# Basic question related to stalk of the structure sheaf of a scheme

Let $X$ and $Y$ be schemes. Suppose I have a morphism of schemes $(\pi, \phi)$, where $\pi: X \rightarrow Y$, thus $\pi$ is continuous, and $\phi: O_Y \rightarrow \pi_*O_X$. Let $p \in X$ and $\pi(p) = q$. I would like to show that the stalk of $\pi_*O_X$ at $q$ is isomorphic to stalk of $O_X$ at $p$, i.e. $$(\pi_*O_X)_q \simeq O_{X,p}.$$ I would appreciate if someone could show me how I can prove this. Thank you very much!

ps On the second thought, I now feel like they may not be isomorphic all the time. I would appreciate if I could verify this with someone. Thanks!

• How can that be true without some further condition on $\phi$ like it being an isomorphism? Mar 14, 2015 at 4:17
• I think that $\phi$ doesn't really play a role here, because I am not asking about $O_{Y,q}$ and $(\pi_*O_X)_q$... Mar 14, 2015 at 4:25
• If we were asking for the local rings $O_{Y,p}$ and $(\pi_*O_X)_q$ as here in the comments, at least for varieties an equivalent condition is that they are birationally equivalent (Hartshorne, exercise I,4.7). Mar 14, 2015 at 5:11

In general, what you claim is false. Imagine this example: let $\pi:\mathbb{A}^2=\text{Spec}k[x,y]\to \mathbb{A}^1=\text{Spec}k[x]$ be the projection on the first component. Fix a closed point $(a,b)$ in the affine plane (corresponding to the maximal ideal $(x-a,y-b)$), we have $\pi(a,b)=a$ in the affine line. Now, let's compare the local ring at $(a,b)$ with the one of the push forward: we have by definition that the local ring in the affine plane is

$$\mathcal{O}_{A^2,(a,b)}=k[x,y]_{(x-a,y-b)}=\lim_{(a,b)\in U}\mathcal{O}_{A^2}(U)$$

and, always applying the definition

$$(\pi_*(\mathcal{O}))_a=\lim_{a\in V} \left( (\pi_*\mathcal{O}_{A^2})(V) \right)=\lim_{a\in V} \left( \mathcal{O}_{A^2}(\pi^{-1}(V) \right) \subseteq \lim_{\pi^{-1}(a)\subseteq U} \left( \mathcal{O}_{A^2}(U) \right)=\lim_{a\times \mathbb{A}^1\subseteq U} \left( \mathcal{O}_{A^2}(U) \right)$$

In this second chain, I put a sign of subset because we don't even know that we are ranging over all possible open sets, but just over those which are preimages of opens of the line (in this specific example, however, we get an equality there, if we think about how the topologies are made). In any case, we only get

$$\lim_{a\times \mathbb{A}^1\subseteq U} \left( \mathcal{O}_{A^2}(U)\right)=k[x,y]_{(x-a)} \subseteq\lim_{(a,b)\in U}\mathcal{O}_{A^2}(U)$$ and the containment is strict in this case. However, it is hard for me to think with certainty to conditions that make what you claim true (of course $\pi$ beign an isomorphism works, but I feel that it can be weakened). I feel that even if we require that the inverse image is just a point we don't have the guarantee that we range over enough open sets via the preimage (not all oopen neighbouroods of the point $p$ are of the form $\pi^{-1}(V)$, where $V$ is a neighbourood of $\pi(p)$).

• Thank you for the detailed answer! Greatly appreciate it. Mar 15, 2015 at 16:09

Given a field $k$, the constant map $\pi:X=\mathbb A^1_k=\operatorname {Spec}(k[T])\to Y=\{q\}=\operatorname {Spec}(k)$ dual to the inclusion map $k\hookrightarrow k[T]$ sends the origin $p=(T)\in \mathbb A^1_k$ to $q \in Y$ but the corresponding morphism of stalks is the inclusion $(\pi_*O_X)_y=k\hookrightarrow O_{X,p}=k[T]_{(T)}$ which is definitely not an isomorphism !

• Thank you very much for this example! It is very helpful! Mar 15, 2015 at 16:09
• I'm glad it helped you, user201886. Mar 15, 2015 at 16:29