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Let $Y_1,\dots,Y_r$ be projective hypersurfaces of $\mathbb{P}^n$. Is it true that $Y_1,\dots,Y_r$ intersect transversally if and only if $\dim (Y_1\cap \cdots \cap Y_r) = n-r$?

PS: The motivation for this question is in my effort to understand why a transversal intersection of hypersurfaces is a complete intersection; see Transversal and complete intersection of hypersurfaces in $\mathbb{P}^{n}$. This is stated by Harris (AG-First Course) below Exercise 17.17. The issue is that i have seen more than one definitions of transversality, and it is unclear to me if they are equivalent or not. For example the definitions of Harris and Shafarevich are different (which makes the answer/comment at the above link particularly unfortunate). At any case Harris defines $Y,X \subset \mathbb{P}^n$ to intersect transversally (if i understand correctly) if $\mathcal{T}_{X,P}+\mathcal{T}_{Y,P}=\mathcal{T}_{\mathbb{P}^n,P}$, for every point $P \in X \cap Y$.

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  • $\begingroup$ What is your definition of transversal intersection? $\endgroup$
    – Mohan
    Oct 3, 2015 at 2:16
  • $\begingroup$ @Mohan: I am sorry that i can not be very precise. I suppose the definition that i am interested in is the one that makes a transversal intersection of hypersurfaces a complete intersection. Do you know what that is? $\endgroup$
    – Manos
    Oct 3, 2015 at 3:35
  • $\begingroup$ Let me just stick to two factors: if I have two subvarieties of something smooth and the intersection has the right dimension and is reduced then the intersection is actually transverse at a general point. This is often all you need. $\endgroup$
    – Hoot
    Oct 3, 2015 at 4:01
  • $\begingroup$ @Hoot: So let us take two affine hypersurfaces $Y_1=Z(f_1),Y_2=Z(f_2) \subset \mathbb{A}^n$. Is it not true that $Y_1 \cap Y_2$ is always reduced, since the ring of regular functions of $Y_1 \cap Y_2$ is $k[x_1,\dots,x_n] / I_{Y_1 \cap Y_2}$ and $I_X$ is always radical for any $X \subset \mathbb{A}^n$? So isn't the "reduced" condition unecessary? $\endgroup$
    – Manos
    Oct 4, 2015 at 22:39
  • $\begingroup$ I'm working scheme-theoretically, so perhaps this is cheating. $\endgroup$
    – Hoot
    Oct 5, 2015 at 0:55

2 Answers 2

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Of course it is not true!
If you take the conic $yz=x^2$ and the line $y=0$ in $\mathbb P^2$, their intersection is of dimension zero but trivially these curves are not transverse at $(0,0)$ since they are tangent at that point.
This is essentially at the level of Euclid, 2300 years ago, who didn't have Harris's book to confuse him...

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  • $\begingroup$ You're welcome, dear Manos. $\endgroup$ Oct 3, 2015 at 19:35
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Consider $Y_1: y^2=x^3$ and $Y_2:y^2=-x^3$ in $\mathbb{P}^2$. We have that $Y_1\cap Y_2=\{[z; x; y]| zy^2-x^3=0=zy^2-x^3\}$, so $x^3=0$, and $zy^2=0$, so our points are $[0, 0, 1]$ and $[1, 0, 0]$, which is finite, and thus $dim_k(Y_1\cap Y_2)=0$, as we want. But $$(\mathcal{O}_{Y_1\cap Y_2})_{[1, 0, 0]}=k[y, x]_{(x, y)}/(y^2-x^3, y^2+x^3)=k[y, x]_{(x, y)}/(y^2, x^3)$$ So we can compute $I([1, 0, 0], \mathcal{O}_{Y_1\cap Y_2})=length_{\mathcal{O}_{[1, 0, 0]}}(\mathcal{O}_{Y_1\cap Y_2})=6$. So the intersection multiplicity is greater than one, so that the varities are not transverse at the origin.

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  • $\begingroup$ Thanks for your answer. Which definition of transversality are you using? $\endgroup$
    – Manos
    Oct 3, 2015 at 3:24
  • $\begingroup$ $X, Y$ is transverse if and only if for all irreducible components $W\subset X\cap Y$, $I(W, X\cap Y)=length_{\mathcal{O}_{W}}(\mathcal{O}_{X\cap Y})=1$. Since you didnt specify if you wanted this or the tangent space defintion, I just rolled with it! $\endgroup$
    – pax
    Oct 3, 2015 at 3:47
  • $\begingroup$ I am sorry, i didn't even know that intersection multiplicity 1 is also know as transversal. $\endgroup$
    – Manos
    Oct 3, 2015 at 4:25
  • $\begingroup$ Ah, I should say though, this is slightly weaker than transverality in general, which states that the tangent bundle is the direct sum of the tangent bundles at points of intersection $\endgroup$
    – pax
    Oct 3, 2015 at 5:24
  • $\begingroup$ Dear Manos, don't worry about length and don't be sorry: "length" is an unnecessary concept to answer your question. $\endgroup$ Oct 3, 2015 at 7:42

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