How to describe the stalk on fiber $X_y$?

Question

Let $f:X\rightarrow Y$ be a morphism of schemes. (May assume more if necessary) Let $x\in X$ and $y\in Y$ such that $f(x) = y$ and $X_y$ be fiber over $y$.

My question: Is there any relation in general between the stalk $\mathcal O_{X_y,x}$ and $\mathcal O_{X,x}$? If the answer is no, what could we say or how to understand the stalk $\mathcal O_{X_y,x}$?

My question is from the following a proof in Qing Liu's book:

Lemma 4.3.20. Let $f:X\rightarrow Y$ be a morphism of finite type between locally Noetherian schemes. Then $f$ is unramified if and only if for every $y\in Y$, the fiber $X_y$ is finite, reduced, and if $k(x)$ is separable over $k(y)$ for every $x\in X_y$.

The author says: 'Indeed, the quotient $\mathcal O_{X,x}/m_y\mathcal O_{X,x}$ remains unaltered when we pass from $X$ to the fiber $X_y$'. I'm confused about the meaning of this statement.

Answer 1

My question: Is there any relation in general between the stalk $\mathcal O_{X_y,x}$ and $\mathcal O_{X,x}$?

Yes. If $\mathfrak m_y$ denotes the maximal ideal of $\mathcal O_{Y,y}$, then $$\mathcal O_{X_y,x}\cong \mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}\,,$$ where $\mathfrak m_y\mathcal O_{X,x}\subseteq \mathcal O_{X,x}$ denotes the ideal generated by the image of $\mathfrak m_y$ under the ring morphism $\mathcal O_{Y,y}\rightarrow \mathcal O_{X,x}$. To see this, we may assume $X\cong\operatorname{Spec} B$ and $Y\cong\operatorname{Spec} A$ are affine, with $x$ and $y$ corresponding to prime ideals $\mathfrak q\subseteq B$ and $\mathfrak p\subseteq A$. Then $X_y\cong \operatorname{Spec}(B\otimes_Ak(\mathfrak p))\cong\operatorname{Spec}B_{\mathfrak p}/\mathfrak p B_{\mathfrak p}$ (here $B_{\mathfrak p}$ denotes the localisation of the $A$-module $B$ at $\mathfrak p$; moreover, $\mathfrak p B_{\mathfrak p}\subseteq B_{\mathfrak p}$ denotes the ideal generated by the image of $\mathfrak p$). Thus the local ring $\mathcal O_{X_y,x}$ is given by the localisation of the ring $B_{\mathfrak p}/\mathfrak p B_{\mathfrak p}$ at its prime ideal $\mathfrak q/\mathfrak pB_{\mathfrak p}$. By elementary properties of localisation, this coincides with $B_{\mathfrak q}/\mathfrak p B_{\mathfrak q}\cong \mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}$.

Also note that $\mathcal O_{X_y,x}$ has the same residue field as $\mathcal O_{X,x}$. Indeed, under the isomorphism $\mathcal O_{X_y,x}\cong \mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}$, the maximal ideal of $\mathcal O_{X_y,x}$ corresponds to $\mathfrak m_x/\mathfrak m_y\mathcal O_{X,x}$, and $$(\mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x})/(\mathfrak m_x/\mathfrak m_y\mathcal O_{X,x})\cong \mathcal O_{X,x}/\mathfrak m_x\cong k(x)\,.$$

The author says: 'Indeed, the quotient $\mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}$ remains unaltered when we pass from $X$ to the fiber $X_y$'. I'm confused about the meaning of this statement.

This is somewhat sloppy language and I'll try to rephrase it in a super pedantic way. Let $Y'=\operatorname{Spec} k(y)$ and let $X'=X\times_YY'\cong X_y$. Let $x'\in X'$ denote the unique point lying over $x\in X$ and let $y'\in Y'$ be the unique point (which automatically lies over $y$). Then what the author is trying to say is that $$\mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}\cong \mathcal O_{X',x'}/\mathfrak m_{y'}\mathcal O_{X',x'}\,.$$ Indeed, the left-hand side equals $\mathcal O_{X',x'}$, as seen above, and on the right-hand side we have $\mathfrak m_{y'}=0$ since $Y'=\operatorname{Spec} k(y)$ is the spectrum of a field.

For the sake of completeness, let me elaborate a bit more on Liu's proof, since I think quite some things get swept under the rug. Assume first that all fibres $X_y$ are finite over the respective residue field $k(y)$. By the structure theory of artinian rings (see [Stacks Project, Tag 00J4]), $X_y$ has only finitely many points $x_1,\dotsc,x_n$, and $X_y\cong \coprod_{i=1}^n\operatorname{Spec} \mathcal O_{X_y,x_i}$. If $X_y$ is reduced, then each of the artinian local rings $\mathcal O_{X_y,x_i}$ must be reduced. But a reduced artinian local ring must be a field, whence $\mathcal O_{X_y,x_i}\cong k(x_i)$. By the discussion above, this also implies $\mathcal O_{X,x_i}/\mathfrak m_y\mathcal O_{X,x_i}\cong k(x_i)$. In particular, if each $k(x_i)$ is separable over $k(y)$, then $f$ is indeed unramified.

