the number of intermediate fields in a simple field extension of degree $n$

suppose that $K|F$ is a simple field extension with degree $n$,prove that the number of intermediate fields is less or equal $2^{n-1}$.

i've done this:

assume $K=F(a)$ and $L$ is a intermediate field .consider $f(x)\in F[x]$ the minimal polynomial of $a$ over $F$ and $g(x)\in L[x]$ the minimal polynomial of $a$ over $L$. we have $g|f$ ,i want to make a surjective correspondence between the irreducible polynomials that divides $f$ and the intermediate fields.

is it a good idea?

any hint is welcomed!

• If you've done some Galois Theory, you can work with the groups instead of the fields, and show that a group of $n$ elements has at most $2^{n-1}$ subgroups. – Gerry Myerson Oct 16 '14 at 9:10
• @Gerry Myerson no i don't know Galios theory. – user115608 Oct 16 '14 at 9:11
• Then you have something wonderful to look forward to. – Gerry Myerson Oct 16 '14 at 9:13
• it is my homework and the professor expect us to solve it. – user115608 Oct 16 '14 at 9:20
• salam shoma soale 9 ro hal kardid ? :) – Fardad Pouran Oct 17 '14 at 19:19

If $f(X)$ is the minimal polynomial for $a$ over $F$, then $f(x)$ factors into irreducibles in $F(a)[X]$: $f(X) = (X-a)f_1(X)\cdots f_k(X)$, where $k\leq n-1$.
If K is an intermediate field, $F \subsetneq K \subsetneq F(a)$, then the minimal polynomial of $a$ over $K$ is a product of some collection of these irreducibles. Since this product always contains $(X-a)$ we can ignore it in our count. So we get a map from the set of intermediate fields into subsets of $\{1,2,\cdots ,k\}$, with $k\leq n-1$.