Strange reasoning of unramified extensions having same residue fields are the same. Let $K$ be a complete with discrete absolute value, assuming $K$ and its residue field are both perfect,$A$ its valuation ring, now let $L$ be a separable algebraic extension of $K$, $B$ the integral closure of $A$ in $L$.
Now if $K'$, $K''$ are unramified extensions of $K$ contained in $L$ with same residue field $k'$, it's well known that this implies $K'=K''$.
But isn't this obvious? I mean if $k'=B'/\mathfrak B'$, then $K'=K''=$ field of fractions of $B'$, why the proofs I saw almost in every books treat it as something needs some explanations? I must misunderstand something...
Some reference(J.S. Milne's Algebraic Number Theory):


 A: The issue here is Hensel's Lemma doesn't hold in the general case of Dedekind domains, the lifting you have is dependent upon the field being Henselian. A toy case:  let $K/F$ be a finite degree field Galois extension. If we have a prime $(p)$ of $F$ which splits completely in $K$, as $\mathfrak{p}_1\mathfrak{p}_2\ldots\mathfrak{p}_r$ then

$$\mathcal{O}_F/(p) \cong \mathcal{O}_F/\mathfrak{p}_1$$

but the rings upstairs are obviously not the same. In your case splitting is not behavior you ever witness because there is always a unique maximal ideal, and that's where your assumptions fall apart. And this is a very general case, we know there is a positive density of totally split primes in any Galois number field extension (indeed in any global Galois extension).
In the theorem you quote, the assumption is $L/K$ is an unramified extension, but if eg. $K=\Bbb Q$ or any field with class number $1$, there are no non-trivial, unramified extensions to choose from! So this cannot be some general fact about rings just based on their residue fields, you need Henselian to help you out. Milne's theorem is true (obviously), but it's not applicable to your case without being in the necessary context. This is not just dependent on residue fields, you need the big fields to have more structure (in this case complete DVR with perfect residue field).
