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 $(I+h)\cap A=\langle x\_2^2,x\_1-x\_3\rangle$.



General hyperplane sections and projection from a point
**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.


