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Question: I know how to detect ramified or unramified fields in the case of number fields. But I have no feelings about how to do this for local fields. How should I find that $\mathbb{Q}_p(\sqrt[3]{2})=\mathbb{Q}_p[x]/(x^3-2)$ is ramified or unramified? I can answer that $\mathbb{Q}_p(\sqrt[3]{2})=\mathbb{Q}_p[x]/(x^3-2)$ is (totally) ramified, because $x^3-2$ is Eisenstein at $2$ (This is not correct. See the comments). But then how should I realize that $\mathbb{Q}_p(\sqrt[3]{2}, \zeta_3)=\mathbb{Q}_p[x]/(x^3-2, x^2+x+1)$ is ramified or unramified? How should I find that $\mathbb{Q}_p[x]/(x^5-x-2)$ is ramified or unramified? How can I find the uniformizer in the local field $\mathbb{Q}_p[x]/P(x)$, where $P(x)$ is an irreducible monic polynomial, and how can I find the ramification index?

My attempt: This is not actually an attempt. I just want to say what I know in the case of number fields, and I have some questions and solutions (probably unsuccessful solutions) based on it. In the case of number fields I used the discriminant. Also Dedekind-Kronecker theorem was very helpful to find the ramification indices, $e_i$'s, and reside degrees $f_i$'s for almost all primes. Do we have a version of Dedekind-Kronecker theorem for local fields?

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    $\begingroup$ $X^3 -2$ is Eisenstein at $2$ but not at $p$ for $p\ne 2$. So in general $\mathbb Q_p(\sqrt[3]2)$ is unramified! $\mathbb Q_p(\sqrt[3]2)$ is ramified at exactly the primes $p$ that $\mathbb Q(\sqrt[3]2)$ is ramified (so $p=2, 3$). But the local question is usually much easier! For example, you don't have to worry about primes splitting. If $d = ef$ is the degree of $K/\mathbb Q_p$ and if $\mathcal O_K = \mathbb Z_p[\alpha]$, then $f$ is exactly the degree of $\mathbb F_p(\alpha)$ , which is much simpler than Dedekind's theorem. You also have Hensel's lemma at your disposal. $\endgroup$
    – Mathmo123
    Jan 20, 2021 at 21:44
  • $\begingroup$ @Mathmo123 Thank you very much for your enlightening comment, this comment is very helpful to me. But yet I do not know how could I compute the degree of $\mathbb F_p(\alpha)$. Let $\alpha$ be a root of $P(x)=x^6+x^5+x^4+x^3+x^2+x+1$. I do not know how can I compute $\mathbb F_2(\alpha)$ in practice. (By another way I can show that $\mathbb Q_2(\alpha)$ is an unramified extension of degree $3$, so $\mathbb F_2(\alpha)$ is of cardinal $2^3$. But that way works just for some cyclotomic polynomials, and I do not know what should I do in general.) $\endgroup$
    – Tireless and hardworking
    Jan 20, 2021 at 21:55
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    $\begingroup$ Computing the degree of $\mathbb F_p(\alpha)$ amounts to factorising $P(x)$ modulo $p$, which is exactly what you need to do to apply the Dedekind-Kronecker theorem. Of course, this might be extremely tedious to do by hand (but is completely doable by a computer)! In your case, the factorisation is $(x^3 + x^2 + 1)(x^3 + x+1)\pmod 2$ and both factors are irreducible, so generated the unique cubic extension of $\mathbb F_2$. $\endgroup$
    – Mathmo123
    Jan 20, 2021 at 22:17
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    $\begingroup$ About extensions of degree $n$, the story is much more complicated. Quadratic extensions are automatically Galois, for one thing; not so for general $n$. And the utility of $K^*/{K^*}^n$ depends on whether all $n$-th roots of unity are in $K$. While there are only finitely many extensions of $\Bbb Q_p$ of each degree, I don’t believe there’s a general rule for describing them. It’ll depend on $p$ and $n$. $\endgroup$
    – Lubin
    Jan 21, 2021 at 2:02
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    $\begingroup$ The tower rule is your friend! It's not too hard to understand $\mathbb Q_p(\zeta_3) = \mathbb Q_p(\sqrt{-3})$. Using Hensel's lemma, you can show that it is just $\mathbb Q_p$ if $p \equiv 1 \pmod 3$. Otherwise, it is a quadratic extension and is ramified iff $p = 3$. When $p\equiv 1\pmod 3$, $\mathbb Q_p(\sqrt[3]2,\zeta_3) = \mathbb Q_p(\sqrt[3]2)$, which you now know how to work with. When $p\ne 2, 3$ and $p\not\equiv 1\pmod 3$, can you show that $2$ has a cube root in $\mathbb Q_p$? If so, then it has all its roots in $\mathbb Q_p(\sqrt[3]2)$. That just leaves the ramified primes $2, 3$. $\endgroup$
    – Mathmo123
    Jan 21, 2021 at 10:31

