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I'm partially studying the book Algebraic geometry of Daniel Perrin, and I have a doubt on proposition 1.11 of chapter 4.

For some reasons I'm reading this section without reading the previous one about sheaves and varieties so I'm trying to prove the results changing "algebraic variety" by "algebraic affine set".

I've managed to do it until corolary 1.10 but on proposition 1.11 that states that the dimension of an irreducible algebraic variety is the same as any open subset of it I'm stuck.

So, my question is if it is possible to prove only using theory on algebraic affine sets and the previous results on chapter 4 of the book (but only for algebraic affine sets) that given $A$ an algebraic affine set and $U$ an open, non-empty, subset of $A$ then the dimension of both sets is the same.

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Eparoh
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    what definition of dimension are you working with? If it's length of a maximal chain of irreducible closed subspaces, I think you may be able to get it just using point set topology and not any algebraic geometry. – hunter Apr 21 '21 at 16:21
  • @hunter Yes, that is the definition I'm using to define dimension. – Eparoh Apr 21 '21 at 16:22
  • perfect -- in that case, see here: https://math.stackexchange.com/questions/721888/if-a-chain-of-distinct-irreducible-closed-subsets-of-a-quasi-affine-variety-y?rq=1 – hunter Apr 21 '21 at 16:23
  • @hunter I have a doubt about the question you linked. In the answer it is proven that we have a chain of irreducible closed subspaces of lenght $n$ that cannot be extended but I don't see how does that imply that it is a maximal chain. Is it true that you can always from a non maximal chain obtain a maximal one? Because I don't really see how. – Eparoh Apr 21 '21 at 16:55
  • my apologies -- your skepticism is warranted! It is true that for an algebraic variety any two maximal chains have the same length, but this is NOT a fact about point set topology, as the two counterexamples listed here show: https://math.stackexchange.com/questions/1707151/dimension-of-a-noetherian-topological-spaces – hunter Apr 21 '21 at 17:48
  • however, it is true for varieties; references here: https://math.stackexchange.com/questions/112218/maximal-chains-have-the-same-length – hunter Apr 21 '21 at 17:49

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Question: "So, my question is if it is possible to prove only using theory on algebraic affine sets and the previous results on chapter 4 of the book (but only for algebraic affine sets) that given A an algebraic affine set and U an open, non-empty, subset of A then the dimension of both sets is the same."

Answer: I do not have access to this book, but the following result may be found in Matsumuras book "Commutative ring theory":

Thm. Let $k$ be a field and let $B$ be a finitely generated $k$-algebra which is an integral domain. It follows $krdim(B)=trdeg_k(K(B)),$

where $krdim(B)$ is the Krull dimension of $B$ and $trdeg_k(K(B))$ is the transcendence degree of the quotient field $K(B)$ over $k$. If $X:=Spec(B)$ and $U:=Spec(A)\subseteq X$ is an open affine subset it follows we may define $K(B)\cong \mathcal{O}_{X,q}$ where $q\in X$ is the generic point. It follows $q\in U$ and hence we get

$$K(B) \cong \mathcal{O}_{X,q}\cong \mathcal{O}_{U,q} \cong K(A)$$

since the local ring does not change when passing to $U$, hence

$$krdim(A)=trdeg_k(K(A))=trdeg_k(K(B))=krdim(B).$$

Hence $A$ and $B$ have the same krull dimension. This proves the claim using field theory for any affine open subscheme. For a general $U \subseteq X$ it follows $dim(U)=dim(\overline{U })=X$ by Hartshorne Prop. I.1.10. The proof uses the above theorem on fields and a result on catenary rings.

hm2020
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