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.
Best Answer
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 $$(\mathcal O_{X,x}/\mathfrak m_y\mathcal O_{X,x})/(\mathfrak m_x/\mathfrak m_y\mathcal O_{X,x})\cong \mathcal O_{X,x}/\mathfrak m_x\cong k(x)\,.$$
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.