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I wish to prove that if $K/E$, $E/F$ are algebraic separable field extensions then $K/F$ is separable. I tried taking $a\in K$ and said that if $a\in E$ it is clear, otherwise I looked at the minimal polynomials over $E,F$ and called them $f(x),g(x)$ respectively, then I said that $f(x)$ is separable and since there exist $h(x)$ s.t $f(x)h(x)=g(x)$ it remains to see that $h(x)$ is separable — here I am stuck.

Can someone please help proving this claim or can say how to continue ?

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I think this is most naturally proved using the notion of the separable degree of an extension. One shows that a finite extension is $K/E$ separable $\Leftrightarrow$ $[K : E]_s = [K : E]$. Have you heard of this? –  Dylan Moreland Jun 6 '12 at 17:28
    
@Dylan - I wonder, how did you make the big "-" sign ? –  Belgi Jun 6 '12 at 17:29
    
@DylanMoreland - no, I do not know about separable degree. I can tell you that this was homework but the "proof" my partner submitted said $h(x)$ splits in $E$, I only discovered this now when I wanted to gain a better understanding of the matirial (so an argument without separable degree can be made since the lecture didn't teach this...) –  Belgi Jun 6 '12 at 17:31
    
Regarding the formatting (which I hope is alright): on a Mac I just hit Shift, Option, then "-". Not sure how to do it in general. As for the proof, I'll have to think of a way that avoids the concept. I agree that it should be doable. –  Dylan Moreland Jun 6 '12 at 17:31
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An irreducible polynomial $p$ is inseparable iff $p' = 0$. Now $g$ is irreducible in $F[X]$ and $$g' = f' h + f h'$$ Since $f'(a) \ne 0$, $f(a) = 0$, we obtain $g'(a) = f'(a) h(a)$. All that's missing now is an argument for why $h(a) \ne 0$. –  Sam Jun 6 '12 at 18:04

1 Answer 1

up vote 3 down vote accepted

This is only problematic in the context of positive characteristic, so assume that all fields are of cahracteristic $p$.

Suppose that $K/E$ is separable, and $E/F$ is separable. Let $S$ be the set of all elements of $K$ that are separable over $F$. Then $E\subseteq S$, since $E/F$ is separable.

Note that $S$ is a subfield of $K$: indeed, if $u,v\in S$ and $v\neq 0$, then $F(u,v)$ is separable over $F$ because it is generated by separable elements, so $u+v$, $u-v$, $uv$, and $u/v$ are all separable over $F$. So $S$ is a field.

I claim that $K$ is purely inseparable over $S$. Indeed, if $u\in K$, then there exists $n\geq 0$ such that $u^{p^n}$ is separable over $F$, hence there exists $n\geq 0$ such that $u^{p^n}\in S$. Therefore, the minimal polynomial of $u$ over $S$ is a divisor of $x^{p^n} -u^{p^n} = (x-u)^{p^n}$, so $K$ is purely inseparable over $S$.

But since $E\subseteq S\subseteq K$, and $K$ is separable over $E$, then it is separable over $S$. So $K$ is both purely inseparable and separable over $S$. This can only occur if $S=K$, hence every element of $K$ is separable over $F$. This proves that $K/F$ is separable.

Added. Implicit above is that an extension is separable if and only if it is generated by separable elements. That is:

Lemma. Let $K$ be an extension of $F$, and $X$ a subset of $F$ such that $K=F(X)$. If every element of $X$ is separable over $F$, then $K$ is a separable extension of $F$.

Proof. Let $v\in K$. Then there exist $u_1,\ldots,u_n\in X$ such that $v\in F(u_1,\ldots,u_n)$. Let $f_i(x)\in F[x]$ be the irreducible polynomial of $u_i$ over $F$; by assumption, $f_i(x)$ is separable. Let $E$ be a splitting field over $F(u_1,\ldots,u_n)$ of $f_1(x),\ldots,f_n(x)$. Then $E$ is also a splitting field of $f_1,\ldots,f_n$ over $F$, and since the $f_i$ are separable, $E$ is separable over $F$. Therefore, since $v\in F(u_1,\ldots,u_n)\subseteq E$, it follows that $v$ is separable over $F$. $\Box$


Added. Despite the OP's acceptance of this answer, it is clear from the comments that he does not actually understand the answer, which is rather frustrating. Equally frustrating is to be told, in drips, what it is the OP does and does not know about separability, in the form of "Just explain why this is true", only to be find out the facts that underlie that assertion are also unknown.

