# Every algebraic extension of a perfect field is separable and perfect

I am trying to prove this statement in the characteristic $p>0$ case.

Every algebraic extension of a perfect field is separable and perfect.

This is stated as a corollary of Proposition V.6.11 in Lang.

Let $k$ be a field and let $K$ be a normal extension of $k$. Let $G$ be the group of automorphisms of $K$ that fix $k$. Let $K^G$ be the fixed field of $G$ acting on $K$, and $K_0$ the maximal separable subextension of $K$. Then $K^G$ is purely inseparable over $k$, $K$ is separable over $K^G$, $K_0\cap K^G =k$ and $K_0K^G=K$.

Lang doesn't write many details in the proof except that every finite extension is contained in a normal extension.

This is what I have so far. Suppose, $k=k^p$. Let $E$ be an algebraic extension of $k$. Let $\alpha \in E$. Now, $\alpha$ is contained in the field, say $K$ generated by the roots of the minimal polynomial of $\alpha$ over $k$ (and is hence normal as it's the splitting field of the minimal polynomial of $\alpha$). At this point, I don't know how to proceed. I am tempted to use the Frobenius map, but it does not fix every element of $k$ although it does stabilize it in this perfect case. Any hints will be appreciated. Thanks.

Suppose that $k$ is of char. $p$ and that $x \mapsto x^p$ is surjective. Let $\alpha$ be an element of $\overline{k}$, with minimal polynomial $f(x)$. Then $f(x)$ is irreducible (being a minimal polynomial). We may write $f(x) = g(x^{p^d}),$ where $g$ is irreducible and separable and $d \geq 0$. However, since every element of $k$ is a $p^d$th power, by assumption, we may find a polynomial $h(x)$ (the $i$th coefficient is the $p^d$th root of the $i$th coefficient of $g$) such that $g(x^{p^d}) = h(x)^{p^d}$.

Since $g(x^{p^d})$ is supposed to be irreducible, we find that $p^d = 1$, i.e. that $d = 0$ and that $f(x) = g(x)$ is separable.

Consequently every algebraic element over $k$ is separable, and so every algebraic extension of $k$ is separable. The fact that such an extension is perfect is also easy to see directly: we may suppose that it is finite (since any algebraic extension is a union of finite ones). Denote the extension by $l$, and suppose that $[l:k] = d$. Then, since raising to the $p$th power is injective, we see that $[l^p:k^p] = d$. Now (and here is where we use perfectness of $k$) we have that $k^p = k$, and so $l^p$ contains $k$ and $[l^p:k] = d$. Since $l^p \subset l$, we conclude that $[l:l^p] = 1$, i.e. that $l = l^p$.

Remember the basic example of a non-perfect field extension is the extension $l = \mathbb F_p(t)$ over $k = \mathbb F_p(t^p)$. The minimal polynomial of the element $t$ over $k$ is $x^p - t^p$, so in the notation of the above proof $f(x) = x^p - t^p$ and $g(x) = x - t^p$. Now in this case $t^p$ is not a $p$th power in $k$, and so we can't write $f(x)$ as a $p$th power. This illustrates the exact way that the above argument breaks down in the non-perfect setting.

More generally, this shows that if $k$ is not perfect, i.e. if there exists $a \in k$ which is not a $p$th power, then $k[x]/(x^p - a)$ is a purely inseparable extension of $k$, and hence $k$ admits algebraic but non-separable extensions. (So the original statement is an if and only if. This also gives another proof that finite extensions of a perfect field are perfect: namely, since finite extensions of finite extensions are again finite extensions, we see that any finite extension of a finite extension $l$ of a perfect field $k$ is separable over $k$, hence separable over $l$, and hence perfect by the preceding remark.)

• Thanks a lot for the elaborate answer and the addendum. It's always good to have an alternate point of view.
– BMI
Oct 6, 2011 at 21:43
• I really love your answers. I am glad I took a photo with you at the recent ICM! Apr 13, 2015 at 18:12
• The $p$-th power map from $\bar{k}$ to itself is surjective. Let $f(x)$ be minimal (hence irreducible) polynomial of $\alpha\in\bar{k}$ over $k$. Then you wrote that in polynomial $f(x)$, powers of $x$ appear to be some $p$-th powers; why this is so?(For example, why $x^2+ax+b$ can't be an irreducible polynomial of some $\alpha$ with $p$ odd prime?) Is question clear? Jun 1, 2018 at 12:40
• @Beginner in your example, $d = 0$, that is $p^d = 1$.
– user279515
Sep 26, 2018 at 15:08

Here is how the claim follows from Lang Proposition V.6.11:

Suppose $k=k^p$. Let $E/k$ be any algebraic extension, let $\alpha\in E$, and let $K/k$ be the finite normal extension generated by $\alpha$ and its conjugates (not necessarily contained inside $E$, of course). To prove that $E/k$ is separable, it suffices to prove that $k(\alpha)/k$ is separable for all $\alpha$, and to do this it suffices to prove that $K/k$ is separable for all $\alpha$ (because $K\supseteq k(\alpha)$).

