10
$\begingroup$

This might be a very simple question but I can't seem to find a precise definition of the degree of an algebraic curve, if such can even be defined. In the plane, the degree of an algebraic curve is clear; it is simply the degree of the defining polynomial (a curve in the plane is defined by the equation $f(x,y) = 0$, for some polynomial $f$). In higher dimensions, say $n$, an algebraic curve is defined by $n-1$ polynomial equations $f_1 =0, f_2 =0, \cdots, f_{n-1} = 0$. However, the degree of the curve is no longer clear.

Can the degree of an algebraic curve be defined in higher dimensions? If so, how is it defined?

Thanks

$\endgroup$
3
  • $\begingroup$ I'd go with the degree of whichever of the $f_i$ has the largest degree. $\endgroup$
    – J. M. ain't a mathematician
    Aug 18, 2011 at 15:34
  • 2
    $\begingroup$ It can be shown that the degree of a projective plane curve is equal to the degree of its hyperplane divisor; in simpler terms, the degree is equal to the number of intersections with a generic line. This has an obvious generalisation to higher dimensions, but I don't know whether this is the accepted definition. $\endgroup$
    – Zhen Lin
    Aug 18, 2011 at 15:44
  • 1
    $\begingroup$ For a purely algebraic definition, see I.7 of "Hartshorne - Algebraic Geometry" for the definition of degree of a projective variety. $\endgroup$
    – Andrea
    Aug 18, 2011 at 16:03

1 Answer 1

Reset to default
14
$\begingroup$

As Zhen Lin says, a geometric way of defining the degree is to take the number of points of intersection of a general hyperplane with the curve. Roughly, general encodes the condition that you do not want the hyperplane to contain the curve, and you also do not want it to be tangent to the curve. This definition is certainly accepted, though it can be difficult to work with it, since often you want to deal with "degenerate" cases, and this definition becomes tricky to handle. The reference to Hartshorne is certainly a good place to start.

I also wanted to make a few comments on some of the statements made. First, it is not true in general that a curve in $\mathbb{P}^n$ is defined by $n-1$ equations: certainly you need at least $n-1$ equations to define it, but normally you would need more than that. An easy example is the twisted cubic in $\mathbb{P}^3$. Second, referring to the comment of J.M., if a curve turns out to be defined by $n-1$ equations in $\mathbb{P}^n$, then its degree would be the product of the degrees of the equations, rather than the maximum.

$\endgroup$
2
  • $\begingroup$ I see. So the intersection of a sphere and an ellipsoid would be a quartic for instance? $\endgroup$
    – J. M. ain't a mathematician
    Aug 21, 2011 at 12:33
  • $\begingroup$ If a codimension $k$ variety is defined by more than $k$ polynomials, would I expect its degree to be higher or lower than the product of the degrees? Can something general be said at all? $\endgroup$
    – M. Winter
    Aug 13 at 22:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .