Let $Y\subseteq \mathbb{P}^n$ be a projective variety. Let $I(Y)=\langle f_1,\dots,f_k\rangle$. Let $f\in k[x_0,\dots, x_n]$ be an irreducible homogeneous polynomial such that $f\notin I(Y)$. Let $Z(f)$ be the zero set of $f$. Let $Y'=Y\cap Z(f)$. Is it true that $\dim Y'=\dim Y-1 $ ?
Reduction of Dimension of a projective variety by intersection with another variety
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algebraic-geometry
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0I think some more conditions are needed here. For example, if $Y=Z(f)$ then clearly the dimensional equality is not true. – 2017-02-09
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0@awllower Thanks for pointing out the mistake. Changed it. – 2017-02-09
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1@awllower Added one more condition. – 2017-02-09
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0Won't you need some kind of transversality condition? – 2017-02-09
2 Answers
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Yes, it is true. Let $Z=Z(f)$. Hartshorne AG I.7.2 says that the dimension of each irreducible component of $Y\cap Z$ is at least $\dim Y-1$. Suppose there were some component $A$ of $Y\cap Z$ such that $\dim A = \dim Y$. Then $A$, being a closed subset of $Y$ of the same dimension, is equal as a set to $Y$. In particular, $Y$ is a closed subvariety of $Z$. But then $f$ vanishes along $Y$, a contradiction to $f\notin I(Y)$.
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It is true, and in fact you don't even need $f$ to be irreducible. Generally, if $A$ is Noetherian and $f$ is a nonzerodivisor of $A$, then $A/f$ has dimension one less than that of $A$.