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I came across the definition in the book titled Elliptic Curves by Anthony W Knapp, couldn't understand it so looked online, which just confused me more. I'm looking for an explanation in the context of curves in projective plane/space.

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    $\begingroup$ In affine geometry you define your space by an equivalence relation $x\sim x+b$ in projective gemoetry instead we take $x\sim ax$ $\endgroup$ – tired Jun 29 '16 at 8:39
  • $\begingroup$ How do we define space using an equivalence relation? That's what's unclear to me. $\endgroup$ – DpS Jun 29 '16 at 8:49
  • $\begingroup$ If the concept of "projective space" is not familiar to you, then please allow me to suggest that "elliptic curves" is a subject too complicated for you to study at your present level of mathematical understanding. $\endgroup$ – Alex M. Jul 6 '16 at 10:09
  • $\begingroup$ Can you suggest me a good book or article or anything to read up projective space first then? $\endgroup$ – DpS Jul 6 '16 at 10:16
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You get a projective plane from an affine plane if you consider “points at infinity” as regular elements of your plane. This simplifies a number of situations, for example two distinct lines will always intersect in a unique point, the special case of parallel lines vanishes. Parallel lines simply intersect at infinity.

Expressed in coordinates, you add one coordinate. So a point $(x,y)$ in the usual Cartesian plane would be represented as $[x,y,1]$ or any multiple thereof. That's called a homogeneous coordinate vector. So in fact you are no longer dealing in vectors, but strictly speaking in equivalence classes of vectors. Most of the time authors will use the same notation for vectors and for equivalence classes, and rely on context to tell you which is which in those cases where it makes a difference. To convert back, a homogeneous coordinate vector $[x,y,z]$ corresponds to a Cartesian vector $(x/z,y/z)$. If $z=0$, this would be undefined; those are the points at infinity. The vector $[0,0,0]$ has to be excluded, since it would otherwise belong to all equivalence classes. The null vector does not represent any point in the projective plane.

In terms of curves, you might want to make certain that all your equations are homogeneous, i.e. have the same degree in each term (for each geometric object). So for example the equation $7x^2 + 5y = 3$ would not be homogeneous: if you have a vector which solves this, then take twice that vector, you may end up with another representative of the same point which fails the equation. $7x^2 + 5yz = 3z^2$ on the other hand would be a homogeneous equation.

See also Difference between Projective Geometry and Affine Geometry which discusses that difference from a different point of view.

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  • $\begingroup$ This was very helpful, thank you. $\endgroup$ – DpS Jun 30 '16 at 4:39
  • $\begingroup$ From what I gather, when we talk about curves in a projective plane/space, we're operating on classes of vectors rather than vectors, which also why we leave out the 0 element. What I don't understand is what are the advantages of doing this? It just seems to complicate matters. $\endgroup$ – DpS Jul 6 '16 at 9:48
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    $\begingroup$ @DeepishaSolanki: It complicates some things, but simplifies others. While the coordinates become more complicated, the operations and the axioms become simpler, both thanks to fewer special cases. It also provides nice generalizations, where you can derive some knowedge in an easy setup but apply it to a much broader range of situations simply because you know that your statement is invariant under projective transformations. Projective transformations themselves are very hard to express in a non-projective plane, but really easy in a projective one. $\endgroup$ – MvG Jul 6 '16 at 10:35

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