The discriminant for the plane cubic curve

10 coefficients determine a degree 3 homogenous polynomial in $k[x,y,z]$. I understand that there is a degree 12 polynomial in these coefficients, called the discriminant, with 2040 terms, which vanishes precisely when the curve is singular. I'm working on some generalizations to tropical geometry, and I would like to understand the classical case. Could someone point me to a reference where that shows how to construct this and check that it has the desired properties? I know that Gelfand, Kapranov, and Zelevinski have an important book about discriminants, but glancing through the table of contents it looks like it has a lot of prerequisits and may not be as concrete as I would like. Google searching has been fruitless so far.

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Getting a simple answer that you can use right now is good, but if you're serious about working in this area, you should read GKZ while you're a grad student and still have time. – Alexander Woo Sep 19 '12 at 0:36

Go to http://www.ma.utexas.edu/cnt/cnt-frames.html and click on "jac_cubic" on the right hand panel. It gives a link to download a pari-gp script that computes the Weierstrass equation of the jacobian of a general cubic and, from that, the discriminant, if you want. Artin, Tate and Villegas have a paper explaining the theory too. It's on Villegas's webpage too.

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The link on Villegas's website is broken, but fortunately the published version is "Open Archive" sciencedirect.com/science/article/pii/S0001870805001775 – j.c. Feb 26 at 3:37

For a less abstract and more computational approach you can for instance look at Example 5.48 in the book "Introduction a la resolution des systemes polynomiaux" by Elkadi and Mourrain (in French). It gives the resultant of three conics as the determinant of an explicit matrix. Apply this to the partial derivatives of your cubic and you will get the discriminant. The method goes back to Sylvester. Also, if you have JSTOR access and don't mind reading rather old fashioned algebra, you can look at this paper by Morley.

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