User lee - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T01:29:02Z http://mathoverflow.net/feeds/user/2102 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6901/two-questions-about-discriminants-of-polynomials-in-qx/6913#6913 Answer by Lee for Two questions about discriminants of polynomials in Q[x] Lee 2009-11-26T20:44:02Z 2009-11-26T20:53:56Z <p>Kristal, yes, yes, sorry. There are at least 2 mistakes in the original post:</p> <ul> <li>I forgot to mention that the polynomial $f$ is assumed monic</li> <li>The sentence "But it seems clear that there is no cubic having discriminant <code>$2^{2} \times 3$</code>" should say "...no cubic having $D = \sqrt{\Delta} = 2^{2} \times 3$<code>". I reach this conclusion just by looking at the roots as points on a line and the numbers </code>$(\alpha_{i} - \alpha_{j})$` as 'oriented distances between the points'. </li> </ul> <p>For your example, <code>$3x^3 - x = 3x(x^{2} - 1/3) = 3x(x - 1/\sqrt{3})(x + 1/\sqrt{3})$</code></p> <p>so that $\Delta = 3^{2(3-1)} [(0 + 1/\sqrt{3})(0 - 1/\sqrt{3})(1/\sqrt{3} + 1/\sqrt{3})]^{2} = 3^{4} \dot (1/3)^{2} \dot (2/\sqrt{3})^{2} = 12$, just as you claimed. But my question is about the square root of the discriminant, which is the interesting quantity for me because the Galois group of $f$ is a subgroup of $A_{n}$ if and only if D(f) is rational!</p> <p>It appears I can't edit my original post (because I didn't register to make it). Thank you all very much for the references, especially David Brown</p>