Does there exist a field $K$ and a finite-dimensional $K$-division algebra $D$ possessing two maximal separable subfields of different dimensions?


Remark: If $D$ is separable ($Z(D)$ a separable field extension of $K$), then all such subfields have the same dimension.



Dimension of maximal tori in division algebras
No, there are no such examples, but I don't know any way to attack this by methods of ring theory. The theory of linear algebraic groups gives a very illuminating insight into this matter, by recasting the problem in a wider context. 


To explain this, let's work more generally for the moment with finite-dimensional associative $K$-algebras $C$ with identity such that the center $Z$ is a finite reduced $K$-algebra (i.e., a product of finitely many finite extension fields of $K$). Note that $Z$ is a field if $C$ is simple, and we specifically avoid limiting ourselves to division algebras because that way our setup is preserved under *separable* extension of the ground field (which is not the case for division algebras). 


Note that separable extension on $K$ will typically make the center decompose into several more factor fields, which is why we do not assume $Z$ is a field either (but just a finite product of such). In particular, 
$$C \otimes\_K K\_s = C \otimes\_Z (Z \otimes\_K K\_s) = \prod (C \otimes\_Z Z\_i)$$
where $\prod Z\_i$ is the decomposition of the nonzero finite reduced $K\_s$-algebra $Z \otimes\_K K\_s$ into a product of fields. Each $Z\_i$ is separably closed, being finite over $K\_s$, so each $C \otimes\_Z Z\_i$ is a matrix algebra over $Z\_i$ (though not over $K\_s$!) since Brauer groups of separably closed fields are trivial.


The crucial thing is to relate separable (commutative) subfields of $C$ over $K$, or more generally finite etale (commutative) $K$-subalgebras of $C$, to $K$-tori in the associated "algebraic $K$-group of units" of $C$. By definition, this linear algebraic group represents the functor on $K$-algebras defined by $R \rightsquigarrow (C \otimes\_K R)^{\times}$. More concretely, if we consider the affine $d^2$-space $\underline{C}$ over $Z$ associated to $C$ then $G$ is the 
Weil restriction ${\rm{R}}\_{Z/K}(\underline{C}^{\times})$ of the affine open subspace $\underline{C}^{\times} \subset \underline{C}$ of "units over $Z$": the complement of the zero-locus of the reduced norm ${\rm{Nrd}}:\underline{C} \rightarrow \mathbf{A}^1\_Z$ that is a polynomial map over $Z$. 


If $A \subset C$ is a finite etale $K$-subalgebra then the algebraic group of units $T\_A := \underline{A}^{\times} \subset G$ 
(defined functorially on points valued in a $K$-algebra $R$ by $(A \otimes\_K R)^{\times} \hookrightarrow (C \otimes\_K R)^{\times}$) is a $K$-torus, and conversely if $T \subset G$ is a $K$-torus then the $K\_s$-subalgebra $K\_s[T(K\_s)] \subset C\_{K\_s}$ is ${\rm{Gal}}(K\_s/K)$-stable and a direct product of copies of $K\_s$ (here one uses that $T\_{K\_s}$ is a split torus) so it descends to a $K$-subalgebra $A\_T \subset C$ that is $K$-etale. These two procedures are inclusion-preserving inverse bijections between the set of such $A$ and the set of $K$-tori in $C$, inclusion-preserving in both directions. By considerations over $K\_s$ one sees that $\dim\_K A = \dim T\_A$, so equality of the $K$-dimensions of maximal $K$-etale commutative subalgebras of $C$ is the same as the equality of the dimensions of
all maximal $K$-tori in $G$.


By construction, the formation of $G$ commutes with any separable extension on $K$. In particular, $G\_{K\_s} = \prod {\rm{R}}\_{Z\_i/K\_s}({\rm{GL}}\_{n\_i}))$ where $n\_i^2 = \dim\_{Z\_i} (C \otimes\_Z Z\_i)$, so $G$ is a smooth connected affine $K$-group. If $Z$ is $K$-etale then $G$ is even reductive (as finite etale Weil restriction preserves reductivity), but this *always fails* when $Z$ has a factor field not separable over $K$ (though $G$ is always pseudo-reductive, but we don't need this). In what follows the link to reductive groups won't be used; nonetheless, it does clarify the significance of $Z$ being $K$-etale or not. 


In case $C$ is a division algebra, a $K$-etale subalgebra cannot have nontrivial idempotents (as $C$ has no non-zero zero-divisors) and hence must be a field, moreover separable over $K$ by the etale condition.
Thus, we see that our task is a special case of one that has nothing at all to do with division algebras: show that every maximal $K$-torus in $C$ has the same dimension. 


More generally, if $G$ is *any* smooth affine group over a field $K$ then we claim that all maximal $K$-tori in $G$ have the same dimension. Over algebraically closed fields it is a basic fact in the theory of algebraic groups that all maximal tori are conjugate and so have the same dimension. Thus, it is enough to prove that if $T \subset G$ is a maximal $K$-torus then $T\_{\overline{K}}$ is a maximal $\overline{K}$-torus in $G\_{\overline{K}}$. The centralizer $Z\_G(T)$ is a smooth closed $K$-subgroup of $G$ by the general theory of linear algebraic groups (this is part of the Corollary to Theorem 9.2 in Borel's book "Linear Algebraic Groups", though that might not be apparent from reading the formulation there), and $H := Z\_G(T)/T$ certainly contains no nontrivial $K$-tori, so it suffices to show that $H\_{\overline{K}}$ contains no nontrivial $\overline{K}$-tori. 


Finally, it suffices to show that if $H$ is a smooth affine $K$-group then there exists *some* $K$-torus $S \subset H$ that remains maximal over $\overline{K}$ (as then if $H$ has no nontrivial $K$-tori the same must hold for $H\_{\overline{K}}$, as desired). This final assertion is an important theorem of Grothendieck proved as 18.2(i) in Borel's book "Linear Algebraic Groups" (a simplification on the original proof in SGA3).


