2
$\begingroup$

Suppose that I have a prime ideal $p$ in $\mathbb{Q}$. Then it ramifies in some extension of $\mathbb{Q}$, namely in $\mathbb{Q}(\sqrt p)$. This seems like it should be true for an arbitrary number field replacing $\mathbb{Q}$. Namely, suppose that I have a prime ideal $\mathfrak{p}$ in a number field $K$. Should there be some algebraic extension $L$ of $K$ (or even better, a quadratic extension of $K$) in which $\mathfrak{p}$ ramifies? The only criterion of a prime ramifying (which involves the prime dividing the relative discriminant of the extension) involves the assumption that $\mathcal{O}_L$ is a free $\mathcal{O}_K$-module, but this isn't true even in the case of a quadratic extension (I believe Keith Conrad has an example written up in a paper), hence my difficulty. But mostly I feel that I'm missing something trivial.

$\endgroup$
7
  • 6
    $\begingroup$ Just take the square root of an element of $K$ with valuation $1$ at the prime. $\endgroup$ Oct 12, 2011 at 15:15
  • 6
    $\begingroup$ Take an element $f$ of $O_K$ which generates $\mathfrak p$ in the local ring $O_{K, \mathfrak p}$ and consider the extension $L=K(\sqrt{f})$. It is quadratic and ramifies at $\mathfrak p$ (look at the ramification index). The ramification is a local phenomena and doesn't have anything to do with the freeness of $O_L$ over $O_K$. $\endgroup$
    – Qing Liu
    Oct 12, 2011 at 15:19
  • $\begingroup$ Thanks, that makes sense. But why would one generally require the freeness when they state the ramification criterion? For example, Milne's Algebraic Number Theory (Thm 3.35) asserts this condition. $\endgroup$
    – user3860
    Oct 12, 2011 at 15:21
  • 3
    $\begingroup$ J, you are referring in your parenthetical remark to www.math.uconn.edu/~kconrad/blurbs/gradnumthy/notfree.pdf, but this issue isn't relevant. A prime in a number field ramifies in a finite extension iff it divides the discriminant ideal of the extension, and there is no need for an initial hypothesis that the top ring of integers is a free module over the bottom ring of integers. (However, in the proof of the theorem one may localize at the prime of interest and the localized ring of integers becomes a PID, so leading to a free module as a technical convenience, not a hypothesis.) $\endgroup$
    – KConrad
    Oct 12, 2011 at 17:47
  • 2
    $\begingroup$ J, you are missing something trivial: see Milne's Remark 3.39(b), where he indicates how to remove the freeness assumption by using localization. $\endgroup$
    – KConrad
    Oct 12, 2011 at 17:52

1 Answer 1

1
$\begingroup$

Why not try to "solve a harder problem", and produce Eisenstein polynomials of whatever degree you want, over any number field? These give totally ramified extensions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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