Degree of an affine variety

The definition of degree is given for projective varieties, is there a definition also for affine varieties? If an affine variety has dimension $k$, is it possible to find at least an upper and a lower bound for the number of points in the intersection with a general linear space of codimension $k$?

• The concept of linear subspace is not invariant under automorphisms of affine space, so the question is ill-formed. It appears however, that you are thinking about a specific embedding of $X$ into $A^n$, then you can take the closure of $X$ under standard $A^n\subset P^n$ -- its degree is the upper bound you seek. Lower bounds appear more subtle, even in algebraically closed field case. – Lev Borisov Sep 6 '14 at 13:39
• Yes, the upper bound is obvious. So you think it is difficult to provide a lower bound? – user46071 Sep 7 '14 at 6:35
• I think that it can be subtle. For example, for conics in $A^2$, parabola intersects every line in at least one point, while the nondegenerate ones have lines that are tangent to it at infinity and thus have zero intersection points in $A^2$. – Lev Borisov Sep 7 '14 at 11:54

Let $X\subset\mathbb{A}^n$ an affine variety of dimension $k$, and let $L\subset\mathbb{A}^n$ a general linear subspace of dimension $n-k$. Consider the projective closures $\overline{X},\overline{L}\subset\mathbb{P}^n$, where $\mathbb{A}^n = \mathbb{P}^n\setminus H$, and $H\subset\mathbb{P}^n$ is an hyperplane. Assume $k$ algebraically closed. Then $$\#(X\cap L) \leq \#(\overline{X}\cap\overline{L}) = deg(\overline{X}).$$ Furthermore, if $\overline{L}$ is such that $(\overline{L}\cap H)\cap \overline{X} = \emptyset$ we have $$deg(X)=\#(X\cap L) = \#(\overline{X}\cap\overline{L}) = deg(\overline{X}).$$