Say we have an affine variety $V \subset \mathbb{C}^n$, and suppose we intersect $V$ with a hyperplane $H$, possibly not in general position. Is it possible for the degree of $V \cap H$ to be larger than the degree of $V$?
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3$\begingroup$ How do you define the degree of $V$ of $V$ is not a hypersurface? $\endgroup$– YCorFeb 13, 2020 at 23:32
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2$\begingroup$ The definition I'm using is that if $V$ has dimension $d$, then the degree of $V$ is equal to the number of intersection point of $V$ with $d$ hyperplanes in general position. (Including intersections at infinity and intersection multiplicity). $\endgroup$– PowerspawnFeb 13, 2020 at 23:36
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$\begingroup$ you mean $n-d$ hyperplanes $\endgroup$– Abdelmalek AbdesselamFeb 13, 2020 at 23:43
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1$\begingroup$ @AbdelmalekAbdesselam The hyperplanes have codimension $1$, so each intersection reduces the dimension of $V$ by $1$, so we need to intersect with $d$ hyperplanes total. $\endgroup$– PowerspawnFeb 13, 2020 at 23:49
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$\begingroup$ oops you're right got mixed up between what is dimension and what is codimension $\endgroup$– Abdelmalek AbdesselamFeb 13, 2020 at 23:55
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1 Answer
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Yes, it is possible for the degree to increase. Say $V \subset \mathbb{C}^3$ is reducible: a union of a curve of degree $d$ plus one more point that doesn't lie on the curve. Then $V$ has degree $d$. But if $H$ is any plane through that extra point, then the intersection of $V$ with $H$ has degree $d+1$: the one point, plus $d$ points of intersection with the curve.
More generally, degree only sees the highest-dimensional component of $V$, but intersection with a hyperplane can bring up lower-dimensional components, raising the degree.
answered Feb 14, 2020 at 2:40