Probably a stupid question, but..
Why is the splitting field of a separable polynomial necessarily separable?
Thanks.
Follow up question
Show that if $F$ is a splitting field over $K$ for $P \in K[X]$, then $[F:K] \leq n!$
I've proven this by induction on $n$, but I'm convinced there's a more algebraic approach ($n!$ screams $S_n$).
If I knew $P$ were separable, then I'd know $F$ was Galois, so $ |\mbox{Gal}(F/K)| = [F:K] $. Considering the action of $ \mbox{Gal}(F/K) $ on the roots of $P$, we'd get an injective homomorphism into $ S_n $, giving the result.
But $P$ is not given to be separable, so $P$ could have repeated roots when factorised in $F$. If $G$ is any finite group of $K$-automorphisms of $L$, then I know that $[F:K] \leq |G| $. Considering the action of $G$ on the set of roots of $P$, say $\Omega$, we get an injective homomorphism of $G$ into $S_{|\Omega|}$, where $|\Omega| \leq n$ (I think). I have a feeling I'm wrong here, since I think $G$ is forced to be $\mbox{Aut}(F/K)$. Any advice would be appreciated.
Best Answer
New Answer
[New version of the new answer. Thank you very much to KCd! --- See his comment.]
The simplest, I think, is to prove the Fundamental Theorem of Galois Theory (FTGT) for a normal extension $E/F$ generated by finitely many separable elements, and to obtain the fact that all the elements of $E$ are separable over $F$ as a corollary.
The point is that the FTGT can be given a very short proof, which will be described below. (I of course believe that this proof is complete. Thanks for correcting me if I'm wrong.)
We are taking for granted
$(1)$ the universal property of a simple algebraic extension, and
$(2)$ the existence and universal property of our splitting field $E$,
which we briefly recall:
$(1)$ If $a$ is algebraic over a field $K$, then the $K$-embeddings of $K(a)$ in an extension $L$ of $K$ are in natural bijection with the roots in $L$ of the minimal polynomial of $a$ over $K$.
$(2)$ Any root of $p$ in any extension of $E$ is in $E$, and $E$ is generated over $F$ by the roots $a_i$ of $p$. Moreover, if $S/F$ is a sub-extension of $E/F$, then any $F$-embedding of $S$ in $E$ extends to an $F$-automorphism of $E$.
Going back to the assumptions of the FTGT, we claim:
$(3)$ If $S/F$ is a sub-extension of $E/F$, then $[E:S]=|\text{Aut}_S E|$.
$(4)$ If $H$ is a subgroup of $G$, then $|H|=[E:E^H]$.
Proof that (3) and (4) imply the FTGT. Let $S/F$ be a sub-extension of $E/F$ and put $H:=\text{Aut}_S E$. Then we have trivially $S\subset E^H$, and $(3)$ and $(4)$ imply $$ [E:S]=[E:E^H]. $$ Conversely let $H$ be a subgroup of $G$ and set $\overline H:=\text{Aut}_{E^H}E$. Then we have trivially $H\subset\overline H$, and $(3)$ and $(4)$ imply $|H|=|\overline H|$.
To prove that any element $a$ of $E$ is separable over $F$, put $H:=\text{Aut}_{F(a)}E$, and note that the set of $F$-embeddings of $F(a)$ in $E$ is in natural bijection with the set $G/H$, whose cardinality is, by the FTGT, equal to $[F(a):F]$.
Proof of (3). The statement follows from the fact that any $F$-embedding of $S_i:=S(a_1,\dots,a_i)$ in $E$ has exactly $[S_{i+1}:S_i]$ extensions to $S_{i+1}$.
Proof of (4). In view of (3) it is enough to check $|H|\ge[E:E^H]$. Let $k$ be an integer larger than $|H|$, and pick a $$ b=(b_1,\dots,b_k)\in E^k. $$ We must show that the $b_i$ are linearly dependent over $E^H$, or equivalently that $b^\perp\cap(E^H)^k$ is nonzero, where $?^\perp$ denotes the vectors orthogonal to ? in $E^k$ with respect to the dot product on $E^k$. Any element of $b^\perp \cap (E^H)^k$ is necessarily orthogonal to $h(b)$ for any $h \in H$, so $b^\perp \cap (E^H)^k = (Hb)^\perp \cap (E^H)^k$, where $Hb$ is the $H$-orbit of $H$. We will show $(Hb)^\perp \cap (E^H)^k$ is nonzero. Since the span of $Hb$ in $E^k$ has $E$-dimension at most $|H| < k$, $(Hb)^\perp$ is nonzero. Choose a nonzero vector $x$ in $(Hb)^\perp$ such that $x_i=0$ for the largest number of $i$ as possible among all nonzero vectors in $(Hb)^\perp$. Some coordinate $x_j$ is nonzero in $E$, so by scaling we can assume $x_j = 1$ for some $j$. Since the subspace $(Hb)^\perp$ in $E^k$ is stable under the action of $H$, for any $h$ in $H$ we have $h(x) \in (Hb)^\perp$, so $h(x)-x \in (Hb)^\perp$. Since $x_j = 1$, the $j$-th coordinate of $h(x) - x$ is $0$, so $h(x)-x = 0$ by the choice of $x$. Since this holds for all $h$ in $H$, $x$ is in $(E^H)^k$.
Old Answer
This is more a hint than an answer. I think the key point here is to understand the equivalence between various definitions of a separable extension. I'll just try to emphasize some of the basic facts.
Let $K$ be a field, $A/K$ an algebraic closure, and $L/K$ an extension of degree $d < \infty$ contained in $A$. We claim that the following conditions are equivalent:
(1) $L/K$ is generated by separable elements,
(2) each element of $L$ is separable over $K$,
(3) there are exactly $d$ embeddings of $L$ in $A$ over $K$.
We say that $L/K$ is separable if it satisfies these conditions. Let's sketch a proof of the equivalence.
Say that the number, denoted by $[L:K]_s$, of $K$-embeddings of $L$ in $A$ is the separable degree of $L/K$.
Let $M$ be a finite degree extension of $K$ contained in $A$, and $L$ a field between $K$ and $M$. We claim:
(4) $[M:K]_s=[M:L]_s\ [L:K]_s$.
This is an easy, but instructive, exercise.
Claim: (1) and (3) are equivalent when $K$ is generated by a single element $\alpha$. Indeed, if $f\in K[X]$ is the minimal polynomial of $\alpha$, then $[L:K]$ is the degree of $f$, whereas $[L:K]_s$ is the number of distinct roots of $f$ in $A$.
In view of (4), the above claim implies that (1), (2) and (3) are equivalent in the general case.
As a corollary, we see that a finite degree extension $L/K$ contains a largest separable sub-extension $S$, and that $[S:K]=[L:K]_s$.
As a corollary to the corollary, we get the fact suggested by the notation that the integer $[L:K]_s$ does not depend on the choice of an algebraic closure of $L$.