# If $L\mid K$ is a finite extension of fields then K is perfect iff L is perfect

Problem: Let $$L\mid K$$ be a finite extension of fields. Then $$K$$ is perfect $$\iff$$ $$L$$ is perfect.

The implication $$\implies$$ is quite easy and it has already been discussed here.

I'm interested in the implication $$\impliedby$$ because of the following application:

Suppose $$\text{char}(k)=p>0$$. Then any finitely generated field extension $$K\mid k$$ that is perfect has $$\text{tr.deg}_k(K)=0$$.

For the proof of this recall that a purely trascendental extension is never perfect as $$t_1$$ is never a $$p$$-power in the field $$k(t_1,\dots,t_n)$$. Then if $$K$$ is a finite extension of a purely trascendental extension it wouldn't be perfect as well.

That can be restated in a neat way in the language of algebraic geometry:

If $$\text{char}(k)>0$$ then any $$k$$-variety with perfect function field must have dimension $$0$$.

Also it would be interesting to see if the equivalence in the problem is true if we change finite extension by algebraic extension. Again the implication $$\implies$$ is not so hard and has been discussed here.

• If you take any nonperfect field, and you consider its algebraic closure, you get a perfect field (since the polynomial $x^p-a$ always splits). This gives you an algebraic extension that is pefect of a field that is not pefect. – Arturo Magidin Feb 3 '19 at 22:54

Suppose $$K$$ is not perfect. Then some $$a\in K$$ has no $$p$$th root in $$K$$. I claim that the polynomial $$f(x)=x^{p^n}-a$$ is irreducible over $$K$$ for any $$n\in\mathbb{N}$$. To prove this, let $$b$$ be a $$p^n$$th root of $$a$$ in an extension field of $$K$$ and note that $$f(x)$$ factors as $$(x-b)^{p^n}$$. Let $$g$$ be the minimal polynomial of $$b$$ over $$K$$. Then $$g$$ is the minimal polynomial of every root of $$f$$, so $$f=g^m$$ for some $$m$$. Since $$\deg f=p^n$$, $$m$$ must be a power of $$p$$; say $$m=p^d$$. We then have $$g(x)=(x-b)^{p^{n-d}}=x^{p^{n-d}}-b^{p^{n-d}}$$ and thus $$b^{p^{n-d}}\in K$$. If $$d>0$$, we see that $$(b^{p^{n-d}})^{p^{d-1}}=b^{p^{n-1}}$$ is a $$p$$th root of $$a$$ in $$K$$, which is a contradiction. Thus $$d=0$$ and $$m=1$$ so $$f=g$$ is irreducible.

Now if $$L$$ is a perfect extension of $$K$$, then $$a$$ must have a $$p^n$$th root in $$L$$ so $$f$$ must have a root in $$L$$. Since $$f$$ is irreducible, this means $$[L:K]\geq \deg f=p^n$$. Since $$n$$ is arbitrary, this means $$[L:K]$$ must be infinite.

• Great argument. Thanks! – Walter Simon Feb 3 '19 at 23:40

In the answer of darij grinberg the following equality is given $$[L:K]=[L^p:K^p]$$ (for the proof notice that $$K$$-basis of the vector space $$L$$ correspond to $$K^p$$-basis of $$L^p$$ under Frobenius). Using this I think we can answer the problem in a really short way:

As we have the tower of extensions $$K^p\hookrightarrow K\hookrightarrow L=L^p$$ we have that $$[L^p:K^p]=[L:K][K:K^p]$$ but as $$[L:K]=[L^p:K^p]$$ we conclude $$[K:K^p]=1$$.

• Very nice! I should have thought of that... – darij grinberg Feb 4 '19 at 14:01

Here is an argument that is simpler than @EricWofsey's, although it may well be equivalent to it in some way.

Assume that $$L$$ is perfect. We must prove that $$K$$ is perfect.

