Is there a generalized method to find the projective closure of an affine curve? For example, I read that the projective closure of $y^2 = x^3−x+1$ in $\mathbb{P}^2$ is $y^2z = x^3−xz^2+z^3$.


If I want to find the the closure of another affine curve, what method should I employ? I can't seem to find an adequate description in any text.


Thanks



Projective closure of affine curve
When the curve is a plane curve of degree $d$, the formula is simple (and in fact, this works for any hypersurface) you take $f(x,y)=0$ and replace it by $z^d f(x/z,y/z)=0$. This will be homogeneous of degree $d$, and when $z\neq 0$, you recover your curve.


Now, if you have a more general affine variety, given by $\langle f\_1,\ldots,f\_k\rangle$ in $k[x\_1,\ldots,x\_n]$, then you first compute a [Groebner Basis](http://en.wikipedia.org/wiki/Gr%25C3%25B6bner_basis), which you can learn how to do using Cox, Little and O'Shea's "Ideals, Varieties and Algorithms" (assumes no background whatsoever) and then you do this same trick with each polynomial in the Groebner basis.


Why can't you just use any basis? Look at the twisted cubic. $t\mapsto (t,t^2,t^3)$. This is cut out, in affine space, by $y-x^2$ and $z-x^3$, though $y-x^2$ and $xy-z$ are better to use. But still, $yw-x^2$ and $xy-zw$ don't give the twisted cubic! Bezout tells us that they give something of degree four containing the cubic, and it's not a hypersurface, so you get the twisted cubic plus a line. To get the cubic itself, you have to use a third quadratic polynomial that you get in the ideal, $y^2-xz$, which is not in the ideal given by the homogenized generators. These three, however, form a Groebner basis for the ideal, and then homogenization gives the homogeneous ideal of the projective twisted cubic.


