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Usually in algebraic geometry,we define the dimension of a topological space as follows:

$\dim(X)=\sup\{ m:\text{ there exists a descending chain of irreducible closed sets of length } m\}$

and in a commutative ring $R$ ,we define the Krull dimension as follows:

$\dim(R)=\sup\{m': \text{ there exists an ascending chain of closed sets of length } m'\}$

I know that there exists a correspondence between irreducible closed sets of $\mathbb A^n_K$ and the prime ideals of $K[X_1,X_2,...,X_n]$ for an algebraically closed field $K$, so via the correspondence we can get one from the other. But what motivated one of the definitions is not clear to me. Can someone help me with this?

KCd
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2 Answers2

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We want to say $K[x_1,\ldots,x_n]$ for a field $K$ has dimension $n$. It has the chain of $n+1$ distinct prime ideals $$ \{0\} \subset (x_1) \subset (x_1,x_2) \subset \cdots \subset (x_1,\ldots,x_n). $$ and you can't fit any prime ideals inside any step of this chain. This suggests in a commutative ring $R$ looking at chains of distinct prime ideals $$ \mathfrak p_0 \subset \mathfrak p_1 \subset \cdots \subset \mathfrak p_n $$ and calling the largest such $n$ the dimension of $R$. That is the Krull dimension of $R$.

KCd
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Maybe another take at the question.

Because of the existence of "wild automorphisms" of the complex numbers $ℂ$, naturally it would not be easy to recover the topology of $^2_ℂ$ from the polynomial ring $ℂ[x, y]$ (or the Zariski topology on $^2_ℂ$).

Nevertheless, we can use the property that (roughly speaking) "in $ℂ[x_1, …, x_n]$, you generally need $n$ equations to uniquely determine a set of solution" in order to determine the dimension.

So, consider $n$ equations like $f_i(x_1,..., x_n) = 0$ for $i ∈ \{1, …, n\}$. Then, generally speaking you would expect $V((f_1, …, f_i))$ to have dimension $n-i$, thus the chain of vanishing set strictly decreases each time a polynomial is added.

Once the vanishing set is reduced to a point, there is no more nonempty closed set strictly contained within it.

user202729
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  • Can you elaborate on your second paragraph? What do you mean by "wild automorphisms"? – tolUene Aug 19 '23 at 02:08
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    @tolUene Apart from the identity and complex conjugation, there exist (assume axiom of choice) field automorphisms of the complex numbers that are not continuous https://math.stackexchange.com/questions/412010/wild-automorphisms-of-the-complex-numbers – user202729 Aug 19 '23 at 02:10