The following is taken from Hungerford's treatment of separability.

Definition. Let $F$ be a field and $f(x)\in F[x]$ a polynomial. The polynomial is said to be separable if and only if for every irreducible factor $g(x)$ of $f(x)$, there is a splitting field $K$ of $g(x)$ over $F$ where every root of $g(x)$ is simple.

Definition. Let $K$ be an extension of $F$, and let $u\in K$ be algebraic over $F$. Then $u$ is said to be separable over $F$ if the minimal polynomial of $u$ over $F$ is separable. The extension is said to be separable if every element of $K$ is separable over $F$.

Theorem. Let $K$ be an extension of $F$. The following are equivalent:

  1. $K$ is algebraic and Galois over $F$.
  2. $K$ is separable over $F$ and $K$ is a splitting field over $F$ of a set $S$ of polynomials in $F[x]$.
  3. $K$ is the splitting field over $F$ of a set $T$ of separable polynomials in $F[x]$.

Proof. (1)$\implies$(2),(3) Let $u\in K$ and let $f(x)\in F[x]$ be the monic irreducible polynomial of $u$. Let $u=u_1,\ldots,u_r$ be the distinct roots of $f$ in $K$; then $r\leq n=\deg(f)$. If $\tau\in\mathrm{Aut}_F(K)$, then $\tau$ permutes the $u_i$. So the coefficients of the polynomial $g(x) = (x-u_1)(x-u_2)\cdots(x-u_r)$ are fixed by all $\tau\in\mathrm{Aut}_F(K)$, and therefore $g(x)\in F[x]$ (since the extension is Galois, so the fixed field of $\mathrm{Aut}_F(K)$ is $F$). Since $u$ is a root of $g$, then $f(x)|g(x)$. Therefore, $n=\deg(f)\leq \deg(g) = r \leq n$, so $\deg(g)=n$. Thus, $f$ has $n$ distinct roots in $K$, so $u$ is separable over $F$. Now let $\{u_i\}_{i\in I}$ be a basis for $K$ over $F$; for each $i\in I$ let $f_i\in F[x]$ be the monic irreducible of $u_i$. Then $K$ is the splitting field over $F$ of $S=\{f_i\}_{i\in I}$, and each $f_i$ is separable. This establishes (2) and (3).

(2)$\implies$(3) Let $f\in S$, and let $g$ be an irreducible factor of $f$. Since $f$ splits in $K$, then $g$ is the irreducible polynomial of some $u\in K$, where it splits. Since $K$ is separable over $F$, then $u$ is separable, so $g$ is separable. Thus, the elements of $S$ are separable. So $K$ is the splitting field over $F$ of a set of separable polynomials.

(3)$\implies$(1) Since $K$ is a splitting field over $F$, it is algebraic. If $u\in K-F$, then there exist $v_1,\ldots,v_m\in K$ such that $u\in F(v_1,\ldots,v_m)$, and each $v_i$ is a root of some $f_i\in S$, since $K$ is generated by the roots of elements of $S$. Adding all the other roots of the $f_i$, $u\in F(u_1,\ldots,u_n)$, where $u_1,\ldots,u_n$ are all the roots of $f_1,\ldots,f_m$; that is, $F(u_1,\ldots,u_n)$ is a splitting field over $F$ of the polynomial $f_1\cdots f_m$.

If the implication holds for all finite dimensional extensions, then we would have that $F(u_1,\ldots,u_n)$ is a Galois extension of $F$, and therefore there exist $\tau\in \mathrm{Aut}_F(F(u_1,\ldots,u_n))$ such that $\tau(u)\neq u$. Since $K$ is a splitting field over $F$, it is also a splitting field over $F(u_1,\ldots,u_n)$, and therefore $\tau$ extends to an automorphism of $K$. Thus, there exists $\tau\in\mathrm{Aut}_F(K)$ such that $\tau(u)\neq u$. This would prove that the fixed field of $\mathrm{Aut}_F(K)$ is $F$, so the extension is Galois. Thus, we are reduced to proving the implication when $[K:F]$ is finite. When $[K:F]$ is finite, there is a finite subset of $T$ that will suffice to generate $K$. Moreover, $\mathrm{Aut}_F(K)$ is finite. If $E$ is the fixed field of $\mathrm{Aut}_F(K)$, then by Artin's Theorem $K$ is Galois over $E$ and $\mathrm{Gal}(K/E) = \mathrm{Aut}_F(K)$. Hence, $[K:E]=|\mathrm{Aut}_F(K)|$.

