Definition of dimension of a projective variety

Question

The title is quite self-explanatory: I'm an undergraduate maths student who hasn't attended (yet) a course in Algebraic Geometry (but I did a course on commutative algebra), however I have to write a short essay on projective varieties for a -difficult to explain- exam.

I'm struggling with the definition of dimension of a projective variety $X \subset \mathbb{P}^n(k)$: I don't have any recommended textbook and all the material I found (which obviously doesn't involve more advanced tools like schemes) is quite confusing.

I'd be very glad if someone could spend a minute writing the correct definition (or the definitions, if there are many equivalent), and maybe where to find some material about this subject.

I apologize for this question, I know it's a bit vague, but I hope someone will be so kind as to help me. Thanks in advance.

Answer 1

I assume you know what is the dimension of an affine variety. You don't need to know more : if $X$ is irreducible and projective, define $\dim X = \dim U$ where $U \subset X$ is an affine open. If $X$ is projective and not irreducible, define $\dim X = \sup_i \dim X_i$ where $X_i$ are the irreducible components.

Without the langage of schemes, nice references are the notes by Gathman, or the book by Harris, "A first course in algebraic geometry".

Answer 2

If you want to do calculations, then Nicolas Hemelsoet's answer works perfectly well. I think though that maybe the "right" definition is simply to define the Krull dimension of a topological space. Namely, the the Krull dimension of a topological space $X$ is the supremum of the length of chains $X_0\subset\dots\subset X_n$ of irreducible closed subsets of $X$. The chain that I wrote down has length $n$. You can prove now that if you have a nonempty open subset $U$ in an irreducible space $X$, then $\dim U=\dim X$. From this, you recover Nicolas Hemelsoet's definition.

Answer 3

Here's another useful result in calculating the dimension of a projective scheme. Let $A_\bullet$ be a homogeneous (graded) algebra over a field $k$. Of course, we can also forget the grading and view $A$ as an ordinary algebra over the same field $k$. Then $$\dim \operatorname{Proj} A_\bullet = \dim \operatorname{Spec} A - 1.$$ For proof, see Liu's Algebraic Geometry and Arithmetic Curves Corollary 2.5.21.

This means that we can easily calculate the dimension of projective varieties if we know how to calculate the dimension of affine varieties; we just subtract one! This lines up with the geometric intuition/most basic example that $$\dim \mathbb{A}_k^n = \dim \operatorname{Spec} k[x_1, \dotsm, x_n] =n, \; \text{and} \; \dim \mathbb{P}_k^{n-1} = \dim \operatorname{Proj} k[x_1, \dotsm, x_n] = n-1.$$

For example, consider $A_\bullet = (k[x,y,z]/(x^3-xz^2-y^2z))$ with the usual grading. The projective variety $$V=\operatorname{Proj}(k[x,y,z]/(x^3-xz^2-y^2z))$$ has dimension $$\dim V=\dim(\operatorname{Spec}(k[x,y,z]/(x^3-xz^2-y^2z))-1=3-1-1=1.$$ This dimension calculation uses:

1. $\dim(A)=\dim(\operatorname{Spec}A)$ where the dimension on the left is the Krull dimension, and the dimension on the right is the dimension as a topological space.
2. For any domain $A$ which is a finitely generated algebra over a field $k$ and ideal $I\subset A$, we have $\dim(A/I)=\dim(A)-\text{ht}(I).$
3. Krull's Hauptidealsatz: For a general commutative ring $A$ and a nonzero, noninvertible element $a\in A$, the principal ideal $I=(a)$ of $A$ has height at most 1. Moreover, $\text{ht}(I)=1$ iff $a$ is a non-zero divisor.
4. The polynomial $x^3-xz^2-y^z$ is clearly not a zero divisor of $k[x,y,z]$.

The variety $V$ is a projective nodal cubic, and this dimension calculation shows that $V$ is a curve.