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S. Carnahan
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R[x,y]/(y^2-x^3-\pi x^2) gives the coordinate ring of a bad cubic curve, where \pi is a uniformizer in R. Remove the origin (which is the one singular point), and you should get what you wantprojectivize the curve by adding a point at infinity, so your group has an identity.

The generic fiber (treating \pi as a unit) is then the smooth part of a nodal cubic, yielding Gm, and the special fiber (setting \pi to zero) is the smooth part of a cuspdal cubic, yielding Ga.

R[x,y]/(y^2-x^3-\pi x^2) gives the coordinate ring of a bad cubic curve, where \pi is a uniformizer in R. Remove the origin, and you should get what you want.

R[x,y]/(y^2-x^3-\pi x^2) gives the coordinate ring of a bad cubic curve, where \pi is a uniformizer in R. Remove the origin (which is the one singular point), and projectivize the curve by adding a point at infinity, so your group has an identity.

The generic fiber (treating \pi as a unit) is then the smooth part of a nodal cubic, yielding Gm, and the special fiber (setting \pi to zero) is the smooth part of a cuspdal cubic, yielding Ga.

Source Link
S. Carnahan
  • 45.7k
  • 6
  • 114
  • 220

R[x,y]/(y^2-x^3-\pi x^2) gives the coordinate ring of a bad cubic curve, where \pi is a uniformizer in R. Remove the origin, and you should get what you want.