Let $A:=V(F_1,...,F_k)\subset\mathbb{P}^n$ with $F_j\in k[X_0,...,X_n]$, a projective algebraic set. Let $C(A)\subset \mathbb{A}^{n+1}$ the affine cone over $X$. Show that $\dim A=\dim B$, where $B$ is the cone $C(A)$ making one of the variable $X_i=1$, when $X_i$ appears on at least one of the $F_j$.

I'm having troubles understanding the concept that relates the "new cone" with the original projective algebraic set, because I thought the cone has the same dimension that the original projective algebraic set. Any suggestion is appreciated.


1 Answer 1


The cone over a projective variety $A$ of dimension $n$ will have dimension $n+1$. One way to convince yourself of this is as follows: the natural map $\pi \colon C(A) \setminus \{0\} \to A$ is surjective, and for every $p \in A$, $\pi^{-1}(\{p\})$ is a line (missing a point) in $\mathbb A^{n+1}$ which is a variety of dimension $1$. Intuitively, you have $1$ more "degree of freedom" in $C(A)$ than in $A$, so the dimension of $C(X)$ should be $1$ greater than that of $A$.

In fact this can be made rigorous: you can then use results on the dimension of fibres of a morphism to show that $\dim C(A) = \dim A +1$. Alternately, you can do this algebraically by computing the Krull dimension of the coordinate ring of $C(A)$, and the Krull dimension of the coordinate ring of an affine piece of $A$.

You can think about "setting a variable equal to one" in several ways. On the one hand, this corresponds to intersecting $C(A)$ with a hyperplane in $\mathbb A^{n+1}$ (the hyperplane $X_i - 1 = 0$). Since codimensions add under suitably nice intersections, $\{X_i = 1 \} \cap C(A)$ will have dimension $(n+1)-1=n,$ equal to the dimension of $A$.

But you can also see this by noting that $\{X_i =1 \} \cap C(A)$ is naturally identified with the open affine set $A \cap U_i$ of $A$, where $U_i \subset \mathbb P^{n}_{X_0,\dots,X_n}$ is the open affine chart $\{X_i \neq 0\}.$


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