Conversely, assume $f$ is unramified. We must show that each fibre $X_y$ is finite and reduced over $k(y)$ (and that all residue fields $k(x)$ are separable over $k(y)$, but this is automatic from the assumption that $f$ is unramified). Since $f$ is unramified, $\mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x}\cong k(x)$ holds for all $x\in X_y$. Thus, all local rings $\mathcal O_{X_y,x}\cong k(x)$ are fields. In particular, they are all 0-dimensional, which implies that $X_y$ must be zero-dimensional. This implies again that $X_y$ has only finitely many points $x_1,\dotsc,x_n$ and that $X_y\cong \coprod_{i=1}^n\operatorname{Spec} \mathcal O_{X_y,x_i}$ (see [Stacks Project, Tag 0AAX]; the disjoint union must be finite since $X_y$ is quasi-compact). Since all $\mathcal O_{X_y,x_i}\cong k(x_i)$ are reduced and finite-dimensional over $k(y)$, we see that $X_y$ too must be reduced and finite over $k(y)$, as claimed.

Answer 2

Question: "My question: Q1. Is there any relation in general between the stalk $O_{X_y,x}$ and $O_{X,x}$? Q2. If the answer is no, what could we say or how to understand the stalk $O_{X_y,x}$?"

Answer: The following seems to hold: If $\phi: X:=Spec(B) \rightarrow Y:=Spec(A)$ and $\mathfrak{p}\subseteq A$ is a prime ideal corresponding to the point $y\in Y$ we get canonical maps

$$ A \rightarrow B \rightarrow B/\mathfrak{p}B\cong A/\mathfrak{p}\otimes_A B.$$ If $Z:=Spec(A/\mathfrak{p})$, we get $X_Z:=\phi^{-1}(Z)\cong Spec(B/\mathfrak{p})$. If $\mathfrak{p}$ is a maximal ideal corresponding to the closed point $y$, it follows $Z\cong Spec(\kappa(\mathfrak{p}))$ hence $X_y\cong Spec(B/\mathfrak{p}B)$.

Let $\mathfrak{m}\subseteq B$ be a maximal ideal in $B$ corresponding to a point $x \in X$ with $\mathfrak{p} \subseteq \mathfrak{m}\cap A$, hence $x\in X_y$ - the fiber at $y$. It follows $\mathfrak{p}B \subseteq \mathfrak{m}$ . Define $\overline{\mathfrak{m}}:=f(\mathfrak{m}) \subseteq B/\mathfrak{p}B$ where $f$ is the canonical map. We get by standard properties of localization a canonical exact sequence

$$ 0 \rightarrow (\mathfrak{p}B)_{\mathfrak{m}} \rightarrow B_{\mathfrak{m}} \rightarrow^i (B/\mathfrak{p}B)_{\overline{\mathfrak{m}}} \rightarrow 0$$

but the map $i$ is not an isomorphism in general - its kernel is the ideal

$$(\mathfrak{p}B)_{\mathfrak{m}}.$$

When we pass to the residue fields we get an isomorphism

$$ \kappa(\mathfrak{m})\cong \kappa(\overline{\mathfrak{m}}).$$

Question Q1: There is when $y$ is a closed point a canonical surjective map

$$M1.\text{ }i^{\#}_x: \mathcal{O}_{X,x} \rightarrow \mathcal{O}_{X_y,x},$$

with kernel described in the above comment: If we choose open affine subschemes $x\in Spec(B), y\in Spec(A)$ it follows the kernel equals

$$ ker(i^{\#}_x)=(\mathfrak{p}B)_{\mathfrak{m}} \subseteq B_{\mathfrak{m}}.$$

Hence the map $i^{\#}_x$ is not an isomorphism in general.

Question Q2: The map $i^{\#}_x$ induce an isomorphism at residue fields, hence the two local rings in M1 have the same residue field.

To the inclusion map $i: X_y \rightarrow X$ there is a canonical map of sheaves

$$i^{\#}: \mathcal{O}_X \rightarrow i_*\mathcal{O}_{X_y}$$ and passing to the stalk at $x$ you recover the map

$$ i^{\#}_x: \mathcal{O}_{X,x}\rightarrow \mathcal{O}_{X_y,x}.$$

It seems the kernel of the map $i^{\#}_x$ is the ideal $(\mathfrak{p}B)_{\mathfrak{m}}$. Note that $\mathcal{O}_{X,x}$ is a local ring with maximal ideal $\mathfrak{m}_x$ hence $\mathfrak{m}_y\mathcal{O}_{X,x} \subseteq \mathfrak{m}_x$.

Note 1: There is a canonical map

$$f^{\#}_y: \mathcal{O}_{Y,y} \rightarrow \mathcal{O}_{X,x}$$

and you may consider the ideal

$$\mathfrak{m}_y\mathcal{O}_{X,x}\subseteq \mathcal{O}_{X,x}.$$

The ideal $\mathfrak{m}_y\mathcal{O}_{X,x}$ is not a prime-ideal in general. If $X_y$ is an integral scheme it follows $\mathcal{O}_{X_y,x}$ is an integral domain, hence there cannot be an isomorphism

$$ \mathcal{O}_{X_y,x} \cong \mathcal{O}_{X,x}/\mathfrak{m}_y\mathcal{O}_{X,x}$$

in general as is being claimed above: This is beacuse the quotient $\mathcal{O}_{X,x}/\mathfrak{m}_y\mathcal{O}_{X,x}$ is not an integral domain. An integral domain cannot be isomorphic to a non-integral domain.