# the fibre of $f$ over $y$ is $\operatorname{Spec} \kappa(y)$ given $f^{\#}(\mathfrak{m}_y) O_{X,x} = \mathfrak{m}_x$

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!

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 $$f^{-1}(y)\cap \operatorname{Spec} B_d = \{x\}$$.
• 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 Aug 3 '20 at 9:17
• $x$ 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. Aug 3 '20 at 9:26
• 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 $x \in X_y$? Aug 3 '20 at 10:20