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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    $\begingroup$ Dear Nathan, you said nothing false but algebraically closed fields are automatically infinite. The proof is a variation of Euclid's argument that there are infinitely many primes. Namely, if you have a finite set of elements $a,b,... \in k$, the polynomial $(X-a)(X-b)...+1 $ has a zero $\in k$ , necessarily different from $a,b,...$ $\endgroup$ Feb 16, 2011 at 8:42
  • $\begingroup$ Thanks for pointing this out! I've changed the assumptions accordingly. $\endgroup$ Feb 16, 2011 at 16:51

3 Answers 3


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.


I will use the notation that if $S\not\subseteq\{x\}$ then $\overline{\phi}(S):=\overline{\phi(S\setminus \{x\})}$.

Assume that $X\neq \{x\}$ and it is not a union of lines through $x$. See below for an explanation that this is equivalent to $X\cap H\not\subseteq \{x\}$. In this case the statement is true.

A general hyperplane through $x$ is the same as the closure of the preimage via the projection map of a general hyperplane $L\subset \mathbb P^{n-1}$. Then we have that $H=\overline{\phi^{-1}L}$ and since $L$ is general, its intersection with $\phi(X\setminus\{x\})$ is dense in its intersection with $\overline{\phi(X\setminus\{x\})}$ and then ${\overline{\phi}(X\cap H)}={\overline{\phi}(X)\cap L}$.

Remark: if $X\cap H\subseteq \{x\}$, then the statement cannot be true as stated. In that case $\overline{\phi((X\cap H)\setminus \{x\})}=\emptyset$ while $\overline{\phi(H\setminus \{x\})}$ is a hyperplane in $\mathbb P^{n-1}$ and $\overline{\phi(X\setminus \{x\})}$ is a nonempty closed subset so their intersection is non-empty.

However, this condition can only hold if $X=\{x\}$ or it is a union of lines: Assume that $X\neq\{x\}$ and replace $X$ with any of its irreducible components. In order for $X\cap H\subseteq \{x\}$, $X$ has to be a curve. In that case $x$ has to be a point of multiplicity $\deg X$ on $X$ and for $\deg X>1$ the above condition could only hold if $H$ contains the tangent space of $X$ at $x$. That would mean that $H$ is not general, so the only possibility allowed by the assumptions is $\deg X=1$, so $X$ is a line.

Notice that Laurent's complaints are covered by the above assumption and the fact that $H$ is general. In particular, quim's extra argument is not necessary.

Notice also that the generality of $H$ is needed so its image would intersect the image of $X$ in a set that's dense in the intersection of $H$ and the closure of the image of $X$.


You have to explain what you mean by these images since $\phi$ is not everywhere defined. For example, what if $X=\{x\}$?

A slightly less stupid example: if $X$ is a line through $x$, not contained in $H$, what is $\phi(H\cap X)$?

And another one: $n=2$ and $X$ is a cubic having a cusp at $x$ with tangent $H$.

My guess is that by $\phi(Z)$ you mean $\phi(Z\setminus\{x\})$, viewing $\phi$ as a morphism from $\mathbb{P}^n \setminus\{x\}$. In this case the answer is probably yes if no component of $X$ is a curve through $x$.

  • $\begingroup$ This is exactly what I meant; I've changed the statement of the problem accordingly. $\endgroup$ Feb 16, 2011 at 16:57
  • $\begingroup$ Laurent, it seems to me that $X=\{x\}$ and $X$ a union of lines through $x$ are the only counter-examples if $H$ is general. $\endgroup$ Feb 17, 2011 at 11:02
  • $\begingroup$ @Sándor: Right, in fact I had skipped "general" on first reading. $\endgroup$ Feb 18, 2011 at 8:32

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