Consider the following theorem due to Minkowski (originally conjectured by Kronecker iirc):

For any proper finite extension $K$ of $\mathbb Q$, some prime in $\mathbb Q$ ramifies in $K$. Equivalently, the discriminant of $K$ has absolute value greater than $1$.

The proof of this result is a well-known consequence of Minkowski's convex body theorem, and indeed this was how he has originally proven this result. I was wondering whether, after all these years, a significantly different proof has been devised. More precisely, is there a proof of this theorem which avoids the use of geometry of numbers? If not, can we possibly prove that there are only finitely many unramified extensions (a special case of Hermite-Minkowski theorem)?

I'd like to point out this question on MO which is closely related, but doesn't address this specific theorem. As answers there point out that Minkowski's convex body theorem is, in a way, a sharpening of pigeonhole principle, so I can imagine that some clever application of this theorem might be able to provide us with a proof of some bound on the discriminant strong enough to give the desired result.

  • $\begingroup$ We need to show that $|d_K|>1$ for the absolute discriminant. Minkowski gives $|d_K|\ge\frac{\pi}{3}(\frac{3\pi}{4})^{n-1}>1$; perhaps this can be achieved by a "strengthening" of the pigeon principle, as Felipe says. $\endgroup$ Nov 28, 2016 at 20:57

2 Answers 2


Landau has given an arithmetic proof in Der Minkowskische Satz über die Körperdiskriminante (Minkowski's Theorem on the field discriminant) Gött. Nachr. 1922, 80-82, where he replaced Minkowski's geometry of number by Dirichlet's pigeonhole principle. Here's an English translation.


The Stark-Odlyzko discriminant bounds give a purely analytic proof of Minkowski's theorem, using the functional equation of $\zeta_K(s)$. See this paper:


As in Minkowski, the basic point is to show that $|\Delta_K| > 1$. The paper above doesn't give the explicit bounds in low degree necessary to conclude this for all degrees, but for that one can see:


(gives a sufficiently good bound for $[K:\mathbf{Q}] \ge 8$)


(gives sufficiently good bounds for degrees $2$ to $8$.)



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