Fixing a field $K$, we work in the category of $\underline{\textsf{Fields} \leftarrow K}$ of "fields over $K$", i.e.
- the objects are (necessarily injective) field homomorphisms $F \leftarrow K$ (often notated as just $F$; if really necessary, I'll refer to the injection as $\iota_{F \leftarrow K}$, or even just $\iota_F$)
- and the morphisms $\text{Hom}_{\underline{\textsf{Fields} \leftarrow K}}(F,E)$ (typically denoted Greek letters like $\eta, \sigma, \tau$) are maps $\eta: F \to E$, that satisfy the property that the obvious triangle (involving $F,E, K$ and $\eta, \iota_F, \iota_E$) commute:
$$\require{AMScd}
\def\diaguparrow#1{\smash{
\raise.6em\rlap{\scriptstyle #1}
\lower.26em{\mathord{\diagup}}
\raise.52em{\!\mathord{\nearrow}}
}}
\begin{CD}
F @>{\eta}>> E\\
@A\iota_E AA \diaguparrow{\iota_F}& \\
K &
\end{CD}$$
We could also work with fields containing $K$ (set theoretically), but this is a bit philosphically weird (algebra in general just doesn't really care about the underlying sets exactly; we care much more about things up to isomorphisms). If this above doesn't make sense, ignore it. It's just a bit of stage-setting.
I will follow this nice article by Brandenburg.
Lemma 2.1: Let $L,M$ over $K$ (in the sense above). Let $\alpha \in L$ be algebaic over $K$ with minimal polynomial $f\in K[t]$ (in the sense that $\iota_{L\leftarrow K}f \in L[t]$ is the minimal degree monic polynomial that has $\alpha \in L$ as a root). Then, we have a bijection
$$\text{Hom}_{\underline{\textsf{Fields} \leftarrow K}}(K(\alpha), M) \cong \{m\in M: \iota_{M\leftarrow K}f(m)=0\}$$
given by $\eta \mapsto \eta(\alpha)$.
Proof: see link; "follows from $K(\alpha) \cong K[t]/(f)$ and universal properties of quotient and polynomial algebras".
Lemma 2.3 (lifting lemma): suppose we have the following diagram
$$\require{AMScd}
\begin{CD}
L && M\\
@A\iota_L AA \diaguparrow{\iota_M}& \\
K &
\end{CD}$$
If we furthermore assume that
- $L$ is an algebraic extension of $K$ (with some generating set $\mathscr G\subseteq L $ say, i.e. $L = \text{minField}(\iota_L K, \mathscr G)$),
- and $M$ satisfies that for all $\beta\in \mathscr G \subseteq M$, and any intermediate subfield $L \supseteq E \supseteq \iota_L(K)$ with map $\iota_{EM}:E\to M$, the minimal polynomial $\text{Irr}(\gamma,E)\in E[t]$ of every generator $\beta\in \mathscr G$, pushed forward via $\iota_{EM}$, has a root in $M$. (This is true for example, if $M$ is algebraically closed. Or if $M =L\supseteq K$ is a splitting field.)
Then, there exists some $\eta: L\to M$ completing the above triangle (called an "extension" or "lifting" of $\iota_M$).
Proof: see link; in finite case, follows from Lemma 2.1 and induction. In general case, use Zorn's lemma.
In the language of "extensions/lifting", Lemma 2.1 says $\iota_M: K\to M$ can "lift" to $K(\alpha) \to M$ by choosing any root of $\iota_M \text{Irr}(\alpha, K)\in M[t]$ to send $\alpha$ to. This is a huge upgrade in "data": with just the data of a single root in $M$, we get a whole field homomorphism. A fantastic deal!
This Lemma 2.1 is enough to show the following:
ThmA: Suppose $L$ is an algebraic extension of $K$, s.t. the minimal polynomial $\text{Irr}(\alpha, K) \in K[t]$ of any $\alpha \in L$, pushed forward to $\iota_L\text{Irr}(\alpha, K) \in L[t]$ has $\geq 2$ distinct roots in $L$ (so definitely $\alpha$, and some other $\beta$). Then can lift $\iota_L: K \to L$ to a field homomorphism $K(\alpha) \to L$ mapping $\alpha \mapsto \beta$.
To extend/lift fully to a field homomorphism $L\to L$ (necessarily an automorphism by algebraicity; see Lemma 2.5 in Brandenburg), we need to use Lemma 2.3 to lift from $K(\alpha)$ to $L$.
(Ok, executive decision, let's make things set theoretically nice: identify $K, K(\alpha)$ with their copies inside $L$, so that $\alpha \in L \supseteq K(\alpha) \supseteq K$)
How do we make sure we meet the preconditions of Lemma 2.3?
One way is to assume that in fact all $\gamma \in L$ (or at least $\gamma$ in a generating set of $L$ over $K$) split in $L$ (i.e. every $\text{Irr}(\gamma, K) \in K[t] \subseteq L[t]$ splits into a product of linear factors in $L[t]$).
So, using nothing more than the lifting lemma and some wishful thinking, we have shown
ThmB: Suppose $L$ is an algebraic extension of $K$ s.t. all $\alpha \in L \smallsetminus K$ (or at least $\gamma$ in a generating set of $L$ over $K$) have minimal polynomial $\text{Irr}(\alpha,K)\in K[t] \xrightarrow{\iota_L} L[t]$ splitting into $\geq 2$ distinct linear factors.
Then, $L^{\text{Aut}(L/K)}=K$. (i.e. for every $\alpha \in L \smallsetminus K$, can find some $K$-automorphism of $L$ moving $\alpha$ off itself)
The converse is true as well, by using $L^{\text{Aut}(L/K)}=K$ to "average" over all “co-roots” of $\alpha$ to land in $K[t]$: for all $\alpha \in L\setminus K$, can define
$$f_\alpha (t):= \prod_{\sigma \in \text{Aut}(L/K)} (x-\sigma(\alpha)) \in K[t].$$
$\text{Irr}(\alpha, K)$ divides $f_\alpha$, and since $f_\alpha$ splits in $L$, so must $\text{Irr}(\alpha, K)$.
To go from $\geq 2$ distinct linear factors to ALL linear factors (of a minimal poly) distinct, see Normal field extension separable over its fixed field. I think of this as a "Buy one get everything else free" kind of result (it's actually "buy 2 get everything else free", but whatever).
This basically proves half the Galois correspondence (main tool: just the lifting lemma!!!). The other half is done in Brandenburg, with basically nothing more than a quite simple combinatorial ("Baire property-like") fact: Lemma 3.3. A field cannot be written as the union of finitely many proper subfields.
Sidenote: I'm being such a pedant about the maps $\iota$ because they're actually important. Consider $K = \mathbb F(t)$ and $L = \mathbb F(t^{1/p}) \supseteq K$, i.e. $L = K(t^{1/p})$. Consider $\iota_1, \iota_2: K \to L$ where $\iota_1$ maps $t \mapsto t$, but $\iota_2$ maps $t \mapsto t^{1/p}$. Then we do NOT get any lift of $\iota_2$ to an hom $L\to L$!
