Empty set in projective space has negative dimension?

Question

I've seen some notes online that show that a closed set in the projective space is nonempty by showing that it has dimension $ \geq 0 $. This seems to rely on the convention that the empty set has dimension $ -1 $. Is it just that, a convention? While the convention is natural in light of $ \dim{S(Y)}=\dim{Y}+1 $, I feel that this argument isn't satisfying because it relies on a convention. Am I missing something?

Answer 1

It's hard to ease your dissatisfaction without seeing the argument. But a typical statement that is used in this way is the following:

If $X$ is a closed subset of $\mathbb P^n$ of positive dimension and $F$ is a homogeneous polynomial, then $\dim\left(X\cap V_+(F)\right)\ge \dim(X)-1$.

By definition of dimension, this is equivalent to: if $X$ is a closed subset of projective space such that there exists a non-empty chain $X_0\subsetneq X_1\subsetneq\cdots\subsetneq X_d$ of irreducible closed subsets of $X$, then there exists a chain $Z_0\subsetneq Z_1\subsetneq\cdots\subsetneq Z_{d'}$ in $X\cap V_+(F)$ of irreducible closed subsets of length $d'\ge d-1$. Note that the translation is in terms of the existence of some subsets, which in particular will imply the existence of some points.

For example, if we want to show that a closed subset of projective space is non-empty, then we might just show that it is defined by fewer than $n$ homogeneous forms, and apply the above result. But this can be phrased either in terms of dimension, or in terms of existence of a chain of irreducible closed subsets. The phrasing doesn't matter, due to our convention that $\dim(\varnothing) = -\infty$.

Answer 2

I think that what these proofs really rely on is that all the (relevant) things whose dimensions are $\ge 0$ have the property that they're nonempty, "things" here being something like "linear subspaces of an affine space, projectivized." That fact is not so much a convention as a computation; because the empty set doesn't have the property of being nonempty, it's convenient to assign it a dimension that's different from those others, so that "nonempty" and "nonnegative dimension" become synonyms. Why define its dimension at all? So that every proof doesn't have to have a "case" statement every time it wants to mention dimension.

Answer 3

Most texts define the dimension only of nonempty sets, but implicitly use a slightly broader definition. The dimension of a topological space $X$ is the length of the longest chain of irreducible or empty closed subsets $C_{-1} \subsetneq C_0 \subsetneq \cdots \subsetneq C_n$.

Equivalently, we could use just irreducible sets, but take into account the empty chain.

This convention is similar to the convention that $\sum_{k=0}^{-1} a_k = 0$, as it is regarded to be a sum with zero terms.

Some of this complexity arises from the fact that we'd rather not regard $\emptyset$ as an irreducible set, just as we'd rather not regard $R$ as a prime ideal of $R$, or $0$ as an integral domain (or $1$ as a prime). But this is just superficial complexity.