After reading this question on MO on the "degree" of an affine variety, I did not see why they said it was hard to give a lower bound on the number of intersection points of an affine variety $X \subseteq \mathbb{A}^n$ with an $(n – \mathrm{dim}(X))$-dimensional linear subspace in general position.
In particular, I would expect these to be the same in the affine and projective case.
Could anyone explain this?
Details
Let us define the "degree" $\deg(X)$ of an affine variety $X$ (with implicit embedding $X \to \mathbb{A}^n$) as the number of points in the intersection of $X$ with an $(n – \dim(X))$-dimensional linear subspace of $\mathbb{A}^n$ in general position to $X$. This is analogous to the projective case.
Then I would expect that $\deg_{\mathbb{A}^n}(X) = \deg_{\mathbb{P}^n}(\overline{X})$, where $\overline{X}$ is the projective closure of $X \subseteq \mathbb{A}^n \subseteq \mathbb{P}^n$.
Proof Sketch Clearly the projective closure $\overline{X}$ of $X$ does not have any irreducible components lying in the hyperplane at infinity $H := \mathbb{P}^n \setminus \mathbb{A}^n$. Hence, the projective variety $Y := \overline{X} \cap H \subseteq H \subseteq \mathbb{P}^n$ has dimension $d – 1$, where $d = \dim(\overline{X}) = \dim(X)$.
Now, a $(n – d)$-dimensional linear subspace $L$ of $\mathbb{P}^n$ in general position clearly does not intersect $Y$, as $\dim(L) + \dim(Y) = n – 1 < n = \dim(\mathbb{P}^n)$ (the statement $L$ intersects $Y$ is a nontrivial, algebraically closed condition on $L$ in $\mathrm{Gr}(d – n, \mathbb{P}^n))$.
However, this should give
\begin{align*}
\deg_{\mathbb{P}^n}(\overline{X}) &= \#(L \cap \overline{X}) = \#(L \cap Y) + \#(L \cap X) \\
&= \#(L \cap X) = \#((L \cap \mathbb{A}^n) \cap X) = \deg_{\mathbb{A}^n}(X)
\end{align*}
since $L \cap \mathbb{A}^n$ is clearly a $(n – d)$-dimensional linear subspace of $\mathbb{A}^n$ in general position. $\square$
I know that I am somewhat imprecise with my use of “general position'', but intuitively, this seems to make sense. However, the MO question seems to indicate that the above does not hold in general, and we just have
\begin{equation*}
\deg_{\mathbb{A}^n}(X) \leq \deg_{\mathbb{P}^n}(\overline{X})
\end{equation*}
Can anyone tell me where/whether the statement or my proof sketch are wrong?
Best Answer
Your proof sketch looks fine to me - from a close reading of the comments at the MO post, it looks like the commenter is concerned about answering the question "what are the bounds for $\#(L\cap X)$ depending on $X,L$", not "what happens for general $L$". I'd still caution you that there's a reason people don't use "degree in $\Bbb A^n$" - it's much less nice than the projective version.