For a division ring $D$, does $[D:C_D(a)]_{right}$ vary when $D$ is enlarged?

Question

In a commutative field $K$, the Zariski dimension of an algebraic subset of $K^n$ over $K$ does not vary if one enlarges $K$ if I understood well. In particular, for two Zariski-closed vector spaces $V(K)\subset W(K)\subset K^n$, if one considers an extension field $L/K$, one has $$[V(K):W(K)]=[V(L):W(L)].$$

Let $D$ be a division ring with infinite centre $k$, infinite dimentionnal over $k$. Let $a$ be a non-central element of $D$. Its centraliser $C_D(a)$ in $D$ is defined by the simple linear equation $$ax-xa=0.$$ The ring $D$ is a right vector space over $C_D(a)$ having dimension $[D:C_D(a)]_{right}$.

Let $L/D$ be a division ring extending $D$ and still having $k$ for centre.

Question 1 : Is it possible to have $[D:C_D(a)]_{right}$ finite and $[L:C_L(a)]_{right}$ infinite ? And vice-versa ?

Question 2 : Is it possible to have $[C_D(a):k]$ finite and $[C_L(a):k]$ infinite ?

Answer 1

There is a result of Brauer (1932) stating that :

Theorem (Brauer). Let $K$ be a division ring with centre $k$ and $A$ a $k$-algebra. Then the centralizer $A'$ of $A$ in $K$ is again a $k$-algebra and the bicentralizer $A''$ contains $A$. Moreover $$[K:A']_{left}=[A:k],$$ whenever either side is finie, and when this is so, $A''=A$.

In particular, taking $A=C(a)$ one gets a No answer to both question 1 and 2.