the minimal polynomial of $\alpha$ over $M'$. This means that $[L:M']=[L:M]=\deg(g(X))$, and so $M=M'$. We have seen that each intermediate field $M$ is determined by a factor of $f(X)$. Therefore, the number of intermediate fields is at most the number of factors of $f(X)$. Since $\deg(f(X))=d$, this number is at most $2^d$ (because there are $2^d$ subsets of the roots of $\alpha$).
My question:
Is my improved bound (and its proof) correct?
Best Answer
Yes, it is correct; well done! Remark that a minor tweak of Milne's proof gives the same bound if $L$ is already Galois over $K$, because then we can replace $d!$ in his proof with $d$.
Now I ask you: is this best possible?