enter image description here enter image description here


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.


1 Answer 1


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 $O_E/(\pi_E)(\zeta_{q-1})$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.