# If $Y$ is a quasi-affine variety, then dim$Y$ = dim$\overline{Y}$

Reading through the proof of proposition 1.10 in Hartshorne's Algebraic Geometry I found some of it to be unnecessary. Is the following proof correct or can you point out my flawed logic?

Let $Z_0 \subset ... \subset Z_n$ be a sequence of distinct closed irreducible subsets of $Y$. Notice that $\overline{Z_0} \subset ... \subset \overline{Z_n}$ is a sequence of closed irreducible subsets of $\overline{Y}$. This is because we know that if $Z_i$ is an irreducible subset of $Y$, then $\overline{Z_i}$ is irreducible in $Y$ and thus it is also irreducible in $\overline{Y}$. Thus we have that dim$Y$ $\leq$ dim$\overline{Y}$. Now dim$\overline{Y}$ is finite as $Y$ is a non-empty open subset of an irreducible space, so that it is irreducible and dense, yielding $\overline{Y}$ is an affine variety. Thus we can choose a maximal chain of distinct closed irreducible subsets of $Y$, $Z_0 \subset ... \subset Z_n$, with dim$Y=n$. Now $\overline{Z_0} \subset ... \subset \overline{Z_n}$ is a maximal chain of closed irreducible subsets of $\overline{Y}$, this is through a contradiction argument that uses the fact that a non-empty open subset of an irreducible space is irreducible and dense. I omit it here as my proof goes into too much detail.

Now from here Hartshorne goes to prove the dimension is $n$ by converting these to prime ideals and finding a maximum chain of those. I understand his proof and it is correct, but is it not correct to say that $\overline{Y}$ is an affine variety whose dimension is given by the supremum of all integers $n$ such that there exists a chain $A_0 \subset ... \subset A_n$ of distinct irreducible closed subsets of $\overline{Y}$? We proved that $\overline{Z_0} \subset ... \subset \overline{Z_n}$ was maximal so that by definition dim$\overline{Y}$ = $n$ = dim$Y$.

I am having trouble seeing a flaw in my proof and do not understand why it is necessary to go to a chain of prime ideals in order to argue the dimension of $\overline{Y}$ is in fact $n$. Any insight would be most appreciated.

• To emphasize: you can't insert more sets into the chain $\overline{Z}_i$, but how do you know that there isn't some longer chain of closed irreducible sets in $\overline{Y}$? You might say "ah, well, take any chain $W_j$ in $\overline{Y}$ and intersect that with $Y$, but some of those sets might miss $Y$ completely. – Hoot May 22 '15 at 16:21
• Ah that was my problem, I never thought that they may miss $Y$ completely. Thank you very much! – Michael N May 22 '15 at 16:55
• In the text of Hartshorne, the chain $\overline{Z}_0\subset\overline{Z}_1\subset\cdots\subset\overline{Z}_n$ is maximal between the chains that contains $\overline{Z}_0$ as first point, it is sufficient for calculate the height of maximal ideal that corresponds to $\overline{Z}_0$. – Elvis Torres Pérez Dec 17 '19 at 1:05

Suppose, by absurd, that the $$\overline{Z_0}\subset\ldots\subset\overline{Z_n}$$ not be maximal. So, there is a variety $$W$$ such that $$\overline{Z_i}\subsetneq W\subsetneq\overline{Z_{i+1}}$$. Since that every $$Z_j$$ is closed in $$Y$$ we have $$Z_i\subset Y\cap W\subset Z_{i+1}$$. Since that $$n=\textrm{dim }Y$$ we have $$Z_{i+1}=Y\cap W$$ or $$Z_i=Y\cap W$$. In the first case we have $$Z_{i+1}\subset W$$ $$\Rightarrow$$ $$\overline{Z_{i+1}}=W$$ by passing through closure, contradiction. Consider now the case $$Z_i=Y\cap W$$. Since that $$Y$$ is quasi-affine there is an open $$U$$ in tha affine space such that $$Y=\overline{Y}\cap U$$ $$\Rightarrow$$ $$Z_i=W\cap U$$ is a open of $$W$$ so dense in $$W$$ $$\Rightarrow$$ $$\overline{Z_i}\cap W=W$$ $$\Rightarrow$$ $$\overline{Z_i}=W$$, contradiction. Since that $$A(\overline{Y})$$ is universaly catenary (H. Matsumura, Commutative Algebra) $$n=\textrm{ht }\mathfrak{m}=\textrm{dim }\overline{Y}$$ $$\Rightarrow$$ $$n=\textrm{dim }\overline{Y}$$ since that $$A(\overline{Y})/\mathfrak{m}=k$$.
Now the fact that for an integral affine $k$-Algebra $A$ all maximal chains have the same length (that is there are no chains which are maximal but have length $n' < \dim A$) is essentially true because of Proposition 1.8Ab which Hartshorne invokes in the proof. If you use this fact, you are obliged to give a proof of it as a separate lemma.