Let $r$ and $n$ be integers.
Give the Galois correspondance for the extension $F_{p^r} \cap F_{p^n} \leq F_{p^r} \vee F_{p^n}$
From my limited understanding $F_{p^r} \vee F_{p^n}$ has to do with lcm of $m,n$ and their intersection with gcd. There are sub-fields for divisors of both. can't systematically break it down.
For security, I recall the first main result on finite fields (see e.g. Lang's "Algebra", chap. VII, §5, thm. 10): For each prime $p$ and each integer $n\ge 1$ there exists a finite field of order $p^n$, denoted by $\mathbf F_{p^n}$, uniquely determined as a subfield of an algebraic closure $\mathbf {\bar F}_p$ of $\mathbf F_p :=\mathbf Z/p\mathbf Z$. The field $\mathbf F_{p^n}$ is the splitting field in $\mathbf {\bar F}_p$ of the polynomial $X^{p^n}-X$. For any $r, n \ge 1$, it follows straightforwardly that $\mathbf F_{p^r}\cap \mathbf F_{p^n}= \mathbf F_{p^m}$, with $m=min (r,n)$, and $\mathbf F_{p^r}\cup \mathbf F_{p^n}= \mathbf F_{p^s}$, with $s=max(r,n)$.
The Galois correspondence in the extension $\mathbf F_{p^s}/\mathbf F_{p^m}$ is given by the second main result on finite fields (which is a direct consequence of the first): $\mathbf F_{p^s}/\mathbf F_{p}$ is galois, with cyclic Galois group of order $s$ generated by the so called absolute Frobenius automorphism defined by $\Phi (x)=x^p$. It follows obviously that $\mathbf F_{p^s}/\mathbf F_{p^m}$ is also galois, with cyclic Galois group of order $s-m$ generated by the relative Frobenius $\phi=\Phi^m$. Because of the structure of cyclic groups, the Galois correspondence actually states that for any $n$ between $m$ and $s$, $\mathbf F_{p^s}/\mathbf F_{p^m}$ admits a unique subextension $\mathbf F_{p^n}/\mathbf F_{p^m}$ of degree $n-m$ .