Non-algebraically closed field in which every polynomial of degree $My problem is to build, for every prime $p$, a field of characteristic $p$ in which every polynomial of degree $\leq n$ ($n$ a fixed natural number) has a root, but such that the field is not algebraically closed.
If I'm not wrong (please correct me if I am) such a field cannot be finite, by counting arguments. But on the other hand, the union of all finite fields (or of any ascending chain of finite fields) of characteristic $p$, which is what I get if I start with $F_p$ and add a root to each polynomial of degree $\leq n$ in each step, is the algebraic closure of $F_p$, hence algebraically closed. I don't see how I can control this process so that in the end I get a field that is not algebraically closed.
Any hint will be welcome. Thanks in advance.

Let $k$ be a field, $\bar k$ an algebraic closure of $k$.  Fix $n>1$ natural. Consider the family $\mathcal{K}_n$ of fields $K$, $k\subset K\subset \bar k$  with the property: there exists a family of intermediate fields 
$$k = K_0 \subset K_1 \subset \ldots K_s= K$$
so that $[K_{i+1}\colon K_i]< n$ for all $1\le i \le s$. It is easy to check the following:


*

*$K \in \mathcal{K}_n$, $K\subset L \subset \bar k$, $[L\colon K]< n$ implies $L \in \mathcal{K}_n$ 

*$K$, $K'\in \mathcal{K}_n$ implies $K K'\in \mathcal{K}_n$.

*$K \in \mathcal{K}_n$, $k \subset K'\subset K $ implies $K'\in \mathcal{K}_n$.
It is easy to see now that the union of the subfields in $\mathcal{K}_n$ is a subfield $k^{(n)}$ and every polynomial of degree $<n$ with coefficients in $k^{(n)}$ splits completely in $k^{(n)}$. 
Note that the degree over $k$ of every element in $k^{(n)}$ has all its prime factors $<n$. Therefore, if $k$ is such that there exist elements in $\bar k$ whose degree over $k$ is a prime factor $>n$ (many examples here) then $k^{(n)}\ne \bar k$, that is $k^{(n)}$ is not algebraically closed.
Note: for $n=3$ we get $k^{(n)}$ are the constructible elements of $\bar k/k$. 
