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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.

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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".

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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.

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  • $\begingroup$ Nice answer ! (+1) $\endgroup$ – Nicolas Hemelsoet Jun 17 '18 at 21:26
  • $\begingroup$ I think the assertion in the penultimate sentence is untrue. Consider the Sierpinski space: it's irreducible, one-dimensional, and the generic point is a dense open zero dimensional subspace. $\endgroup$ – Arrow Jan 29 at 1:43

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