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My problem: Let $X$ be an affine variety and $U$ an open subset of $X$. Then $dim U = dim X$.

My attempt: If I take a chain of irreducible and closed sets $Z_1\subsetneq... \subsetneq Z_n$ in $U$, then I get a chain of irreducible and closed sets $\bar Z_1\subsetneq ... \subsetneq \bar Z_n$.

I'd like to prove that if the chain in $U$ is a non-refinable chain of irreducible and closed sets, then the chain of its clausures is a non-refinable chain of irreducible and closed sets on $X$. Then, I'd like to build some chain of prime ideal using the fact $I(Z_0)$ maximal and finding that $n$ is the hight.

However. I don't know how can I avance and prove this...

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1 Answers1

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To show the equality of dimensions, it suffices to show that $Z \mapsto \overline{Z}$ and $Z \mapsto Z \cap U$ gives an order preserving bijection between the irreducible closed subsets of $U$ and the irreducible closed subsets of $X$ which intersect $U$. This follows from the definition of irreducibility.

Daniel
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  • A proof of this statment can be found here: https://math.stackexchange.com/questions/1021297/why-are-the-irreducible-components-of-a-subspace-of-noetherian-space-just-the-in – 3nondatur Jun 22 '23 at 11:31