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!

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    $\begingroup$ 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 $X$.) $\endgroup$
    – user276115
    Commented Nov 23, 2016 at 3:53

2 Answers 2


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 $$ \dim C(X)=\dim (X)+1. $$

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    $\begingroup$ 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! $\endgroup$ Commented May 19, 2016 at 1:13
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    $\begingroup$ @Tsang You are right: one must suppose the subvariety irreducible, or at least equidimensional. $\endgroup$ Commented May 19, 2016 at 7:16
  • $\begingroup$ this is a very neat proof that doesn't really use much dimension theory, only $\mathrm{ht}\ \mathfrak p + \dim A/\mathfrak p = \dim A$. $\endgroup$ Commented Mar 16 at 15:35

$\DeclareMathOperator{\dim}{dim}\DeclareMathOperator{\htt}{height}$Here's a way of doing this without codimension (and following Hartshorne, where this is Chapter I Ex. 2.10.).
I assume that we have shown that $Y$ and $C(Y)$ have the same ideal $I$. (In particular, $Y$ is irreducible iff $C(Y)$ is so.)

Special case. Assume first that $Y$ is irreducible.

Then, $I$ is prime and we have $\dim(Y) = n - \htt(I)$. (This follows from Ex. 2.6.)
Similarly, $\dim(C(Y)) = (n + 1) - \htt(I)$, since $C(Y)$ sits inside the $(n + 1)$-affine space.

General case. Note that $Y$ is a Noetherian space and we can write $$Y = \bigcup_{i = 1}^{t} Y_i,$$ where the $Y_i$ are the irreducible components of $Y$.

Then, we have $$C(Y) = \bigcup_{i = 1}^{t} C(Y_i).$$ The above can be checked set-theoretically since unions commute with pre-images.
As noted earlier, $C(Y_i)$ continue to be irreducible. Moreover, since no $Y_i$ was contained in the union of the other $Y_j$, this continues to be true for the cones as well. (Again a set-theoretic argument.)
In other words, $C(Y_i)$ are the irreducible components of $C(Y)$.

Now, note that $\dim(Y) = \max \dim(Y_i)$ and similarly for $\dim C(Y)$.
(Indeed, any chain of irreducible sets is actually contained in an irreducible component; see here, for example.)

Thus, if we can show $\dim(Y_i) + 1 = \dim(C(Y_i))$, then we are done. But this follows from the special case.


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