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?

  • $\begingroup$ Your definition of $\;M'\;$ is odd: if $\;g(x)\in M[x]\;$ then obviously all the coefficients of $\;g\;$ are in $\;M\;$ and thus pretty trivially we get $\;M[a_0,...,a_n]=M\;$...why did you have to argue longer to get this equality? From here that the other equalities are trivial, too. $\endgroup$
    – DonAntonio
    Oct 11, 2013 at 22:38
  • $\begingroup$ @DonAntonio I think $M'$ should be $K[a_0,...,a_n]$ $\endgroup$
    – Cocopuffs
    Oct 11, 2013 at 22:41
  • $\begingroup$ That's seems more accurate wrt what the OP, apparently, meant, @Cocopuffs , but I can't tell. $\endgroup$
    – DonAntonio
    Oct 11, 2013 at 22:43
  • $\begingroup$ @Cocopuffs: Yes, that's what I meant. Fixing. $\endgroup$
    – Gils
    Oct 11, 2013 at 22:49

1 Answer 1


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?

  • 1
    $\begingroup$ It's not the best possible. In my proof, we cannot choose "d-1" of the roots, because then we already have the last root too. So $2^{d-1}$ is a bound too. I tried in several ways to see if this is tight, but with no conclusion so far. Can you give me a hint? $\endgroup$
    – Gils
    Oct 12, 2013 at 11:47
  • 2
    $\begingroup$ @Gils Your new bound is correct, and it's what I had in mind. I'm not sure about a tight bound. For Galois extensions, the problem breaks down to asking for a bound, depending on $d$, on the number of subgroups of a group with $d$ elements. It's probably a difficult problem. $\endgroup$ Oct 12, 2013 at 16:33

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