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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!

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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 $f^{-1}(y)\cap \operatorname{Spec} B_d = \{x\}$.

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  • $\begingroup$ 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 $\endgroup$
    – Johnny T.
    Aug 3 '20 at 9:17
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    $\begingroup$ $x$ 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. $\endgroup$
    – KReiser
    Aug 3 '20 at 9:26
  • $\begingroup$ 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 $x \in X_y$? $\endgroup$
    – Johnny T.
    Aug 3 '20 at 10:20
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    $\begingroup$ You've already asked and I've already answered that question here. $\endgroup$
    – KReiser
    Aug 3 '20 at 10:52
  • $\begingroup$ oh dear... thank you... $\endgroup$
    – Johnny T.
    Aug 3 '20 at 11:40

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