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.
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.