Must algebraic extensions of the same degree have subfields of the same degree? Let $\mathbb F$ be a field and let $\mathbb K_1$ and $\mathbb K_2$ be finite extensions of $\mathbb F$ with the same degree, that is, $[\mathbb K_1:\mathbb F]=[\mathbb K_2:\mathbb F]$. Now, assume that $\mathbb K_1$ contains a subfield of degree $s$ over $\mathbb F$. My question is: Can we conclude that also $\mathbb K_2$  contains a subfield of degree $s$ over $\mathbb F$?
If the field $\mathbb F$ is finite, then this is true, since we can  embedded $\mathbb K_1$ and $\mathbb K_2$ in the matrix ring $M_m(\mathbb F)$, where $m=[\mathbb K_1:\mathbb F]=[\mathbb K_2:\mathbb F]$ and it can be proved that there is an inner authomorphism of $M_m(\mathbb F)$ that apply $\mathbb K_1$ in $\mathbb K_2$. 

Let $F=\mathbb{Q}$. Let $K_1$ be the splitting field of $f_1=x^4+8x+12$ over $\mathbb{Q}$, and let $K_2$ be the splitting field of $f_2=x^{12}+x^{11}+\cdots+x+1$ over $\mathbb{Q}$ (this is just the $13$th cyclotomic polynomial). Then
$$[K_1:F]=[K_2:F]=12$$
However, $\mathrm{Gal}(K_1/F)\cong A_4$ and $\mathrm{Gal}(K_2/F)\cong\mathbb{Z}/12\mathbb{Z}$, and recall that $\mathbb{Z}/12\mathbb{Z}$ has a subgroup of size $6$, while $A_4$ does not. Therefore, by the fundamental theorem of Galois theory, there is an intermediate field $K_2\supset L_2\supset F$ with $[L_2:F]=2$, but there is no intermediate field $K_1\supset L_1\supset F$ with $[L_1:F]=2$.
(For my claim that $\mathrm{Gal}(K_1/F)\cong A_4$, see p.6 of this article by Keith Conrad.)
