Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 P2 is y^2·z = x^3−x·z^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

share|improve this question
add comment

2 Answers

up vote 18 down vote accepted

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, 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.

share|improve this answer
add comment

Let me add a comment with some geometric flavor to Charles's clean answer. Once the field is algebraic closed, then the affine curve $C\subset \mathbb{A}^2$ is never compact. However it admits an embedding into the projective space (remember that we know that $\mathbb{A}^2\subset \mathbb{P}^2$) where we can look for a compactification. The construction above, geometrically speaking, takes a hyperplane in $\mathbb{P}^2$ and compactifies with it. In other words, there is a hyperplane (divisor) $H=[z=0]\subset \mathbb{P}^2$ which tells you which points to add to your curve $C$, in order to get it compact: let's denote it by $\tilde{C}$. At the end of the day, you will have that $\tilde{C}\cap H$ are the points that you have added to $C$.

In a more general setting (related to the Groebner basis), if you have a manifold $M$, you may ask yourself that it would be desirable for a compactification of it to be algebraic. Now, as far as I now, there are two main obstructions for a compactification to be algebraic. There may not be a "nice" compactification and it may not have enough meromorphic functions. By "nice" here, I mean a compactification given by divisors (with normal crossings usually is asked), and by "enough meromorphic functions", there may not be ample line bundles. Questions on -When such obstructions do not cause troubles?- are big theorems in algebraic geometry. Not referring to self-confidence issues... "here, positivity helps a lot!"

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.