Suppose $K_1$ and $K_2$ are two different field extesnions over field $F$, and all the proper intermediate fields in the field extension $K_1$ of $F$ and $K_2$ of $F$ are the same.(here we suppose both of the field extensions have more than $1$ intermidiate field, i.e. $F$ is not the only intermediate field of the extensions above), can we find such fields $K_1,K_2$ and $F$?
Does there exist two different fields over $F$ such that they have the same intermediate fields
abstract-algebraextension-fieldfield-theory
Related Solutions
Based on @JyrkiLahtonen's comment above.
If either $[K_1 : K] = \infty$ or $[K_2 : K] = \infty$, then the inequality is trivially true. So, let both, $K_1/K$ as well as $K_2/K$ be finite extensions. Let $\{ a_1,\dots,a_n \}$ be a basis for $K_1$ over $K$ and let $\{ b_1,\dots,b_m \}$ be a basis for $K_2$ over $K$. Consider $V = \operatorname{span}_K\{ a_i b_j : 1 \leq i \leq n, 1 \leq j \leq m \}$, the vector subspace of $L$ spanned by the vectors $a_i b_j$ over the field $K$. We wish to show that $V$ is a field.
To show that $V$ is a ring, it suffices to show that $a_i b_j \cdot a_k b_l \in V$ for all $1 \leq i, k \leq n$, $1 \leq j, l \leq m$. Now, $$ a_i b_j \cdot a_k b_l = a_i a_k \cdot b_j b_l $$ and $a_i a_k \in K_1$, $b_j b_l \in K_2$. So, we can express them as a $K$-linear combinations of $ a_1,\dots,a_n$ and $b_1,\dots,b_m$, respectively. That is, $$ a_i a_k = \sum_{p = 1}^n c_{ikp} a_p \quad \text{and} \quad b_j b_l = \sum_{q = 1}^m d_{jlq} b_q $$ for some scalars $c_{ikp}, d_{jlq} \in K$. Hence, $$ a_i b_j \cdot a_k b_l = \left( \sum_{p = 1}^n c_{ikp} a_p \right) \cdot \left( \sum_{q = 1}^m d_{jlq} b_q \right) = \sum_{p=1}^n \sum_{q=1}^m (c_{ikp}d_{jlq}) a_p b_q \in V. $$ Hence, $V$ is a ring. In particular, $V$ is an integral domain because it is contained in $L$ which is a field.
Next, we need to show that the multiplicative inverse in $L$ of every non-zero element in $V$ lies in $V$ itself. Let $r \in V$, $r \neq 0$. Since $V$ is spanned over $K$ by a finite set, $V$ is a finite-dimensional vector space over $F$. If $\dim_K V = d$, then the set $\{ 1, r, r^2, \dots, r^d \}$ is a $K$-linearly dependent set. Hence, there exist $c_0,c_1,\dots,c_d \in K$, not all zero, such that $$ c_0 + c_1 r + c_2 r^2 + \dots + c_d r^d = 0. $$ Let $k = \min\{ 0 \leq i \leq d : c_i \neq 0 \}$. Then, $$ c_k r^k + c_{k+1} r^{k+1} + \dots + c_d r^d = 0.\tag{1} $$ It cannot be that $c_i = 0$ for all $i \neq k$ because otherwise we would have $$ c_k r^k = 0 \implies r^k = 0 \implies r = 0, $$ which is a contradiction. Note that here we are crucially using the fact that $V$ is an integral domain. So, we have concluded that $k < d$. Now, from $(1)$ we get that $$ \begin{align} & &c_k r^k + c_{k+1}r^{k+1} + \dots + c_d r^d &= 0 \\ &\implies &r^k(c_k + c_{k+1}r + \dots + c_d r^{d-k}) &= 0 \\ &\implies &c_k + c_{k+1}r + \dots + c_d r^{d-k} &= 0\\ &\implies &r(c_{k+1} + c_{k+2} r + \dots + c_d r^{d-k-1}) &= -c_k\\ &\implies &-c_k^{-1}(c_{k+1} + c_{k+2} r + \dots + c_d r^{d-k-1}) &= r^{-1}. \end{align} $$ So, $r^{-1}$ lies in the $K$-span of $\{ 1 , r, r^2, \dots, r^d \}$ which is a subspace of $V$. Therefore, $r^{-1} \in V$ for all nonzero $r \in V$. Thus, $V$ is a field.
Any field containing both $K_1$ and $K_2$ must contain $a_i b_j$ for all $1 \leq i \leq n$, $1 \leq j \leq m$. Hence, it must also contain the $K$-span of $\{ a_i b_j \}$. But we have just shown that this is a field, so it must be the minimal field containing both $K_1$ and $K_2$. In other words, $L = K(K_1,K_2) = V$. Thus, any basis for $L$ over $K$ can contain no more than $nm$ elements. In other words, $$ [L : K] \leq [K_1 : K] [K_2 : K]. $$ Hence, proved.
Here's an example for $F_1,F_2$ finite extensions of $k$. Take $k = \mathbb{Q}, F_1 = \mathbb{Q}(\sqrt{2}), F_2 = \mathbb{Q}(\sqrt{3}), K = \mathbb{C}$. I claim $F_1F_2$ is not a field. Indeed, $\sqrt{2} = \sqrt{2}\cdot1 \in F_1, \sqrt{3} = \sqrt{3}\cdot1 \in F_2$, but $\sqrt{2}+\sqrt{3} \not \in F_1F_2$ since if $\sqrt{2}+\sqrt{3} = (a+b\sqrt{2})(c+d\sqrt{3})$ for $a,b,c,d \in \mathbb{Q}$, then we must have $bd = 0$; if $b = 0$, then we'd need $ac+ad\sqrt{3} = \sqrt{2}+\sqrt{3}$, impossible, and similarly if $d = 0$.
Best Answer
Consider $F=\Bbb Q$, $K_1=\Bbb Q(\sqrt[4]{2})$ and $K_2=\Bbb Q(\sqrt{2+\sqrt{2}})$. In both cases, the only intermediate field is $\Bbb Q(\sqrt{2})$. (Besides $\Bbb Q$, of course)