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Let $X=V(F_1,...,F_k)\subset \mathbb{P}^n$with $F_i\in k[X_0,...,X_n]$ an projective algebraic set. Let $C(X)\in \mathbb{A}^{n+1}$ the affine cone over $X$, that is

$C(X)=\theta^{-1}(X)\cup \{(0,...,0)\}$

where $\theta:\mathbb{A}^{n+1}\setminus \{(0,...,0)\}\rightarrow \mathbb{P}^n$ is defined by $P\to \bar{P}$

I want to show that every irreducible component of the cone contains the vertex $(0,...,0)$. Any suggestion is appreciated.

Irene
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1 Answers1

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This corresponds to the fact that any ideal generated by homogeneous polynomials is the intersection of homogeneous ideals. That is, if you know this (or see how to prove it), your claim will follow (because if every component is defined by homogeneous ideals, clearly $(0,\ldots,0)$ lies in every component).

Here's another way. Note that $C(X)$ comes with a natural $k^\ast$-action. For any $P \in C(X)$, the point $tP$ also lies in $C(X)$ for $t \neq 0$. Since $C(X)$ is closed, it follows that $0 \in C(X)$.

Now write $C(X) = \cup_{i=1}^n X_i$ as a union of irreducible components. I claim that there is a subgroup $L^\ast \subset k^\ast$ such that $L^\ast$ act on the $X_i$. Now since $k^\ast$ act on $C(X)$, it must permute the $X_i$. But since there are only $n$ components, all the elements $c^n$ must fix $X_i$. Hence the subgroup $\{ c^n \mid c \in k ^\ast \}$ act on each $X_i$ individually.

Each of the irreducible components are closed. Consider $P \in X_0$. Then $\{ tP \mid t \in k \}$ is an irreducible closed set contained in $X$ (it is a line), that intersects $X_0$. Now the $k^\ast$-action restrict to an action on $X_0$, so the whole line must be contained in $X_0$ since $X_0$ is closed.

Remark Note that if $k^\ast$ is a divisible group (for example if $k=\mathbb C^\ast$), then we can take $L=\mathbb C$.

Fredrik Meyer
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  • "But X_0 is an irreducible component, so the whole line must line in X_0" why ? – user44636 Nov 21 '16 at 15:44
  • @user44636 I notice I wasnt very clear when I wrote this. Because of the $k^\ast$-action, we have automatically that ${ tP \mid t \in k^\ast }$ is contained in $X_0$. But since $X_0$ is closed, it follows that it contains $(0,0,0...0)$. – Fredrik Meyer Nov 21 '16 at 21:14
  • Dear Fredrik, it's still not clear for me why line $tP$ should be contained in the $X_0$? Why the line couldn't just intersect $X_0$ at one point? – user44636 Nov 22 '16 at 10:31
  • @user44636 Thank you for the comment again. I have added more details to my answer (in fact, it is probably a different answer now). Do you think it makes more sense? – Fredrik Meyer Nov 22 '16 at 11:33
  • Fredrik, maybe it's annoying, but I'm still confused by the statement: "Now since $k^*$ act on $C(X)$, it must permute the $X_i$", it should be indeed the case, but could you clarify this, please? – user44636 Nov 22 '16 at 19:58
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    @user44636 No, it's not annoying at all, these questions got me thinking. About your last question: multiplying by $c \in k^\ast$ leaves $C(X)$ invariant, and hence the decomposition into irreducible components is the same. Then, since $c \cdot C(X) = \cup c \cdot X_i = C(X)$, and the decomposition is unique, we must have $c \cdot X_i = X_j$ for some $j$. – Fredrik Meyer Nov 23 '16 at 09:44
  • Dear, Fredrik ! Now it seems that everything clear to me. Thank you! – user44636 Nov 23 '16 at 10:58