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I am trying to understand the proof of Theorem 3 in Chapter III. 5 of Mumford's red book. Let $f: X \to Y$ be a morphism of schemes (noetherian), $f(x) = y$ and the induced map of the residue fields is an isomorphism. Suppose $$ f^{\#}(\mathfrak{m}_y) O_{X,x} = \mathfrak{m}_x. $$ Then Mumford states ``In other words, the fibre of $f$ over $y$, near $x$ is just a copy of $\operatorname{Spec} \kappa (y)$.'' I am not quite seeing how this is the case. Any explanation is appreciated. Thank you!

James Moriarty
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Johnny T.
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1 Answers1

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The ideals of $\mathcal{O}_{X,x}$ correspond to germs of closed subschemes of $X$ that pass through $x$ - that is, these ideals correspond to closed subschemes of $X$ passing through $x$ up to the equivalence relation that $Z\sim Z'$ if there's an open neighborhood $U$ of $x$ so that $Z\cap U=Z'\cap U$. The pullback of $\mathfrak{m}_y$ gives the ideal of the germ of the fiber $X_y$ over $y$ as a closed subset passing through $x$. If that's just $\mathfrak{m}_x$, this means that the germ of the fiber is just $x$, or that there's an open neighborhood $U$ of $x\in X$ so that $U\cap X_y=x$.

In the comments, the OP asked for some clarification. One may immediately reduce to the affine case: pick an open affine neighborhood $\operatorname{Spec} A\subset Y$ of $y$, and an open affine neighborhood $\operatorname{Spec} B\subset f^{-1}(\operatorname{Spec} A)$ of $x$. Now $x$ and $y$ correspond to prime ideals $q\subset B$ and $p\subset A$ with $\varphi^{-1}(q)=p$ (where $\varphi:A\to B$ is the map of rings corresponding to $f$). The closure of $f^{-1}(y)$ is given by $V(\varphi(p)B)\subset \operatorname{Spec} B$, and the statement that $f^\sharp(\mathfrak{m}_y)\mathcal{O}_{X,x}=\mathfrak{m}_x$ translates to $(\varphi(p)B)_q=qB_q$. Picking finite sets of generators for both sides by noetherianness, we can see that each set of generators can be expressed as a finite $B_q$-linear combination of each other, and so up to multiplying all the denominators involved we get a single element $d$ so that $(pB)_d=(qB)_d$ (we also note this element is outside $q$ by construction). Now on the affine open $\operatorname{Spec} B_d \subset \operatorname{Spec} B$, we have that $V(\varphi(p)B_d)=V(qB_d)$, and so $\operatorname{Spec} B_d$ is an open neighborhood of $x$ so that $f^{-1}(y)\cap \operatorname{Spec} B_d = \{x\}$.

KReiser
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  • Thank you again! I am struggling to understand ``The pullback of $\mathfrak{m}_y$ gives the ideal of the germ of the fibre $X_y$ over $y$ as a closed subset passing through $x$.'' 1) I guess there is a unique point of $X_y$ that is identified with $x \in U$? 2) When you say the pullback of $\mathfrak{m}_y$ what exactly does this mean? Thank you – Johnny T. Aug 03 '20 at 09:17
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    $x$ is fixed at the start. There was also an instance of unintentional capitalization which might have made things a little more difficult to understand than they should have been - try giving the post another read, otherwise I don't understand what your point 1 is talking about. As for the pullback, I mean the LHS of the equation you've written in the main post - it's not unusual to refer to the map of local rings induced by a map of schemes as a pullback. – KReiser Aug 03 '20 at 09:26
  • Let me clarify about 1. $x$ is a point in $X$. How is it also a point in the fibre $X_y$? I thought fibre product gives a scheme that 'lives' somewhere else. So there is a unique way to identify $x \in X$ with this $x \in X_y$? – Johnny T. Aug 03 '20 at 10:20
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    You've already asked and I've already answered that question here. – KReiser Aug 03 '20 at 10:52
  • oh dear... thank you... – Johnny T. Aug 03 '20 at 11:40
  • @JohnnyT. Is there a reason you've awarded the bounty but not marked the answer as accepted? – KReiser Aug 04 '20 at 09:13
  • I am not quite done going through them yet (thank you for all the answers!) but I am leaving for vacation so I wanted to award the bounty now because they would expire – Johnny T. Aug 04 '20 at 09:58
  • Could you possibly point me to a reference so that I can work out the details of ``The pullback of $\mathfrak{m}_y$ gives the ideal of the germ of the fiber $X_y$ over $y$ as a closed subset passing through $x$''? I'm trying to figure this part out... thank you – Johnny T. Aug 18 '20 at 09:03
  • @JohnnyT. This is just direct reasoning. You may find an example where $X,Y$ are affine instructive. Is there some part in particular you're having issues with? – KReiser Aug 18 '20 at 09:25
  • The issue I am having is that 1) I get that the pullback of $\mathfrak{m}_y$ corresponds to the closure of $f^{-1}(y)$ (maybe this is what you meant by 'as a closed subset'?) 2) When the pullback is $\mathfrak{m}_x$ I get that this corresponds to the closed subset $\overline{{ x }}$ so I haven't been able to conclude $U \cap X_y = {x}$ for some open $U$... – Johnny T. Aug 18 '20 at 10:41
  • Is Mumford assuming that $x$ is a closed point perhaps? – Johnny T. Aug 19 '20 at 08:32
  • @JohnnyT. No, no assumptions about $x$ or $y$ being closed should be made (I did implicitly assume $y$ was, which is why I said $X_y$ instead of it's closure). I've updated the post with an explanation in the affine case. – KReiser Aug 19 '20 at 09:02
  • I see, thank you for your explanation. I am struggling to see why $V(qB_d)$ only has one point, since we only know $q$ is a prime ideal.. how do I deduce this? – Johnny T. Aug 19 '20 at 11:35
  • I also have another question but asked it as a separate question (https://math.stackexchange.com/questions/3796219/fibre-with-only-one-point-is-isomoprhic-to-the-spec-of-a-field ) in effort to not make this too long.. thank you – Johnny T. Aug 19 '20 at 12:30
  • I had more thought about this. If $d$ happens to be a unit in your argument, then $\operatorname{Spec} B_d = \operatorname{Spec} B$ and $\overline{f^{-1}(y)} \cap \operatorname{Spec} B = V(q)$. So I don't see why we can conclude $f^{-1}(y) \cap \operatorname{Spec} B = { [q]}$, since we only know $q$ is prime.. – Johnny T. Aug 20 '20 at 09:24