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Let $X \subset \mathbb{P}^n$ be a nonempty projective variety. Show that the dimension of the cone $C(X):=\{0\} \cup \{(x_0,...,x_n)\in \mathbb{A}^{n+1}:(x_0:...:x_n)\in X\}$ is dim$X+1$.

I know how to prove dim$C(X) \geq$ dim$X+1$: Let $\varnothing \neq X_0 \subsetneq ... \subsetneq X_n\subset X$ be a chain of irreducible closed subsets in $X$, then $\{0\} \subsetneq C(X_0) \subsetneq ... \subsetneq C(X_n) \subset C(X)$ is a chain of irreducible closed subsets in $C(X)$.

However I cannot prove the other side of the inequality, if I let $\varnothing \neq Y_0 \subsetneq ... \subsetneq Y_m\subset C(X)$ be a chain of irreducible closed subsets in $C(X)$, I do not know how to generate a chain of irreducible closed subsets in $X$. If the $Y_i$ are of form $V(S_i)$, where $S_i$ is a homogenous ideal, then I can take projectivization $\mathbb{P} (Y_i):=\{(x_0:...:x_n)\in \mathbb{P}^{n+}:0 \neq (x_0,...,x_n)\in X\}$ and obtain my conclusion easily. Is there a way to assert this? This is equivalent to the claim that every irreducible component of a cone is also a cone, which seems to have some geometric basis.

I have seen some answers on related questions on this site using the tool of fiber dimension and transcendence degree of the coordinate ring, however I do not want an answer with these advanced techniques, as the notes I am reading does not assume the reader has these knowledge; it would be very nice to see a proof using basic constructions such as cone and projectivization defined above.

Any help is appreciated!

Chi Cheuk Tsang
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    Another way may be: $\mathcal{I}(C(X))=\mathcal{I}(X)$. So,$X$ projective variety implies $C(X)$ affine variety. But we know $\dim X+1=\dim S(X)=n+1- ht(\mathcal{I}(X))=n+1-\mathcal{I}(C(X))=\dim C(X)$. ($S(X)$ is the homogeneous coordinate ring of $X$.) –  Nov 23 '16 at 03:53

1 Answers1

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Define the codimension of a subvariety of $\mathbf P^n$ or $\mathbf A^n$ à la Krull, and use the elementary fact from commutative algebra that the sum of the dimension and the codimension of a subvariety is the dimension of the ambient variety. Then, a similar argument as yours shows that $$ \mathrm{codim}(C(X))\geq\mathrm{codim}(X)=n-\dim X. $$ Since $$ \mathrm{dim}(C(X))+\mathrm{codim}(C(X))=\mathrm{dim}(\mathbf A^{n+1})=n+1, $$ one deduces that $$ \mathrm{dim}(C(X))=n+1-\mathrm{codim}(C(X))\leq n+1-n+\dim(X)=\dim(X)+1. $$ Together with your statement that $\dim(C(X))\geq\dim(X)+1$, one has indeed $$ \dim C(X)=\dim (X)+1. $$

Johannes Huisman
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    I believe the fact that sum of dimension and codimension subvariety equals dimension of ambient variety only works when both the variety and the subvariety is irreducible, but I can always use an irreducible decomposition. Thanks for your answer! – Chi Cheuk Tsang May 19 '16 at 01:13
  • @Tsang You are right: one must suppose the subvariety irreducible, or at least equidimensional. – Georges Elencwajg May 19 '16 at 07:16