1 Answer 1

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You said you know about using the discriminant of a $\mathbf Z$-basis of the integers of a number field to detect ramification. You can do the same thing in a local field: if $K$ is a finite extension of $\mathbf Q_p$ then the discriminants of all $\mathbf Z_p$-bases of $\mathcal O_K$ are equal to multiplication by the square of a unit in $\mathbf Z_p$, and $K/\mathbf Q_p$ is ramified if and only if the discriminant of a $\mathbf Z_p$-basis of $\mathcal O_K$ is divisible by $p$ (equivalently, the discriminant is not a unit in $\mathbf Z_p$).

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  • $\begingroup$ Thanks. Yet I have difficulties finding an integral basis for $\mathbb{Q}_p(\sqrt[3]{2}, \zeta_3)$. More generally I confess that I do not know how should I deal with fields of the form $K[x]/(P_1(x), P_2(x))$ to find an integral basis. I know how to find an integral basis for $K[x]/(Q(x)$, when the extension is given by a single polynomial. (This is my way: I compute the discriminant of the polynomial, and the only suspicious primes are that are dividing $disc(Q)$, and I can remove some of these primes with the help of 2.1-2.3 of kconrad.math.uconn.edu/blurbs/gradnumthy/totram.pdf ) $\endgroup$
    – Tireless and hardworking
    Jan 21, 2021 at 9:10
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    $\begingroup$ Yeah, I'm familiar with that document. $\endgroup$
    – KCd
    Jan 21, 2021 at 9:17
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    $\begingroup$ Make sure you can show that if $K$ and $L$ are unramified extensions of $\mathbf Q_p$, then the composite field $KL$ is unramified over $\mathbf Q_p$ and that $\mathbf Q_p(\alpha)/\mathbf Q_p$ is unramified if $\alpha$ is $p$-adically integral and its minimal polynomial has discriminant in $\mathbf Z_p^\times$. Then $\mathbf Q_p(\sqrt[3]{2})$ is unramified if $p$ is not $2$ or $3$ and $\mathbf Q_p(\zeta_3)$ is unramified if $p \not= 3$. Therefore $\mathbf Q_p(\sqrt[3]{2},\zeta_3)$ is unramified if $p$ is not $2$ or $3$, so the only cases left to think about are $p=2$ and $p=3$. $\endgroup$
    – KCd
    Jan 21, 2021 at 9:21
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    $\begingroup$ In those two cases there genuinely is ramification. The number field $\mathbf Q(\sqrt[3]{2})$ is totally ramified at $p=2$ and $p=3$, and similar reasoning carries over to $\mathbf Q_p(\sqrt[3]{2})$ for $p=2$ and $p=3$. $\endgroup$
    – KCd
    Jan 21, 2021 at 9:21
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    $\begingroup$ The Eisenstein condition applies to $x^3 - 2$ over $\mathbf Q_2$ and $(x-1)^3-2$ over $\mathbf Q_3$ in the same way they do over $\mathbf Q$ at $p=2$ and $p=3$. $\endgroup$
    – KCd
    Jan 21, 2021 at 9:42

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