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This comes from the study of Krull's Intersection Theorem, and deriving a geometric meaning.

Let $I \subset R$ be an ideal of a commutative ring (we shall see the case when $R$ is Noetherian). Consider $\bigcap_{n>0} I^n = \underset{\longrightarrow}{\textrm{lim}} \, I^n$ as a colimit. I would like to apply the $\textrm{Spec} \_ = \mathcal{V}(\_)$ functor, and would like to a statement (the second equality) as $$\mathcal{V}(\bigcap_{n>0} I^n) = \mathcal{V}(\underset{\longrightarrow}{\textrm{lim}} \, I^n) = \underset{\longrightarrow}{\textrm{lim}} \mathcal{V}(I^n)$$

However, after some searches on math stackexchange for instance On limits, schemes and Spec functor , it seems the $\textrm{Spec} \_$ functor does not preserve colimits.

For instance, my goal is to geometrically interpret the Krull's Intersection Theorem by applying the $\textrm{Spec} \_$ to the equation $\bigcap_{n>0} I^n = 0$ when the ring is Noetherian and is local or a domain. The hope is get $\underset{\longrightarrow}{\textrm{lim}} \mathcal{V}(I^n) = \mathcal{V}(0) = \mathbb{A}^n$, which I am having doubts if it is even true.

Any comment on the above is appreciated, for instance is it correct that the infinite intersection of ideals a colimit?

Edit:

Consider the diagram

enter image description here

The morphism $I^j \to I^i$ are just the inclusion. Then I get $\bigcap_{n>0} I^n = \underset{\longrightarrow}{\textrm{lim}} \, I^n$ as a direct limit (colimit). Is this somehow wrong? Please explain.

*** Need to consider inverse limit of the diagram instead of the direct limit which yields $I$ ***

metalder9
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    Intersection is a limit, not a colimit. Try making this precise. – Thorgott Mar 01 '24 at 13:06
  • @Thorgott Please see the edit. – metalder9 Mar 01 '24 at 17:42
  • The direct limit of the diagram pictured is just $I$. This is a general fact: if a category $J$ has a terminal object $j_1$, then any diagram $F : J \to D$ admits a colimit, which is simply $F(j_1)$. – diracdeltafunk Mar 01 '24 at 17:45
  • Dear @diracdeltafunk I am not sure I understand, so here the terminal object $j_1$ in the diag is $I$, then $F: J \to D$ is a functor from category $J$ to $D$? What is $F(j_1)$? – metalder9 Mar 01 '24 at 17:58
  • Morphisms out of the diagram you write are the same as morphisms out of $I$, because the maps in the diagram are just inclusions, so the map on $I^2$ is determined as a restriction of the map on $I$, etc.. – Thorgott Mar 01 '24 at 18:00
  • A diagram is a functor, much like how a sequence of real numbers is a function $\mathbb{N} \to \mathbb{R}$. The domain of the functor is called the "indexing category" and the codomain of the functor is where the diagram lives. In this case, $F(j_1)$ is the object $I$. – diracdeltafunk Mar 01 '24 at 18:01
  • @Thorgott So what you are saying is that what I wrote is basically just a system $... \to I \to I \to I$? What would be the correct morphisms I should be looking at to get $\bigcap_{n>0} I^n$? – metalder9 Mar 01 '24 at 18:14
  • No, the system you wrote is the system you wrote, involving all of the powers of $I$. You need to take the inverse limit of this system to get the intersection, not the direct limit. – diracdeltafunk Mar 01 '24 at 21:59
  • @diracdeltafunk Ok I understand what I am confused about now: since I am new to limits and colimits, I thought that for arrows going to the left like this $... \to N_2 \to N_1 \to N_0$ that one has to take colimit (the case of stalks of a sheaf), and for arrows going to the right $M_0 \to M_1 \to M_2 \to ...$, one takes limit (like the completion of a module wrt an ideal). See my edit2. – metalder9 Mar 01 '24 at 22:50
  • You are saying to take the inverse limit of this $... \to N_2 \to N_1 \to N_0$? – metalder9 Mar 01 '24 at 22:55
  • You don‘t have to take colimits or limits depending on the shape of a diagram but you got it the wrong way around: completions are e.g. limits of certain diagrams of the form $\cdots \to N_1 \to N_0$. – Qi Zhu Mar 01 '24 at 23:19
  • @QiZhu I see, well I have to review limits/colimits, I am new to them so. Wiki says for completion of a module as it being an inverse limit of quotient modules https://en.wikipedia.org/wiki/I-adic_topology Are the maps just natural projections in this case $M/I^nM \to M/I^{m}M$ where $n>m$? – metalder9 Mar 02 '24 at 08:31
  • @QiZhu Yes you are right, for completion the diagram is of the form $... \to M/I^2M \to M/IM$ thanks! – metalder9 Mar 02 '24 at 08:43
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    I feel as though the geometric intuition for the Krull intersection theorem is simpler than you think. To me the main point of the Krull intersection theorem is the claim that $\bigcap I^n=0$. If one thinks of the ring of functions on $\mathrm{Spec}(R)$ as being $R$, then saying that $f$ is in $I^n$ means that $f$ vanishes along $V(I)=V(I^n)$ but more so that it and its derivatives up to degree $n$ vanish along $V(I)$. So, the Krull intersection theorem says that if $f$ and all of its derivatives vanishes along $V(I)$ then $f$ is zero. This is an analogue of the theorem from complex analysis – Alex Youcis Mar 02 '24 at 18:41
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    that a holomorphic function is determined by its Laurent expansion at any point. – Alex Youcis Mar 02 '24 at 18:42

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