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!
Best Answer
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 $$ \dim C(X)=\dim (X)+1. $$