Extension of divisible fields 
Assume that $F$ is an infinite subfield of a field $K$ such that its multiplicative group, $F^\times$, is divisible. Also, $a\in K$ and $[F(a):F]<\infty$. Can we conclude that the  multiplicative group of $F(a)$ is a divisible group? 

For example the root closure of $\mathbb Q$ in $\mathbb C$, is a divisible field.
Can we obtain that any finite extension of this field is a divisible field?
 A: There is a finite extension of the root closure of $\mathbb{Q}$ whose multiplicative group is not divisible.
First, even though it is obvious, let us state it: a field $F$ has a divisible multiplicative group if and only if for every $n\in \mathbb{N}$, every element of $F$ admits an $n$-th root.
Recall the definition of the root closure $\mathbb{Q}^r$ of $\mathbb{Q}$.  Define a root extension of $\mathbb{Q}$ to be a sequence of extensions of the form
$$
\mathbb{Q}\subset \mathbb{Q}(a_1) \subset \mathbb{Q}(a_1, a_2) \subset \ldots \subset \mathbb{Q}(a_1, \ldots, a_r),
$$
where a power of $a_1$ lies in $\mathbb{Q}$ and for each $i\in\{2,\ldots r\}$, a power of $a_i$ lies in $\mathbb{Q}(a_1, \ldots, a_{i-1})$.
Let $\mathbb{Q}^r$ be the union of all possible root extensions of $\mathbb{Q}$.  It is not hard to see that $\mathbb{Q}^r$ is a subfield of $\mathbb{Q}^a$ (where $\mathbb{Q}^a$ is the algebraic closure of $\mathbb{Q}$).  Moreover, since some polynomials over $\mathbb{Q}$ cannot be solved by radicals by the Abel-Ruffini Theorem, the field $\mathbb{Q}^r$ is strictly contained in $\mathbb{Q}^a$.
By construction, for every $n\in \mathbb{N}$, every element of $\mathbb{Q}^r$ admits exactly $n$ $n$-th roots in $\mathbb{Q}^r$.  Thus $(\mathbb{Q}^r)^*$ is divisible.
Let $L$ be a finite extension of $\mathbb{Q}^r$ (such an extension exists, since $\mathbb{Q}^r \subsetneq \mathbb{Q}^a$).  We can assume that $L/\mathbb{Q}^r$ is a Galois extension (if it isn't, replace $L$ by its Galois closure).  Let $p$ be a prime number dividing the order of the Galois group $Gal(L/\mathbb{Q}^r)$.  By Cauchy's theorem, there is a subgroup of $Gal(L/\mathbb{Q}^r)$ of order $p$.  Hence, by the Galois correspondence, there is an intermediate field extension
$$
\mathbb{Q}^r \subsetneq L' \subsetneq L
$$
such that $[L:L']=p$.  
Then the multiplicative group of $L'$ is not divisible.  Indeed, assume that it is.  Since $L'$ contains all roots of unity, then for any $n\in \mathbb{N}$, any element of $L'$ admits exactly $n$ $n$-th roots in $L'$.  Now, the Galois group of $L/L'$ has order $p$, and is thus solvable; therefore, if $b\in L\setminus L'$, then the minimal polynomial $P_b$ of $b$ over $L'$ is solvable by radicals.  But this implies that $b\in L'$, a contradiction.
Thus $(L')^*$ is not divisible, and there exists an $a\in L'$ and an $n\in \mathbb{N}$ such that $a$ does not admit any $n$-th root in $L'$.
Therefore, if we take $F=\mathbb{Q}^r$ and $a$ as above, then $F(a)^*$ is not divisible, even though $F^*$ is.

A finite example
The only finite field whose multiplicative group is divisible is $\mathbb{F}_2$.  Thus, taking $F= \mathbb{F}_2$ sitting inside any finite extension $K$ gives us a counter-example.