Thus, it suffices to show that when $K$ is a finite extension of $F$ and is a splitting field of a finite set of separable polynomials $g_1,\ldots,g_m\in F[x]$, then $[K:F]=|\mathrm{Aut}_F(K)|$. Replacing the set with the set of all irreducible factors of the $g_i$, we may assume that all $g_i$ are irreducible.

We do induction on $[K:F]=n$. If $n=1$, then the equality is immediate. If $n\gt 1$, then some $g_i$, say $g_1$, has degree greater than $1$; let $u\in K$ be a root of $g_1$. Then $[F(u):F]=\deg(g_1)$, and the number of distinct roots of $g_1$ in $K$ is $\deg(g_1)$, since $g_1$ is separable. Let $H=\mathrm{Aut}_{F(u)}(K)$. Define a map from the set of left cosets of $H$ in $\mathrm{Aut}_F(K)$ to the set of distinct roots of $g_1$ in $K$ by mapping $\sigma H$ to $\sigma(u)$. This is one-to-one, since $\sigma(u)=\rho(u)\implies \sigma^{-1}\rho\in H\implies \sigma H=\rho H$. Therefore, $[\mathrm{Aut}_F(K):H]\leq \deg(g_1)$. If $v\in K$ is any other root of $g_1$, then there is an isomorphism $\tau\colon F(u)\to F(v)$ that fixes $F$ and maps $u$ to $v$, and since $K$ is a splitting field, $\tau$ extends to an automorphism of $K$ over $F$. Therefore, the map from cosets of $H$ to roots of $g_1$ is onto, so $[\mathrm{Aut}_F(K):H]=\deg(g_1)$.

We now apply induction: $K$ is the splitting field over $F(u)$ of a set of separable polynomials (same one as we started with), and $[K:F(u)] = [K:F]/\deg(g_1)\lt [K:F]$. Therefore, $[K:F(u)]=|\mathrm{Aut}_{F(u)}(K)|=|H|$.

Hence $|\mathrm{Aut}_F(K)| = [\mathrm{Aut}_{F}(K):H]|H| = \deg(g_1)[K:F(u)]=[F(u):F][K:F(u)] = [K:F]$, and we are done. $\Box$

Corollary. Let $F$ be a field, and let $f_1,\ldots,f_n\in F[x]$ be nonconstant separable polynomials. Then any splitting field of $f_1,\ldots,f_n$ over $F$ is separable over $F$.

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"if $u\in K$, then there exists $n\geq 0$ such that $u^{p^n}$ is separable over $F$" - why ? –  Belgi Jun 6 '12 at 18:39
    
@Belgi: If $u$ is separable, take $n=0$. If $u$ is not separable, then its minimal polynomial $f$ has $f'=0$, hence it is of the form $g(x^p)$ for some $g$. Therefore, $u^p$ has minimal polynomial $g(x)$, which is of degree strictly smaller than $f$. Applying an inductive hypothesis, we get that there exists $m$ such that $(u^p)^{p^m}$ is separable over $F$, hence $u^{p^{m+1}}$ is separable over $F$. Alternatively, the minimal polynomial must be a polynomial in $x^p$; find the largest $n$ such that it is a polynomial in $x^{p^n}$, and that's the $n$ you want. –  Arturo Magidin Jun 6 '12 at 18:42
    
You are assuming that we know that $F(u,v)$ is separable if $u$ and $v$ are. But how do you prove this without making use of the separability degree (which Belgi wanted to avoid)? –  Sam Jun 6 '12 at 18:45
    
@SamL. An extension is separable if and only if every element is separable if and only if it is generated by separable elements. You don't need the separable degree. But I'll add that as a lemma. –  Arturo Magidin Jun 6 '12 at 18:48
    
@ArturoMagidin - as far as I know what you wrote in you last comment only holds for finite extensions –  Belgi Jun 6 '12 at 18:51

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