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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): enter image description here enter image description here

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    Your argument seems to apply even if $K$ is a number field, where the result is false. How do you get from $k' = B'/\mathfrak B'$ to $K'' = \mathrm{Frac}(B')$? – 2017-02-28
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    @Mathmo123 By definition of residue field, $B'$ equals to the valuation ring of $K''$, hence $K''$ is equals to the field of fraction of $B'$? – 2017-02-28
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    You have $B'/\mathfrak B'$ = $B''/\mathfrak B''$ where $\mathfrak B'$ and $\mathfrak B''$ are respectively primes of $B'$ and $B''$. That does not tell you that $B' = B''$. For example, $\mathbb Z/5\mathbb Z\cong \mathbb Z(i)/(1+2i)$. – 2017-02-28
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    @Mathmo123 minor quibble: I think you want $\Bbb Z[i]$ there (since rational functions would give you a field) – 2017-02-28
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    @Mathmo123 $\mathfrak B'$ must equals to $\mathfrak B''$, otherwise their residue fields are at best the isomorphism? (In the proof it uses $K'K''$ has the same residue field as $k'$, if they are merely isomorphism, then isn't $k'k''$ bigger than but not equal to $k'$?) – 2017-02-28
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    @CYC the problem states $k'=k''$ with the phrase "with the same residue field." In particular $k''k'=k'$. – 2017-02-28
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    @AdamHughes If $k'$ strictly equals to $k''$, say equals to $B'/\mathfrak B'$, then isn't this implies $K'=K''=$ field of fraction of $B'$? – 2017-02-28
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    @CYC no, and that's the point: equal residue fields do not imply the big rings are isomorphic in general. – 2017-02-28

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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).

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    If it means equal up to isomorphism, is the residue field of $K'K''$ still equals to k'(Isn't it $k'k''$)? – 2017-02-28
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    @CYC I don't see your objection. The assumption is that the residue fields are the same, then because the extension is unramified in a Henselian field, this means $[K''K':K]=[k':k]=[K':K]=[K'':K]$ but then $K''K'/K'$ is a trivial extension and likewise $K''K'/K'$ is trivial so $K'\subseteq K'' \subseteq K'$ proving equality. This is the point of a Henselian fields, and part of what distinguishes them from eg. global fields. – 2017-02-28
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    It uses $K''K'$ also has residue field $k'$ right? But that assumption is based on $K'$ and $K''$ both has residue field $k'$, their residue fields are strictly equal not only isomorphism, so isn't this imply $B'=B''$? – 2017-02-28
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    @CYC I'm a little confused as to your question, you were asking only why things require justification, the answer is: ring quotients being equal does not mean the fraction fields are equal, and in general they aren't, as I illustrate, you can have the same residue field in a non-Henselian context and not the same fraction field for the rings. Look at mathmo's example, there is a unique field of order $5$ and $\Bbb Z/(5)\cong\Bbb Z[i]/(1+2i)$ but the fraction fields are $\Bbb Q$ and $\Bbb Q(i)$ respectively in this case. – 2017-02-28
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    @CYC the assumptions you have does not hold in general fields, your quoted theorem from Milne assumes the existence of a non-trivial, unramified extension, but clearly this is false for, eg. the base field being $\Bbb Q$. So this case does not fit the assumptions of Milne's theorem! – 2017-02-28
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    @CYC in particular, you should read the last paragraph in my edit: what you seem to think is that residue fields correspond to the integer rings in general, which is false as I explain in my highlighted isomorphism in the answer. In short: **you need more than just the quotients are equal in general**. And that is why it is not obvious: because the idea you quote is false: it cannot be done just based on the quotients. – 2017-02-28
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    So if $K$ is a perfect, complete DVR, with perfect residue field, then $K'$, $K''$'s equal(not only isomorphism) residue fields, say equal to $C/\mathfrak C$, imply $K'=K''=$ field of fraction of $C$? – 2017-02-28
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    I found myself falsely assume discrete absolute value as non-archemedian, I think I know the answer now! – 2017-02-28
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    @CYC glad we could clear that up for you. :) – 2017-02-28