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This should be a very simple question but I can't seem to figure it out. If we have a morphism of schemes $f: X \rightarrow Y$, I am trying to understand what fiber is above some $\operatorname{spec} \kappa(f(p))$ for $p$ a point in $X$. Let's just work with affine schemes for now. If we have an $A$-algebra $\phi: A \rightarrow B$ which induces a morphism of schemes $f: \operatorname{spec}B \rightarrow \operatorname{spec}A$. Then firstly, if $p \in \operatorname{spec}B$ is a point, then what does the pullback along the morphism $\operatorname{spec} \mathcal{O}_{f(p)} \rightarrow \operatorname{spec}A$ look like? If $\mathfrak{p}$ is the prime ideal of $B$ corresponding to the point $p$, then obviously the resulting scheme is the tensor product $\operatorname{spec}(B \otimes_{A} A_{\phi^{-1}(\mathfrak{p})})$. But what does this scheme actually correspond to? Is there some simplification of this? I used to wrongly think that this was the scheme $\operatorname{spec}\mathcal{O}_{p}$ but I have realised there is no reason to think that is true.

So once I understand that, I would like to pullback further along the morphism $\operatorname{spec}\kappa(f(p)) \rightarrow \operatorname{spec}\mathcal{O}_{f(p)}$. I know this should be quite easy but I'm having trouble seeing it.

Luke
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  • Are you sure you don't want $p$ to be a point in $\operatorname{Spec} A$? And once you make that change, isn't this covered by your old question here? – KReiser Dec 16 '20 at 04:01
  • I think $p$ should definitely be a point in $\operatorname{spec}B$, since then $f(p)$ is in $\operatorname{spec}A$, right? – Luke Dec 16 '20 at 04:03
  • Oh, sorry, missed the $f(p)$ and read $p$ instead. Still, though, doesn't your former question handle this? You're asking what the fiber product of the maps $\operatorname{Spec} B\to \operatorname{Spec} A$ and $\operatorname{Spec} A_\mathfrak{q} \to \operatorname{Spec} A$ is, where $\mathfrak{q}$ is the prime ideal of $A$ corresponding to $f(p)$. This is just the spectrum of $B\otimes_A A_\mathfrak{q}$, or $B_\mathfrak{q}$: all the prime ideals in $B$ not meeting the image of $A\setminus\mathfrak{q}$, exactly as the first line of the accepted answer reads. – KReiser Dec 16 '20 at 04:05
  • I didn't think it did, but I will read over it and think about it some more, perhaps I am missing something. I feel like I should be concerned that after all this time I am still asking the same basic questions I was 3 years ago :S – Luke Dec 16 '20 at 04:10
  • I think I might be able to get the information I need from that answer, so thank you. Can I ask though, is it normal to take this long to grasp the basic foundations of algebraic geometry? – Luke Dec 16 '20 at 04:20
  • I don't think it's possible for me to give a good answer to that question - it depends too much on you and your particular situation, and I don't really know you. If you're in school, this would be something to talk to an advisor about. (I would also add that never having learned something is perhaps slightly different than having learned something, not using it for a while, and having to check a reference to refresh oneself. Again, I don't know you or your situation!) – KReiser Dec 16 '20 at 04:40

1 Answers1

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The construction of the fiber product is an important construction in algebraic geometry hence this question may have general interest for students.

If $\phi: A \rightarrow B$ is a map of unital commutative rings with $X:=Spec(B), S:=Spec(A)$ and $f: X \rightarrow S$ the induced morphism of affine schemes, we may for any point $\mathfrak{p} \subseteq B$ consider the image under $f$ , $\mathfrak{q}:=\phi^{-1}(\mathfrak{p})$. We get an inclusion of affine schemes (the inclusion of the point $\mathfrak{q}$ in $S$):

I1. $i:Spec(\kappa(\mathfrak{q})) \rightarrow S$

and we may ask: What is the fiber of $f$ over the point $\mathfrak{q}$? The answer is given by an exercise in Atiyah-Macdonald (Exercise 4.21). It says that the fiber is given by the following formula:

I2. $f^{-1}(\mathfrak{q})\cong Spec(B\otimes_A \kappa(\mathfrak{q}))$.

Your question: "But what does this scheme actually correspond to? Is there some simplification of this?"

Answer: You must do Exercise 4.21 in AM. Here you prove that this is the scheme theoretic fiber of the map $f$ at $\mathfrak{q}$. The inverse image scheme is the schematic fiber product of $i$ and $f$, and this is given by the tensor product of the rings $\kappa(\mathfrak{q})$ and $B$ over $A$.

In Hartshorne Theorem II.3.3 they prove that the fiber product $X\times_S Y$ of two schemes $X/S,Y/S$ is unique up to unique isomorphism, and the fiber product is constructed for affine schemes using the tensor product. Hence if $S:=Spec(R), X:=Spec(U), Y:=Spec(V)$, it follows there is a canonical isomorphism

I3. $X\times_S Y \cong Spec(U\otimes_R V)$.

Hence the formula I2

$f^{-1}(\mathfrak{q}):=Spec(B)\times_{Spec(A)} Spec(\kappa(\mathfrak{q}) \cong Spec(B\otimes_A \kappa(\mathfrak{q}))$

calulates the schematic inverse image of the point $\mathfrak{q}$.

More generally: If $\pi: X \rightarrow S$ is any morphism of schemes and if $s\in S$ is any point, let $i:=Spec(\kappa(s)) \rightarrow S$ be the inclusion of the pont $s$ into $S$. The fiber of $\pi$ at $s$ is by definition the fiber product $Spec(\kappa(s))\times_S X$.

Question: "I think I might be able to get the information I need from that answer, so thank you. Can I ask though, is it normal to take this long to grasp the basic foundations of algebraic geometry?"

Answer: If you do exercise 4.21 in AM and understand Theorem II.3.3 in Hartshorne you will get a better understanding of what the fiber product is. For affine schemes it follows the universal property for the fiber product translates into a universal property for the tensor product of commutative unital rings. The exercise in AM establish a 1-1 correspondence between the prime ideals $\mathfrak{p} \subseteq B$ with $\mathfrak{p}\cap A= \mathfrak{q}$ and the set of prime ideals in $B\otimes_A \kappa(\mathfrak{q})$.

hm2020
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