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Let $P$ be the probability that an elliptic curve with a rational point has an infinite number of rational points. From what I understand, the value of P is unknown.

This got me thinking about a similar question related to Galois theory. Let $N$ be the probability that a finite extension of $\mathbb{Q}$ is a normal extension. Is anything known about the value of $N$?.

I wouldn't expect it to be 1 or 0.

My gut feeling is that not much is known as it seems related to the inverse Galois problem. Indeed, if every finite group were the galois group of some finite galois extension of $\mathbb{Q}$ i would imagine that $R$ would be somehow related to the "probability of a subgroup being normal" (this is intentionally vague).

Question: Is anything (if anything) known about $N$?

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Why don't you expect it to be 0? Quadratic fields are misleading. Think about the cubic case. To be Galois requires the discriminant is a perfect square and already that is not so easy to stumble across by chance. Now think about the chance that a random quartic will be Galois. Wouldn't you kind of surprised if you wrote one down and it was actually normal? The book "Field Arithmetic" has a treatment of probabilistic questions about random elements of the absolute Galois group of Q. Maybe something in there addresses your question with precision. –  KConrad Jun 18 '10 at 1:04
Bhargava has theorems on the number of number fields with fixed degree (4 or 5) and increasing discriminant, and I believe that among quartic fields the ones with Galois group S_4 and D_4 contribute positive proportions and the rest (in particular, Galois quartics) contribute density 0. –  KConrad Jun 18 '10 at 1:06
Before Bhargava, there was Andrew Baily, On the density of discriminants of quartic fields, J Reine Angew Math 315 (1980) 190-210, MR 81c:12006. Andy proved (among other things) that the number of Galois quartic number fields with discriminant $d$, $|d|\lt X$, is asymptotic to a constant times $\sqrt X\log^2X$. Before Baily, there were results on cubic fields by H Cohn, and by Davenport and Heilbronn. –  Gerry Myerson Jun 18 '10 at 1:19
Gerry: I knew about Davenport--Heilbronn for the cubic case, but off the top of my head I remembered Bhargava's result for the quartic case so I just went with that. –  KConrad Jun 18 '10 at 2:36
If you fix the degree and count by coefficients of polynomials you get 0. Most of the polynomials have full galois group, preventing their corresponding extension to be normal (van der Waerden). On the other hand, if you count all number fields by their discriminant (possible by Hermite), then the first thing to note is that "most" number fields are in fact quadratic or cubic. This can be seen from the degree of the discriminant. Hence, in this case, the density will actually be positive. Finally, if you fix the degree and count by discriminant, then you can read the comments above. –  Dror Speiser Jun 18 '10 at 7:20

1 Answer 1

The probability is 0, if you count number fields of some fixed degree in order of discriminant. Venkatesh and I prove in this paper (later published in Ann. of Math., 163, no.2, 2006) that there are at most c_n,e X^{3/8+e} Galois extensions of degree n and discriminant at most X, while there are at least c_n X^{1/2} degree-n extensions of discriminant at most X in all. (Theorem 1.1 and Prop 1.3).

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Very good to publish a paper about discriminants in Volume 163 (Volume $-163$ would have been even better, albeit impractical). –  Gerry Myerson Jun 29 '10 at 6:41

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