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.

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    $\begingroup$ 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. $\endgroup$ – 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.

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  • $\begingroup$ Great argument. Thanks! $\endgroup$ – 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 $[K:K^p]=1$.

  • $\begingroup$ Very nice! I should have thought of that... $\endgroup$ – 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.

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

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