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Kristal, yes, yes, sorry. There are at least 2 mistakes in the original post:

• I forgot to mention that the polynomial $f$ is assumed monic
• The sentence "But it seems clear that there is no cubic having discriminant $2^{2} \times 3$" should say "...no cubic having $D = \sqrt{\Delta} = 2^{2} \times 3$". I reach this conclusion just by looking at the roots as points on a line and the numbers$(\alpha_{i} - \alpha_{j})$ as 'oriented distances between the points'.

For your example, $3x^3 - x = 3x(x^{2} - 1/3) = 3x(x - 1/\sqrt{3})(x + 1/\sqrt{3})$

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 relates to 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!

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

1

Kristal, yes, yes, sorry. There are at least 2 mistakes in the original post:

• I forgot to mention that the polynomial $f$ is assumed monic
• The sentence "But it seems clear that there is no cubic having discriminant $2^{2} \times 3$" should say "...no cubic having $D = \sqrt{\Delta} = 2^{2} \times 3$". I reach this conclusion just by looking at the roots as points on a line and the numbers$(\alpha_{i} - \alpha_{j})$ as 'oriented distances between the points'.

For your example, $3x^3 - x = 3x(x^{2} - 1/3) = 3x(x - 1/\sqrt{3})(x + 1/\sqrt{3})$

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 relates to 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!

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