Assume the contrary. Thus, $$K$$ (and therefore $$L$$ as well) has characteristic $$p$$ for some prime $$p$$. Consider this $$p$$. Let $$F$$ denote the Frobenius endomorphism of any field of characteristic $$p$$; this sloppy notation is justified because the restriction of a Frobenius endomorphism to a subfield is always the Frobenius endomorphism of that subfield. Note that $$F$$ is always injective.

We have $$F\left(K\right) = K^p \subsetneq K$$ (since $$K$$ is not perfect). Thus, $$F^i\left(F\left(K\right)\right) \subsetneq F^i\left(K\right)$$ for each $$i \geq 0$$ (since $$F$$ is injective, and thus $$F^i$$ is injective). Hence, for each $$i \geq 0$$, we have $$F^{i+1}\left(K\right) = F^i\left(F\left(K\right)\right) \subsetneq F^i\left(K\right)$$. Hence, we have a chain $$F^0\left(K\right) \supsetneq F^1\left(K\right) \supsetneq F^2\left(K\right) \supsetneq \cdots$$ of fields. Since $$L \supseteq K = F^0\left(K\right)$$, we can extend this chain to a chain $$\begin{equation} L \supseteq F^0\left(K\right) \supsetneq F^1\left(K\right) \supsetneq F^2\left(K\right) \supsetneq \cdots . \label{darij1.eq.chain1} \tag{1} \end{equation}$$

On the other hand, $$F\left(L\right) = L^p = L$$ (since $$L$$ is perfect), and thus (by induction) we see that $$F^i\left(L\right) = L$$ for each $$i \geq 0$$.

Now, let $$d = \left[L : K\right]$$. Also, let $$i = d+1$$. Thus, $$d+1 = i < i+1$$.

From $$d = \left[L : K\right]$$, we see that $$L$$ is a $$d$$-dimensional $$K$$-vector space. Hence, $$F^i\left(L\right)$$ is a $$d$$-dimensional $$F^d\left(K\right)$$-vector space (since $$F$$ is injective, and thus $$F^i$$ is injective). In other words, $$L$$ is a $$d$$-dimensional $$F^i\left(K\right)$$-vector space (since $$F^i\left(L\right) = L$$).

Moreover, from \eqref{darij1.eq.chain1}, we obtain $$\begin{equation} L \supseteq F^0\left(K\right) \supsetneq F^1\left(K\right) \supsetneq F^2\left(K\right) \supsetneq \cdots \supsetneq F^i\left(K\right) . \label{darij1.eq.chain2} \tag{2} \end{equation}$$ This is a chain of fields, and thus is a chain of $$F^i\left(K\right)$$-vector subspaces of $$L$$ (because any field is a vector space over any of its subfields). All these subspaces have dimension $$\leq d$$ (since $$L$$ is a $$d$$-dimensional $$F^i\left(K\right)$$-vector space), and these dimensions must decrease by at least $$1$$ at each $$\supsetneq$$ sign. Thus, the chain \eqref{darij1.eq.chain2} yields the following chain of inequalities: $$\begin{equation} d \geq \dim\left(F^0\left(K\right)\right) > \dim\left(F^1\left(K\right)\right) > \dim\left(F^2\left(K\right)\right) > \cdots > \dim\left(F^i\left(K\right)\right) \end{equation}$$ (where $$\dim$$ stands for dimension as $$F^i\left(K\right)$$-vector spaces). Thus, the $$i+1$$ integers $$\dim\left(F^0\left(K\right)\right),\dim\left(F^1\left(K\right)\right),\dim\left(F^2\left(K\right)\right),\ldots,\dim\left(F^i\left(K\right)\right)$$ must be distinct and must all belong to the set $$\left\{0,1,\ldots,d\right\}$$. Thus, this set $$\left\{0,1,\ldots,d\right\}$$ must contain at least $$i+1$$ distinct integers. But it does not (since its size is $$d+1 < i+1$$). This contradiction shows that our assumption was wrong. Hence, we have shown that $$K$$ is perfect.

• Thanks! this is indeed nice. I reinterpreted this in another solution below. – Walter Simon Feb 4 '19 at 10:45