The degree of an algebraic curve in higher dimensions

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

• I'd go with the degree of whichever of the $f_i$ has the largest degree. Aug 18, 2011 at 15:34
• 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. Aug 18, 2011 at 15:44
• For a purely algebraic definition, see I.7 of "Hartshorne - Algebraic Geometry" for the definition of degree of a projective variety. Aug 18, 2011 at 16:03

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.