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Dec 15, 2010 at 2:08 comment added JSE Per the philosophy of Bhargava (and indeed it seems plausible that this could be proved via his techniques) the number of S_5-extensions with squarefree discriminant between -N and N should be about (13/120)*(6/pi^2)*2N; i.e. each squarefree discriminant has 13/120 A_5-extensions on average.
Nov 15, 2010 at 16:34 comment added Tim Dokchitser @François: I think any quintic $f$ over ${\mathbb Q}$ with Galois group $S_5$ and square-free discriminant $D$ defines an unramified $A_5$-extension of $Q(\sqrt D)$. This is because exactly two roots of $f$ coincide modulo any prime divisor of $D$, so the inertia at $p$ must be exactly $C_2$. E.g. $f(x)=x^5-x^4-x^3+x^2-1$ has $D=1609$, but I don't know whether there are any smaller examples.
Nov 5, 2010 at 13:22 comment added François Brunault By the way, is there a (conjectural) characterization of quadratic fields admitting an $A_5$ unramified extension ? Is $2869$ the smallest possible discriminant ?
Nov 4, 2010 at 12:04 comment added Kevin Buzzard @Denis Serre: is there no computer algebra package on your computer? :-) Yes, 2869 is the discriminant.
Nov 4, 2010 at 11:57 comment added Denis Serre Is $2869$ something like the discriminant of $X^5-X+1$ ?
Nov 4, 2010 at 11:37 history edited Kevin Buzzard CC BY-SA 2.5
sqrt->\sqrt
Nov 4, 2010 at 11:23 history answered Robin Chapman CC BY-SA 2.5