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 Feb 15 revised Using the Residue Theorem for a contour integral along the Riemann sphere added 1 characters in body Feb 15 asked Using the Residue Theorem for a contour integral along the Riemann sphere Feb 2 comment The Process of Choosing Projective Axes to Put an Elliptic Curve into Weierstrass Normal Form it's 5 o'clock, do you know where your children are? Jan 30 accepted Proving a group isomorphism from $(S,+)$ to $(S,+')$ Jan 30 comment Proving a group isomorphism from $(S,+)$ to $(S,+')$ Damn, I don't know if I could have ever figured this out using just the composition law, it would have been really complex. Thanks. Jan 30 comment Proving a group isomorphism from $(S,+)$ to $(S,+')$ @DonAntonio: Ok thanks Jan 30 comment Proving a group isomorphism from $(S,+)$ to $(S,+')$ @DonAntonio: yup, it's from Rational Points on Elliptic Curves by Silverman Jan 30 asked Proving a group isomorphism from $(S,+)$ to $(S,+')$ Jan 21 comment Given an algebraic curve $F(x,y)=0$, why do the partial derivatives of $F(x,y)$ being zero at a point imply the plane curve has a singularity? @Lubin: I mean that you're taking partial derivatives of the function $z=F(x,y)$ which defines a surface in $\mathbb{C}^3$, and looking at points which are in some sense tangent to the $z=0$ portion of $\mathbb{C}^3$, and these are the points at which the partials are zero and the curve defined by the intersection of $z=F(x,y)$ with $z=0$ has a singularity. As an aside, it's not clear to me whether I'm using the words: tangent, plane, surface correctly in the context of $\mathbb{C}$ rather than $\mathbb{R}$ Jan 21 comment Given an algebraic curve $F(x,y)=0$, why do the partial derivatives of $F(x,y)$ being zero at a point imply the plane curve has a singularity? I mean crossing singularities don't satisfy the implicit function theorem, but each line generally has a well defined tangent line, while cusp singularities it's the opposite. Both however satisfy the partial derivative criterion for being a singularity, so maybe I should take that as a definition instead of a theorem to be proven? Or is the best definition that no open set around the point is homeomorphic to $\mathbb{R}$ (for curves)? Jan 21 comment Given an algebraic curve $F(x,y)=0$, why do the partial derivatives of $F(x,y)$ being zero at a point imply the plane curve has a singularity? Maybe I wasn't clear enough in my wording. I'm talking about singularities of algebraic curves here, I'm just looking at surfaces because those are where you take your partial derivatives. As for a definition of singularity well I'm not sure I have a good one, beyond calling it any point which does not have a unique well defined tangent line. Jan 21 accepted Show that the rational conic $F(x,y)=ax^2+bxy+cy^2+dx+ey+f$, subject to a certain condition, is non-singular Jan 21 comment Show that the rational conic $F(x,y)=ax^2+bxy+cy^2+dx+ey+f$, subject to a certain condition, is non-singular After pondering this answer all day, and enriching my understanding of the projective plane considerably in the process, I have fully understood the particulars of how affine and projective singularities relate. Thank you. Jan 21 asked Given an algebraic curve $F(x,y)=0$, why do the partial derivatives of $F(x,y)$ being zero at a point imply the plane curve has a singularity? Jan 20 revised Show that the rational conic $F(x,y)=ax^2+bxy+cy^2+dx+ey+f$, subject to a certain condition, is non-singular deleted 1 characters in body Jan 20 revised Show that the rational conic $F(x,y)=ax^2+bxy+cy^2+dx+ey+f$, subject to a certain condition, is non-singular added 4 characters in body Jan 20 asked Show that the rational conic $F(x,y)=ax^2+bxy+cy^2+dx+ey+f$, subject to a certain condition, is non-singular Jan 11 awarded Tumbleweed Jan 4 revised The Process of Choosing Projective Axes to Put an Elliptic Curve into Weierstrass Normal Form added 154 characters in body Jan 4 asked The Process of Choosing Projective Axes to Put an Elliptic Curve into Weierstrass Normal Form