TITLE: Is the sub-field of algebraic elements of a field extension of $K$ containing roots of polynomials over $K$ algebraically closed?
QUESTION [5 upvotes]: If I have a field $K$ and an extension $L$ of $K$ such that all (non-constant)
polynomials in $K[X]$ have a root in $L$, is the set of algebraic elements 
of $L$ over $K$ (the sub-field of all the elements of $L$ which are roots of a polynomial in $K[X]$) algebraically closed ?
Do you have a counterexample ?

REPLY [5 votes]: This is a standard result that follows by factorization into separable and inseparable extensions. See the post below from Ask an Algebraist. Alas, the thread has since been deleted (due to vandals), but the references below to Isaacs will help you to locate the standard conceptual proof (to be contrasted with the non-conceptual elementary form given below by a student - cf. my remarks following it). See Isaacs: Algebra: a graduate course. Ch. 19: Separability and Inseparability and see esp. the remark preceding Theorem 19.18.

From: Bill Dubuque; Date: July 6, 2010
In reply to: "Thanks & Summary", posted by student on July 6, 2010:
student wrote:  

Thanks for your help, it's all clear now. The proof you gave provides good intuition!
  For my own understanding I have written it up in total and tried to make the paralles 
  between the separable and the inseparable case clear. I post the proof below,
  maybe someone reading this will benefit from it.
BTW you said that this theorem occurs in many textbooks.
  Actually I could only locate it in Isaacs book (pp. 303-304),
  however with a different proof (as you suggested).
  Are there any specific books about algebra in general
  or field theory in particular you would recommend?
OK here comes the proof (hopefully without typos),
  which is basically a summary of the previous posts in this thread:
THEOREM
  Let $L$ and $K$ be fields s.t. $L$ is an algebraic extension of $K$.
  Assume every irreducible polynomial in $K[X]$ has at least one zero in $L$.
  Then $L$ is an algebraic closure of $K$.
REMARK
  If in the situation of the theorem $\operatorname{char}(L) = p$, then for every integer $k \geq 1$
  every element $c \in K$ has a unique $p^k$-th root $d$ in $L$ (i.e. $d^{p^k} = c$).
PROOF (REMARK)
  Existence: The polynomial $X^{p^k} - c$ is in $K[X]$ and hence has a root $d$ in $L$.
  Uniqueness: Assume $d^{p^k} = c = e^{p^k}$. Then $(d - e)^{p^k} = 0$ [freshman's dream]
  and hence $d = e$.
PROOF (THEOREM)
  Suppose $f$ is an irreducible polynomial with coefficients in $K$.
  To prove the theorem we have to show that $f$ splits over $L$.
  The proof is divided into two cases: $f$ separable and $f$ inseparable.
*) 1st case: $f$ separable.
  Let $S$ denote a splitting field for $f$ over $K$
  lying inside a splitting field for $f$ over $L$.
  Being the splitting field for a separable polynomial over $K$,
  $S$ has a primitive element: $S = K(a)$.
  Let $g$ be the minimal polynomial of $a$ over $K$.
  Then $g$ has a root, say $b$, in $L$.
  Since $g$ is irreducible and both $a$ and $b$ are roots of $g$,
  $K(a)$ is isomorphic to $K(b)$.
  [In fact $K(a) = K(b)$:
  Since $S$ (being the splitting field for a polynomial in $K[X]$)
  is normal over $K$, and since $g$ is irreducible and
  has a root in $S$ (namely $a$), $g$ must split over $S$, hence $b$ is in $K(a)$.
  Since $K(a)$ and $K(b)$ both have the same degree (namely $\deg(g)$) over $K$,
  $K(a) = K(b)$.]
  Thus $L \supseteq K(b) = K(a) = S$ contains all roots of $f$.
*) 2nd case: $f$ inseparable.
  Then $\operatorname{char}(K) = p > 0$,
  and $f = g(X^{p^n})$ for a separable irreducible polynomial $g\in K[X]$ and some $n > 0$.
  By the 1st case $g$ splits in $L$, and we may let $a_1,...,a_m$ be the roots of $g$ in $L$.
  Then $S = K(a_1,...,a_m)$ (being the splitting field for a separable polynomial over $K$)
  has a primitive element: $S = K(a)$.
  Let $h$ be the minimal polynomial of $a$ over $K$.
  The polynomial $h(X^{p^n})$ is in $K[X]$ and hence has a root in $L$, say $b$.
  Then $b^{p^n}$ is a root of $h$ in $L$.
  Since $h$ is irreducible and both $a$ and $b^{p^n}$ are roots of $h$,
  $K(a)$ is isomorphic to $K(b^{p^n})$.
  In fact $K(a) = K(b^{p^n})$ [same argument as in the 1st case].
Now let $r$ denote a root of $f$ lying inside a splitting field for $f$ over $L$.
  Then $r^{p^n}$ is a root of $g$, hence $r^{p^n}$ is in $S = K(a) = K(b^{p^n})$.
  Thus $r^{p^n} = F(b^{p^n})$ for some polynomial $F \in K[X]$.
  By the remark, every coefficient of $F$ has a $p^n$-th root in L,
  so $F(b^{p^n}) = G(b)^{p^n}$ for some $G \in L[X]$ [freshman's dream].
  Thus $(r - G(b))^{p^n} = 0$, hence $r = G(b)$ is in $L(b) = L$.
Q.E.D.

This is just an "elementary" form of the standard proof (e.g. cf. Isaacs ref. above).
But, alas, by eliminating the innate structural elements (factorization into separable
and inseparable extensions) in favor of "simpler" calculations with elements,
the innate structure of the proof is greatly obfuscated. The elementary proof
has been constructed by directly inlining the lemmas leading up to Isaacs proof
vs. invoking them by name. This is not good pedagogically. Decomposing algebraic 
extensions into their separable and inseparable parts is an essential tool required 
to study general algebraic extensions. Such key ideas should not be obscured such
as above - esp. when, as here, the structural form is just as short and simple as is
the elementary form. Please read the section of Isaacs leading up to the proof to
better understand the innate structure. You've a much better chance of remembering
the structural proof years later because it has been abstracted into simple steps
using key structures/lemmas that will often be reused. But few if any students
would probably remember the non-conceptual elementary form as presented above.