Let $K/k$ be our finite normal extension. By the proposition, $K=K^GK_0$ and $K^G\cap K_0=k$, and $K^G/k$ is purely inseparable, and $K_0/k$ is separable. This means that for every $\beta\in K^G$, the minimal polynomial for $\beta$ over $k$ is of the form $x^{p^n}-b$ for some $b\in k$ (see Lang's definition of purely inseparable). But because the Frobenius map is surjective on $k$, there is some $c\in k$ such that $c^{p^n}=b$. Hence the minimal polynomial for $\beta$ is $$x^{p^n}-b=x^{p^n}-c^{p^n}=(x-c)^{p^n}$$ so that $\beta=c\in k$. Thus $K^G=k$, so that $K=K^GK_0=K_0$ is separable over $k$.

EDIT: See Matt E's answer for an explanation of why $K$ is perfect (and a more enlightening, comprehensive discussion of the whole situation than my answer, certainly).

I take the definition of a perfect field to be one for which every algebraic extension is separable.

Suppose $K$ is a perfect field and $L/K$ is an algebraic extension. For any algebraic extension $F/L$, the extension $F/K$ is also algebraic, hence separable because $K$ is perfect. But this implies that $F/L$ is separable (because for any $\alpha\in F$, we have that $\text{Irr}(\alpha,L)\mid \text{Irr}(\alpha,K)$, and $F/K$ separable $\implies$ $\text{Irr}(\alpha,K)$ has no repeated roots, which therefore implies that $\text{Irr}(\alpha,L)$ doesn't either). Hence any algebraic extension of $L$ is separable, so $L$ is perfect. Finally, $L/K$ is separable, of course, because $K$ is perfect.

• Thanks for the answer. The definition lang uses is that a field of characteristic $0$ is perfect and any field $k$ of characteristic $p$ for which $k=k^p$ is perfect. I was hoping to see why the proof follows from the proposition of Lang and assuming this definition of perfect.
– BMI
Oct 6, 2011 at 3:14
• @BMI: That's completely valid, I will think about how to answer your question (or, I'm sure, someone else will be able to). Oct 6, 2011 at 3:17
• @BMI: I think I've got at least part of it, I've edited my answer. Oct 6, 2011 at 3:55
• Thanks for modifying your answer. I will select this as my preferred answer since it uses the proposition in question.
– BMI
Oct 6, 2011 at 21:42
• Very nice answer! +1. Frobenius map is surjective because field is perfect right?
– ZFR
Mar 4, 2019 at 19:33

Part of the question depends upon what your definition of a perfect field is. One of the common definitions is that a field is perfect iff every algebraic extension field is separable. In this case half of what you want is true by definition and the other half follows easily from the fact that if $$M/L$$ is separable and $$L/K$$ is separable, then $$M/K$$ is separable. (See also Zev Chonoles's answer...)

This makes me think you are using the other common definition of a perfect field: namely, a field is perfect iff it has characteristic $$0$$ or it has positive characteristic $$p$$ and the Frobenius endomorphism $$x \mapsto x^p$$ is surjective. That this second definition holds iff every algebraic extension is separable is shown in $$\S 6.1$$, Proposition 36 of these notes.

It is enough to prove that every finite extension of a perfect field is perfect and separable. Thus let $$K$$ be a perfect field and $$L/K$$ a finite extension and consider $$\Phi:L\to L,x\mapsto x^p$$.

$$\mathbf{L}$$ is perfect. We endow $$L$$ with two $$K$$-vector space structures: the standard one (using the product of $$L$$) and the following $$K$$-vector space structure: $$\forall c\in K,\forall x\in L,~~ c.x \overset{\text{def}}= c^p\times x$$ Let's write $$L'$$ for $$L$$ with this $$K$$-vector space structure. Every linearly independent family in $$L$$ remains linearly independent when viewed as a family in $$L'$$, and $$K$$ being perfect, every spanning family of $$L$$ remains so when viewed as a family in $$L'$$. In other words: every $$K$$-basis of $$L$$ is a $$K$$-basis of $$L'$$, so that $$\dim_K L=\dim_KL'.$$ Now $$\Phi$$ is a one to one $$K$$-linear mapping $$L\to L'$$. Since both $$L$$ and $$L'$$ have the same finite dimension over $$K$$, $$\Phi$$ is an isomorphism, and so $$L^p=\mathrm{Im}(\Phi)=L$$ and $$L$$ is perfect.

$$\mathbf{L/K}$$ is separable. Let $$x\in L$$ have minimal polynomial $$\mu$$. Let $$d\geq 0$$ be maximal with the property that there exists some polynomial $$Q=\sum_{i=0}^{n}a_i X^i\in K[X]$$ with $$\mu=Q(X^{p^d})$$. Note that $$Q$$ is separable over $$K$$. Let us prove that $$d=0$$ (so that $$\mu=Q$$ is separable). Since $$K$$ is perfect we can find $$\alpha_0,\dots,\alpha_n\in K$$ with $$\alpha_i^{p^d}=a_i$$ for $$i=0,\dots,n$$, thus $$\mu=R^{p^d}$$ where $$R=\sum_{i=0}^n\alpha_iX^{i}\in K[X]$$. Since $$\mu$$ is irreducible over $$K$$, $$p^d=1$$ i.e. $$d=0$$ and $$\mu=R=Q$$ is separable.