TITLE: Why is the Galois Correspondence intuitively plausible?
QUESTION [53 upvotes]: The Galois Correspondence Theorem says that for any Galois extension of fields $K/F$, there is a one-to-one inclusion reversing correspondence between the intermediate fields $K \supseteq E \supseteq F$ and subgroups of the Galois group $\text{Gal }K/F$. I have two questions about this:

Why this is theorem important? When I first learned it, I told myself that it was just a computational tool: it converts problems about fields, which are hard to understand, into problems about groups, which are easier to understand. But clearly, it is not just there for computational convenience, it is an important result in its own right. So I ask: is there a non-utilitarian reason why the theorem is profound? In other words, what deep "underlying truth" about the algebraic structure of fields does this theorem reveal?

Why is this theorem intuitively plausible? The way I have understood it, if you place the subfield lattice of $K/F$ and the subgroup lattice of $\text{Gal }K/F$ side-by-side (with the latter flipped upside-down), then the diagrams are the same. This seems very out-of-blue to me. Why is it intuitively plausible that these two diagrams should look the same? Why should we expect that by studying the symmetries of a field extension, we can recover the structure of whole field itself? An explanation with a specific example of a field extension, would be great.


Thanks for the help!

REPLY [9 votes]: Why is this theorem intuitively plausible?

In addition to looking for abstract reasons for plausibility and heuristics, you should also always look for good, memorable examples. The finite fields do the trick here.
The intermediate fields between $\mathbb{F}_{p^n}$ and $\mathbb{F}_p$ are $\mathbb{F}_{p^k}$ for $k \mid n$. The automorphisms of $\mathbb{F}_{p^n}$ over $\mathbb{F}_p$ are $\{1, \phi, \phi^2, \ldots, \phi^{n-1}\}$ where $\phi$ is the Frobenius automorphism $\phi(x) = x^p$, and the automorphism group is cyclic of order $n$. The subgroups are hence cyclic groups of order $k$ for $k \mid n$. So, we have the same number of subfields and subgroups--is there a bijection?
The inclusion lattices are clearly the same, namely linear. The subgroup $\{1, \phi^k, \phi^{2k}, \ldots\}$ is the automorphism group of $\mathbb{F}_{p^n}$ over $\mathbb{F}_{p^k}$, so there's some order-reversing going on. In the other direction, the fixed field of $\{1, \phi^k, \phi^{2k}, \ldots\}$ is $\mathbb{F}_{p^k}$, and taking the fixed field is also an order-reversing operation. It's initially quite plausible that these operations will interchange meets and joins as well, though this example doesn't really illustrate that.
A wide-eyed optimist might then blithely conjecture that the Galois correspondence always holds based on this one example. That optimist would do well to stress-test the conjecture, which of course requires more hypotheses. A good next example to work through would be $\mathbb{Q}(2^{1/4}, i)$ over $\mathbb{Q}$. It turns out the Galois group is the dihedral group of order 8, whose subgroup lattice is not symmetric, though it's small enough to not be too irksome to work with by hand. Of course, that optimist would soon have to reckon with non-normal extensions and seeing their dream of a completely universal correspondence blown away, but c'est la vie.