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[Question]

I know that $K'\cdot K''$ is an unramified extension of $K$ but I don't know why $K'\cdot K''$ have a residue field $k'$.

is it always true that $K_1\cdot K_2$ have a residue field $k_1 \cdot k_2$? (where $k_1,k_2$ are residue fields of $K_1, K_2$)

I think that if we prove the proposition 7.50 , then we can use " $K_1\cdot K_2$ have a residue field $k_1 \cdot k_2$" in this situation.

However, we can't use that fact while proving this proposition.

How can I prove this?

Thank you for your attention.

reference(J.S. Milne's Algebraic Number Theory) and this post 1: Strange reasoning of unramified extensions having same residue fields are the same.

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For $K/\Bbb{Q}_p$ a finite extension then $F/K$ is unramified iff $F=K(\zeta_n)=K(\zeta_{q-1})$ with $p\nmid n$ and $q= |O_F/(\pi_F)|$. This is the main application of Hensel lemma.

When $E/K,E'/K$ are ramified then it is not always the case that the residue field of $EE'$ is the smallest field contained those of $E,E'$, try with $E=\Bbb{Q}_2(2^{1/3}),E'=\Bbb{Q}_2(\zeta_3 2^{1/3})$.

When $E'/K$ is unramified then $EE'=E(\zeta_{q-1})$ has residue field $O_E/(\pi_E)(\zeta_{q-1})$.

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  • $\begingroup$ Thank you. May I ask you something? If the base field is $\mathbb{F}_p((t))$ ( not $\mathbb{Q}_p$, then still this hold? (Maybe, you assumed that characteristic is $0$) $\endgroup$
    – hew
    Jan 4, 2021 at 12:44
  • $\begingroup$ Also, is $n$ any natural number and $q$ prime number such that $p\nmid n$ and $q= |O_F/(\pi_F)|$ ? Then, you mean, there are infinitely many $n$ such that $F=K(\zeta_n)$. Probably, it seems to be misunderstood. So, may I ask you the reference with proof of this "the main application of Hensel lemma" ? Finally, I don't understand why this imply that $K'\cdot K''$ have a residue field $k'$. Sorry and thank you. $\endgroup$
    – hew
    Jan 4, 2021 at 12:54
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    $\begingroup$ It works the same way for finite extensions of $\Bbb{F}_p((t))$ and any complete discretely valued fields. Unramified means that $[F:K]=[O_F/(\pi_F):O_K/(\pi_K)]$, take a generator of the residue field extension, lift its minimal polynomial to $O_K[x]$ which stays irreducible, Hensel lemma gives a root $a$ of it in $O_F$ then $[K(a):K]=[O_F/(\pi_F):O_K/(\pi_K)]$ thus $K(a)=F$. $\endgroup$
    – reuns
    Jan 4, 2021 at 13:56
  • $\begingroup$ $\zeta_{q-1}$ is when the residue field is finite. $\endgroup$
    – reuns
    Jan 4, 2021 at 13:56

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