= Solution
Assume $K\subseteq L^p$. Every $a\in A$ has a unique $p$th root $r(a)$ in $L$: existence is the <field> inclusion and uniqueness follows from injectivity of the <Frobenius endomorphism> in a <field>. Since $r(a)^p=a$, this root satisfies the monic polynomial $T^p-a$ over $B$. Because $B$ is an <integrally closed domain>, this gives $r(a)\in B$. The identities
$$
(r(a)+r(a'))^p=a+a',\qquad (r(a)r(a'))^p=aa'
$$
and uniqueness show that
$$
\eta_V:A\longrightarrow B,\qquad a\longmapsto r(a)
$$
is a <ring homomorphism>.
On $D(a)$ the rule sends a fraction to the fraction of its unique roots. Since $r(a)^p=a$, inverting $a$ is equivalent to inverting $r(a)$, so the rule localizes. Also, for a prime $\mathfrak q\subset B$, $r(a)\in\mathfrak q$ exactly when $a\in\mathfrak q$; hence the induced point map agrees with that of $f$. The localized rules consequently glue to a morphism $h:X\to Y$. Its pullback followed by the <Frobenius endomorphism> sends $a$ to $r(a)^p=a$, which proves
$$
\boxed{f=h\circ F_X.}
$$
If another morphism satisfies this equation, it has the same point map because the <Absolute Frobenius morphism> fixes points, and its pullback of $a$ must be a $p$th root of $a$ in the <normal domain> $B$. That root is unique. Thus $h$ is unique.
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