[Math] General hyperplane sections and projection from a point

ac.commutative-algebraag.algebraic-geometryelimination-theory

Let $k$ be an algebraically closed field, and consider some subscheme $X\subset \mathbb{P}_k^n$. Let $x$ be a closed point of $X$, and $H$ a general hyperplane containing $x$. There is a regular map $\phi:\mathbb{P}^n\setminus{\{x\}}\rightarrow \mathbb{P}^{n-1}$ gotten by projecting from the point $x$.

My question: Is $\overline{\phi(H\setminus x)}\cap\overline{\phi(X\setminus x)}=\overline{\phi((X\cap H)\setminus x)}$?

This can be rephrased in terms of commutative algebra: Let $A=k[x_1,\ldots,x_n]$, $B=A[x_0]$, $I$ be a homogeneous ideal of $B$, and $h$ a general linear form of $A$. The question above is equivalent to the following: Is $(I\cap A)+h=(I+h)\cap A$?

More generally, what if we replace $H$ and $h$ by a general hypersurface/general polynomial of degree $d$?

If $h$ or $H$ isn't general, equality doesn't hold, even for $d=1$. Consider for example $n=3$ and take $I=\langle x_0x_1,x_2^2+x_0x_3\rangle$, and consider $h=x_3-x_1$. Then $I\cap A=\langle x_1x_2^2 \rangle$, so $(I\cap A)+h=\langle x_1x_2^2,x_1-x_3\rangle$, but $(I+h)\cap A=\langle x_2^2,x_1-x_3\rangle$.

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EDITED to match clarifications in the question and in Sándor's answer.

The question is equivalent to asking whether the tangent cone at $x$ of the hyperplane section coincides with the hyperplane section of the tangent cone:

Blow up $x$, and denote $\tilde{\mathbb{P}}^n$, $\tilde X$, $\tilde H$ the resulting variety and the proper transforms of $X$ and $H$. Now $\phi$ extends to $\tilde \phi$, defined on the whole $\tilde{\mathbb{P}}^n$, and for every subvariety $Z\subset \mathbb{P}^n$, $\overline{\phi(Z\setminus \{x\})}=\tilde\phi(\tilde Z)$. Thus, $\overline{\phi(X)} \cap \overline{\phi(H)}= \tilde\phi(\tilde X) \cap \tilde\phi(\tilde H)=\tilde\phi(\tilde X\cap \tilde H)$, because $\tilde H=\tilde \phi^{-1}(\tilde \phi(\tilde H))$ as in Sándor's computation. On the other hand, $\overline{\phi(X \cap H)}=\tilde \phi(\widetilde{X\cap H})$.

So, whenever $\tilde X\cap \tilde H=\widetilde{X\cap H}$, your statement holds. Now we can work componentwise: assume $X$ is irreducible. It is clear that $\widetilde{X\cap H}=\tilde X \cap \tilde H$ unless $\tilde X \cap \tilde H$ has a component contained in the exceptional divisor $E\cong \mathbb{P}^{n-1}$. Since $\tilde X$ meets the exceptional divisor properly, the only case when $\tilde X \cap \tilde H$ can have a component contained in the exceptional divisor arises if $\tilde H$ does not meet $\tilde X \cap E$ properly. But if $H$ is general, so is $\tilde H \cap E=L$, and then it does meet $\tilde X \cap E$ properly!

In your example with non-general $H$, $\tilde X \cap E$ is a point which gives trouble because your $\tilde H$ happens to contain it.

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[Math] Blowing-up an ordinary double point, then contracting the exceptional locus to a curve

I guess the answer in no when $\deg(X) \geq 3$.

In fact, in this case Griffiths showed that the two rulings $L_1$ and $L_2$ in $E$ are numerically equivalent in $\widetilde{X}$ (although not algebraically equivalent in general), so the corresponding $K_{\widetilde{X}}$-negative rays $R_1$ and $R_2$ in the Mori cone of $\widetilde{X}$ coincide.

It follows that the contraction of $R_1$ is actually the contraction of the whole quadric, that is it is not possible to contract the two rulings separately.

The situation is very subtle: in fact, the completed local ring $\widehat{\mathcal{O}}_{X,o}$ is not factorial, but the fact that $L_1$ is numerically equivalent to $L_2$ implies that $\mathcal{O}_{X,o}$ is factorial.

See [Debarre, Higher Dimensional Algebraic Geometry, p. 160] and the references given therein.

[Math] How to find the generic initial ideal

You should apply a generic linear coordinate transform to the ideal and then compute the initial ideal. The matrices for which the result is the generic initial ideal is a (Zariski) open subset (Lemma 2.6, in the book). In particular the complement is of lower dimension.
Leaving distribution issues aside, if you 'pick a random coordinate transform', then you'll always find the generic initial ideal. In practice you'll find it with high probability. There is also an implementation in Macaulay2

Edit: Here is one way to computer-prove the statements from the book. To do this I make a ring in which the coefficients of $f,g$ are extra variables. There are 10 degree 2 monomials in four variables, therefore we need a ring with 10+10+4 variables. Here is the Macaulay2 code:

S = QQ[a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,b1,b2,b3,b4,b5,b6,b7,b8,b9,b10, x1,x2,x3,x4]
ba = {x1^2, x1*x2, x1*x3, x1*x4, x2^2, x2*x3, x2*x4, x3^2, x3*x4, x4^2}
a={a1,a2,a3,a4,a5,a6,a7,a8,a9,a10}  -- coefficients of f
b={b1,b2,b3,b4,b5,b6,b7,b8,b9,b10}  -- coefficients of g
f = sum (for i to 9 list a#i*ba#i)
g = sum (for i to 9 list b#i*ba#i)
I = ideal (f,g)
monomialIdeal I  -- returns the initial ideal

Running it computes the revlex initial ideal because revlex is the default order. The output is $(a_1 x_1^2,b_1 x_1^2,a_2 b_1x_1 x_2,a_5 b_1 x_1 x_2^2,a_1 a_5 b_2 x_1 x_2^2,a_5^2 b_1^2 x_2^3)$ and thus making the $a_i$ and $b_i$ invertible we find the claimed result. For lex you have to change the variable and monomial order on $S$. One has to be a bit careful with the ordering of the variables, because M2 does not know that the $a_i$ and $b_i$ are invertible. Putting the $x_1$ last does the right thing for revlex(this needs a quick check). The Macaulay2 package that I linked is for a concrete ideal (not the generic complete intersection) like in the example.

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Related Solutions

[Math] Blowing-up an ordinary double point, then contracting the exceptional locus to a curve

[Math] How to find the generic initial ideal

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